58 Astronautics, Space & Astrodynamics

Introduction[1]

Humanity has long sought to fly into space. Eventually, there is a height above Earth’s surface at which the atmosphere begins to thin, making it impossible for aircraft to fly at that altitude. This height marks the beginning of what is known as outer space, often called space. The fastest and highest air-breathing propelled aircraft ever flown, the SR-71 Blackbird, achieved a speed of 3,540 kph (2,200 mph) at an altitude of 85,000 ft (about 26 km or 16 miles) before it ran out of atmosphere to fly as an aircraft. The rocket-powered North American X-15 achieved its maximum speed of 7,273 km/h (4,520 mph or Mach 6.7) and 107.8 km (67.1 miles), putting the aircraft well into the edge of space.

Why would humans, as firmly and well-established terrestrial beings, want to go up and outward into space and visit the stars? Space is a very unfriendly, if not hostile, environment for humans, as it lacks breathable air and experiences extreme temperature fluctuations. Distances are enormous; even the Moon, Earth’s closest natural satellite, is about 384,400 km (239,000 miles) away on average. Many will argue that human ventures into space waste time, money, and resources. Are proposed human missions to Mars just a fantasy? Perhaps limited resources are better spent on Earth-bound activities? Such arguments, however, are flawed and entirely contrary to human nature and the pursuit of scientific understanding.

Stephen Hawking, the renowned astrophysicist and former Lucasian Professor of Mathematics at the University of Cambridge, delivered a lecture titled “Why we should go into space” as part of NASA’s 50th Anniversary Lecture Series. Professor Hawking, now deceased, was severely disabled by ALS and eventually had to speak using a computer, but listen very carefully to his lecture on why he advocated going into space, returning to the Moon, and eventually attempting to go to Mars.

It soon becomes clear that, had it not been for the development of rockets, satellites, and other space systems, humankind would know much less about its planet. Many “spinoff” technologies used in everyday life would never have been developed. For example, live weather, GPS, Google Maps, satellite television, the internet, battery-powered tools, and many other technologies used daily would not exist. NASA has listed thousands of technological spinoffs from space activities that have benefited humans. The advances made in space exploration have also driven advancements in materials science, computer technology, and medicine, among others. As NASA says, “There’s more space in your life than you think!”

Learning Objectives

  • Appreciate the challenges of humankind reaching into space.
  • Learn about the history of astronomy and the planets.
  • Understand the fundamental laws of satellite motion and spaceflight dynamics.
  • Be able to distinguish between different types of orbits.
  • Use Kepler’s equations to predict the motion of a satellite.
  • Know how to transfer spacecraft between orbits.
  • Understand the challenges of interplanetary travel, such as a mission to Mars.

Definition of Space

Theodore von Kármán reasoned that space must begin where the atmosphere’s density becomes so thin that aerodynamic lift on a flight vehicle is negligible. This height boundary between the world of “aeronautics” and atmospheric flight vehicles and that of “astronautics” and space vehicles has been defined as 100 km (approximately 62 miles or 328,000 ft); it has since become known as the Kármán line. Although there is no clear physical divide between the two, the Kármán line provides a convenient point of reference for aerospace engineers to differentiate between the two realms of flight.

It is often quoted that space is “The Final Frontier.” However, the reality is that it is only the next frontier, and the final frontier of the universe remains largely unknown. So, where does the universe end? Is there even an end that can be identified? What is out there? The answers to these questions pose some of the greatest mysteries of science. The prevailing theory is that the universe is still expanding and has no defined boundary or final frontier. However, this idea remains under investigation; new discoveries could challenge current understanding.

The vastness of space and the universe as a whole is best summed up by a quote about the Earth from the extraordinary scientist, astronomer, and professor, Dr. Carl Sagan: “On it is the aggregate of everything we know since the beginning of time, and a tiny stage in the vastness of the cosmic arena.” The recent photograph of Earth taken from the Orion spacecraft, shown in Figure 1, clearly illustrates the point. Behind this serene photo, we should also remember that as a civilization, we are hurtling through space at about 500,000 miles per hour, spiraling around the center of our galaxy in a helical path as the Sun carries the Solar System through the Milky Way. Despite numerous technological advancements and increased understanding, knowledge of the universe remains limited and enigmatic in many respects. As shown below, the photograph of Earth taken from space is a humbling reminder of Earth’s small place in the cosmos. It inspires humankind to continue exploring and learning more about space and the universe.

The “pale blue dot” (as described by Carl Sagan) of the planet Earth was visible from the far side of the Moon by the Orion spacecraft on December 22, 2022.

What is Out There?

Humans have roamed the Earth for approximately 300,000 years, but civilization has existed for only about 10,000 years. This time is a mere blink of an eye on the universe’s time scale, which is estimated to be over 14 billion years. In other words, human civilization has existed for only about 0.0002% of the Earth’s history. Advances in technology are among the most defining human pursuits, driven by various motivations, including an enduring fascination with the stars and the exploration of the universe. However, only in the last 500 years has humankind begun to make more sense of its place in the universe, thanks to significant advancements in science, astronomy, engineering, and spaceflight. The study of celestial objects, including planets, is called astronomy, whereas the study of space and the universe, collectively known as the cosmos, is called cosmology.

There are billions upon billions of galaxies, each containing billions of planets, a fact well established by observations with space telescopes. Therefore, it is impossible to rationalize the true scale of the universe. But are there other civilizations on those planets as well? It seems likely that humans are not the only living things in the universe. Could humans one day encounter intelligent extraterrestrial life? Will they look like humans? Most likely not, as the randomness of evolution suggests they will be very different from us. Maybe Earth has already been visited by extraterrestrial beings? However, while some believe in extraterrestrial visits based on anecdotal evidence, no solid scientific proof exists to support these claims. Further research and exploration of the universe are necessary to address these questions and deepen understanding of the possibilities for extraterrestrial life and other civilizations.

Geocentrism

The earliest contributors to the science of the universe and space included the Greek philosopher Aristotle, who made extensive observations of the planets. Aristotle’s religious beliefs influenced his views on cosmology, but they were also based on scientific understanding and the celestial observations available at the time. However, his idea of a stationary Earth at the center of the universe, known as geocentrism, was later disproved by Galileo and other scientists who used telescopes to observe the movements of the stars and planets. The astronomer Claudius Ptolemaeus also believed the Sun, Moon, and other planets circled the Earth.

Heliocentrism

It is now well understood that the Earth is not at the center of the universe and orbits the Sun as part of our solar system, a concept known as heliocentrism. However, Aristotelian cosmology and geocentrism dominated astronomy and science for millennia. Advancements in technology and scientific understanding gradually led to its rejection. In the 16th and 17th centuries, astronomers such as Copernicus, Galileo, and Kepler provided additional evidence supporting heliocentrism, suggesting that the Earth and other planets revolved around the Sun.

In 1532, the Polish astronomer Nicolaus Copernicus proposed that the Sun was the center of the solar system, with the planets orbiting it in circular orbits, and that the Earth turned once daily on its axis. Copernicus’s heliocentric model of the universe significantly departed from the widely accepted geocentric model and sparked a paradigm shift in scientific thinking. His work challenged the prevailing views of the time and paved the way for further advancements in astronomy and science.

The Scientific Revolution, brought about by these advancements, led to significant discoveries, theories, and insights into the nature of the universe, laying the foundations for modern science. The works of Galileo, Brahe, Kepler, and Newton, among others, built upon Copernicus’s ideas and expanded our understanding of the solar system and the universe.

Celestial Observations of Galileo

The Italian astronomer and scientist Galileo Galilei supported Copernicus’s theories, which posited that the Earth rotates on its axis once every 24 hours and orbits the Sun once a year. Starting in 1609, Galileo built upon the concept of a spyglass, or monocular, to develop various astronomical telescopes with increasing magnification. Galileo played a pivotal role in the Scientific Revolution by using telescopes to observe the heavens and gather empirical evidence supporting the heliocentric model. He made numerous important discoveries, including the observation of four large moons orbiting Jupiter and the phases of Venus, which provided further evidence for the Copernican system.

Galileo made many vital contributions to science and astronomy and is widely regarded as one of the most outstanding scientists in history. His observations of celestial objects were groundbreaking and challenged long-held notions about the nature of the universe. Galileo discovered that the Moon’s surface was not smooth but was composed of numerous craters formed by asteroid impacts. Additionally, it was not translucent, a common belief at the time. His study of sunspots was also significant and helped establish the idea that the Sun was not perfect and unchanging, as previously thought. Galileo’s use of the telescope to make detailed observations of the sky revolutionized astronomy and paved the way for future discoveries.

For these reasons, Galileo is known as the “Father of Observational Astronomy,” a title first assigned by Albert Einstein. His legacy continues to inspire scientists and astronomers. His book Sidereus Nuncius (Sidereal Messenger), published in 1610, presents the results of Galileo’s early observations of the planets.

Kepler’s Laws of Orbital Motion

The 17th-century German astronomer Johannes Kepler, a student of Tycho Brahe, made the profound observation that the planets orbit the Sun in elliptical paths, rather than in circular orbits, as illustrated in the schematic in Figure 2. This discovery marked a significant departure from prevailing scientific views of the solar system. It helped establish that scientific and mathematical laws could explain the behavior of the universe.

Kepler concluded that the planets moved in elliptical orbits around the Sun, with his observations and scientific reasoning serving as the basis for his three laws of planetary motion.

Kepler’s laws provided a foundation for the work of later scientists, including Isaac Newton, who showed that they could be derived from his laws of motion and the universal law of gravitation.[2] Kepler’s work was a significant step toward a more complete understanding of the solar system and the universe as a whole and continues to influence astronomical research.

What is the closest planet to Earth besides the Moon?

First off, the Moon is not considered a planet but the Earth’s closest natural satellite. But most people will say that the closest planet to Earth is Mars. However, when averaged over time, Venus is typically the closest planet to Earth, with an average distance of about 41 million km (25.2 million miles). The closest distance between Earth and Venus is about 38 million km (23.6 million miles). Mercury is closer to the Sun than Earth and, depending on the relative orbital positions, may at times be closer to Earth than either Venus or Mars. Because planetary orbits are elliptical and continually changing in their relative positions, the planet closest to Earth at any given time can vary.

Kepler’s laws, published between 1609 and 1619, were not immediately accepted. Indeed, Galileo himself questioned Kepler’s observations and largely disregarded his book Astronomia nova.[3] Kepler’s laws were later confirmed by other astronomical observations, particularly transits of Venus and Mercury across the face of the Sun. These observations are now primarily of historical significance.

Kepler’s laws provided the empirical foundation for Isaac Newton’s laws of motion and universal gravitation, which remain central to modern celestial mechanics. Since their formulation, Kepler’s laws have significantly advanced our understanding of the solar system and have been instrumental in space exploration and satellite navigation. Kepler’s Astronomia Nova became one of the most important books in the history of astronomy, and his work remains relevant in modern astrophysics and planetary science.

Kepler’s next book, “Epitome Astronomiae Copernicanae” (Epitome of Copernican Astronomy), was widely read by astronomers during the 17th century. However, its content was considered controversial because it supported the Copernican system and challenged long-standing geocentric doctrine. Nevertheless, many established astronomers and scientists of the era continued to debate Kepler’s theories as a physical basis for explaining planetary motion.

Newton & the “Principia”

Several other physical theories on planetary motions were developed from Kepler’s work. These theories culminated with the publication of Isaac Newton’s definitive Philosophiæ Naturalis Principia Mathematica in 1687. In this book, Newton explained the movement of orbiting planets in terms of the Sun’s gravitational attraction. In addition, he applied his laws of motion to rederive Kepler’s laws of planetary motion from a universal law of gravitation, a mathematical challenge later known as “Kepler’s Problem.” Over the centuries, numerous astronomical observations and experiments have confirmed Kepler’s work and Newton’s Law of Gravitation. Indeed, Kepler’s laws have proven to be a cornerstone of modern astronomy, contributing significantly to our understanding of the cosmos.

Defining Space

The aerodynamic environment in which aircraft fly has been extensively studied, and standardized mathematical models of the atmosphere, extending up to the edges of space, have been developed. The question now is how to describe the space environment. Many would claim that space is a vastness of absolute nothingness with no properties, but this is untrue.

Vastness of Nothingness?

Space is an almost perfect vacuum at extremely low pressures and temperatures, but primarily devoid of matter. Unlike air and other gases, space does not impede the motion of objects, but they experience extremely small but non-zero drag as they move through space. However, small amounts of gases, dust, and other matter still float around in the vastness of the empty regions. Therefore, concepts such as pressure, temperature, and thermal transfer remain relevant in space, and radiation must also be considered. In other regions, matter can accumulate to form planets, stars, and galaxies; the Earth is just one of these fortunate regions, along with a cluster of other planets in the solar system, as shown in the animation in Figure 3.

An animation of our solar system illustrating distances and relative orbital velocities. The Earth is the third planet from the Sun.

Space is also filled with various forms of radiation and electromagnetic fields, including cosmic rays, solar wind, and magnetic fields. These radiation fields can affect the behavior of spacecraft and other space objects and must be considered in spacecraft design and mission planning. The study of the space environment, known as space physics, also involves understanding the interactions between various forms of radiation and matter in space, and how these interactions influence the behavior of objects in the solar system. Space physics enables successful space missions, ranging from communication and navigation to scientific exploration and human spaceflight.

Quantifying Vastness

Space and the universe are vast, extending enormous distances from the Earth. But the vastness of space can be measured! These measurements can be performed using radar because it is known that the speed of light, which in astronomy is given the symbol {c}, has an established value of 299,792,458 m/s (983,571,056 ft/s) or about 300,000 km/s (186,000 miles/s)! To put this into perspective, the speed of light is extremely fast, approximately one million times the speed of sound in air on Earth.

The speed of light can be used to confirm that the Earth is, on average, 384,400 km (239,000 miles) from the Moon and 149.6 million km (93 million miles) from the Sun. That does not sound too far until you realize that the diameter of the Earth is only 12,741.2 km (7,917 miles), so the Moon is about 30 Earth diameters away, and the Sun is 24,378 diameters away!

This is why scientists and engineers measure immense distances in light-years: the distance that light can travel in one year, approximately 10 trillion kilometers (6 trillion miles). Yes, space is most definitely vast. However, the preferred unit of distance in astrometry is the parsec, denoted by the symbol “pc,” defined as the distance at which an object will appear to move one arcsecond of parallax when the observer moves one astronomical unit perpendicular to the line of sight. One parsec is equal to approximately 3.26 light-years.

Planet Earth

Given the vast size and complexity of the universe, the Earth is just one of the many tiny planets among billions of others. The Earth is part of a galaxy, part of an arm of a massive collection of stars swirling through space and throughout the universe. Indeed, our galaxy, the Milky Way, is estimated to be just one of the trillions of galaxies in the universe. Groups of galaxies are arranged in clusters, which then form larger superclusters. These superclusters extend thousands of light-years across the universe, interspersed with dark voids and other features, such as black holes.

Our galaxy contains about 400 billion stars and is about 100,000 light-years from end to end. The neighboring Andromeda Galaxy is more than 220,000 light-years across. It is estimated that there could be as many as 40 billion Earth-sized planets within our galaxy alone. Eleven billion of these estimated planets may be orbiting Sun-like stars, and some of them could potentially harbor intelligent life. Alpha Centauri, the Earth’s nearest Sun-like star system, is 4.37 light-years away.

What exactly is a “light-year”?

A light-year is the distance light travels in one year, given the symbol “ly.” Light travels through space at about 300,000 km per second (186,000 miles per second) and covers approximately 9.46 trillion km (5.88 trillion miles) annually. Light-time is used to measure the vast distances of space; nothing travels faster than light. Earth is about eight light minutes from the Sun, which is very close for a celestial body. A journey to the nearest star system, Alpha Centauri, would take more than four years, even if traveling at the speed of light, which is currently impossible under the prevailing understanding of physics. To put it into perspective, the Parker Solar Probe, the fastest spacecraft ever launched by humanity, takes about 90 years to travel one light-year at its maximum speed. These vast distances, combined with the long time it takes for light to travel from distant objects, pose significant challenges for exploring and understanding the universe beyond our solar system.

What is Astrodynamics?

Astrodynamics, also known as orbital mechanics, is a critical engineering field that encompasses the design and control of spacecraft missions. It involves applying mathematical and physical principles to understand and predict satellite motion, calculate spacecraft trajectories, determine orbits, and control spacecraft. The field is extensive, and this eBook can only begin to introduce the subject.

The study of astrodynamics is crucial to various aspects of space exploration, including the design of interplanetary missions and the maintenance of satellites in orbit for communication, navigation, and other purposes. Astrodynamics principles are used to determine the types of propulsion systems required, the propellant requirements, and the control strategies necessary to place a satellite into orbit, maintain it there, and maneuver it to achieve the mission objectives.

Getting into Space

Rockets or launch vehicles send satellites, space probes, and human-carrying spacecraft into Earth orbit. A representative launch profile of a rocket is shown in Figure 4. At the moment of the initial launch, the thrust produced by the rocket engine(s) will be greater than the vehicle’s weight, so the rocket accelerates away quickly from the pad. The rocket’s weight decreases rapidly because of the high propellant consumption, enabling it to accelerate quickly as it gains altitude. It also begins to pitch toward a more horizontal flight path under the influence of gravity, thereby increasing its translational velocity. Finally, as it exits the atmosphere at approximately 60,000 ft (20 km), it will fly at supersonic speed.

A representative launch profile for a two-stage rocket. The objective of the launch is to deliver the maximum payload to the required orbital altitude with the minimum fuel.

A few minutes into the ascent, staging will occur on the rocket booster. The first stage is jettisoned as it runs out of propellant, and the second stage’s rocket engine(s) are ignited. The empty first stage then falls back toward the surface and either burns up in the atmosphere (depending on the staging altitude) or breaks up and crashes into the ocean. In exceptional cases, the first stage may be recovered; solid rocket boosters are usually recovered by parachute and may be reused.

After reaching orbit, the payload is then on its own, traveling thousands of kilometers per hour as it orbits the Earth. The payload can then be configured to perform its intended mission, whether deploying satellites, conducting scientific experiments, or even transporting astronauts to the International Space Station. The complexity of the launch, ascent, and orbital insertion processes underscores the importance of a highly skilled and knowledgeable engineering team, a thorough understanding of the atmospheric and space environments, and the use of reliable, tested technologies. These factors are crucial to ensure the success of a rocket launch, which is also a costly endeavor.

Gravitational Laws of Attraction

The payload (e.g., a satellite) remains in orbit because it gains momentum and kinetic energy from the chemical energy released by the rocket engines’ propellant. In orbit, the balance of forces can be explained using the principles of statics and dynamics. In an inertial frame, the gravitational force provides the centripetal acceleration required for orbital motion. In a rotating (non-inertial) frame, this effect may be interpreted as a balance between gravitational attraction and an outward-directed centrifugal force. This balance between gravity and centrifugal forces keeps the satellite orbiting Earth, and any perturbation to this balance will cause it to either fall to Earth or escape to interplanetary space.

It is often said that Isaac Newton discovered the law of gravity. The legend is that he was eating his lunch under a tree at his mother’s house in 1666 when an apple fell on his head. Even Newton himself was happy to go along with this story if he were to claim the discovery of gravity. Although Isaac Newton is widely credited with formulating the law of gravity, Robert Hooke had previously proposed the concept of gravitational force and its attraction between bodies. However, Newton provided a robust mathematical framework for understanding gravity and formulated the law of universal gravitation, so most of the credit goes to Newton.

Universal Law of Gravitation

Hooke and Newton understood that the “force of gravity” between two masses must be related to the magnitude of the masses and the distance between them; the larger and closer they are, the stronger the mutual force between them. They both hypothesized that this force of attraction, F_{\rm grav}, will be directly dependent upon the product of the masses of both objects, i.e., m_1 and m_2, respectively, and inversely proportional to the square of the distance that separates their centers of mass, i.e.,

(1)   \begin{equation*} F_{\rm grav} \ \propto \ \frac{m_1 \, m_2}{R^2} \end{equation*}

where R is the distance between the centers of the two masses, as shown in Figure 5.

 

The “force of gravitational attraction” is proportional to the product of the masses and inversely proportional to the square of the distance between them.

Newton went further and determined that there must exist a universal constant proportionality, G, such that

(2)   \begin{equation*} F_{\rm grav} = \frac{G \, m_1 \, m_2}{R^2} \end{equation*}

Equation 40 is called the Universal Law of Gravitation. This force is attractive, i.e., the masses are attracted toward each other. Despite its simplicity, this fundamental work established Isaac Newton as one of the most influential scientists in the history of science.

Gravitational Constant

Newton did not live long enough for him to determine the numerical value of the universal gravitational constant, G. Its value was subsequently determined by Henry Cavendish, who measured the Earth’s average density and mass, M_E, which is 5.97 \times 10^{24} kg, and then deduced the value of G. The accepted modern value of the universal gravitational constant is G = 6.67428 \times 10^{-11} N m{^{2}} kg^{-2}.

Consequently, Eq. 40 is often redefined for the Earth as

(3)   \begin{equation*} F_{\rm grav} = \frac{\mu_E \, m}{R^2} \end{equation*}

where m is the mass of the orbiting body of the Earth. In this case, \mu_E is called the Earth’s gravitational constant, i.e.,

(4)   \begin{equation*} \mu_E = G \, M_E = 398,600.5~\mbox{km${^3}$ s$^{-2}$} \end{equation*}

It is essential to appreciate that the value of \mu will vary from planet to planet.

Check Your Understanding #1 – Your weight on another planet?

You are fortunate to be an astronaut sent to explore planet X. This planet has a mass four times Earth’s and a radius three times Earth’s. How will the “force of gravity” feel to you on planet X compared to when you are on Earth?

Show solution/hide solution.

Using the universal law of gravitation gives

    \[ F_{\rm Earth} = \frac{G \, M_{\rm Earth} \, m_{\rm You}}{R_{\rm Earth}^2} \]

The contribution of planet X’s mass will change your weight compared to Earth’s. Therefore, in this case,

    \[ { F_{\rm X} = \frac{G \, M_{\rm X} \, m_{\rm You}}{{R_{\rm X}}^2} } \]

Substituting the specified information gives

    \[ F_{\rm X} = \frac{G \, 4 M_{\rm Earth} \, m_{\rm You}}{(3 R_{\rm Earth})^2}\]

so that

    \[ F_{\rm X} = \frac{4}{9}  F_{\rm Earth} \]

Therefore, the force of gravity you will experience as an astronaut on planet X would be approximately 0.44 times that on Earth. This means you would weigh significantly less on planet X than on Earth.

Circular Orbits

Consider a satellite in a circular orbit around the Earth. Circular orbits occur when the gravitational force acting on a satellite provides the centripetal force required to maintain uniform circular motion, as illustrated in Figure 6.

Circular orbits arise whenever the gravitational force on a satellite is equal to the centrifugal force needed to move in a uniform circular motion.

The force equilibrium of a satellite of mass m_s is

(5)   \begin{equation*} \frac{m_s \, V_s^2}{R_s} = \frac{G \, M_E \, m_s}{R_s^2} = \frac{\mu_E \, m_s}{R_s^2} \end{equation*}

where V_s is the tangential or orbital speed of the satellite, and R_s is the radius of its orbit as measured from the center of the Earth. Solving for V_s gives

(6)   \begin{equation*} V_s = \sqrt{ \frac{\mu_E}{R_s} } \end{equation*}

so the orbital speed is inversely proportional to the square root of its orbital radius. Notice that this prior result is independent of the satellite’s mass.

It can be concluded from Eq. 6 that satellites that orbit closer to Earth (i.e., the value of R_s is less) experience more substantial effects of gravity, as shown in Figure 7. Therefore, they must travel faster than a satellite orbiting farther from Earth to stay in orbit. In a geosynchronous orbit (GEO), the satellite completes one orbit around Earth in 24 hours, so it appears stationary in the sky relative to an observer on Earth. This behavior is beneficial for communication satellites that require continuous coverage over a specific location on the Earth’s surface. The altitude of GEO is high enough, at approximately 35,786 km (22,236 miles) above the surface, that the satellite is not affected by drag from the Earth’s atmosphere, which would cause its orbit to decay. Note: Satellites at that altitude with no orbital tilt (0 degrees) are called geostationary, and those that have an inclination other than zero are called geosynchronous.

Low Earth orbit (LEO) is about 402 km (250 miles) above the Earth’s surface. Many satellites are in geosynchronous orbit (GEO) at more than 35,786 km (22,236 miles).

Low Earth orbit (LEO) is much closer to the Earth’s surface. Satellites in LEO are subject to atmospheric drag, which requires more energy to maintain their orbits and limits their lifetimes. For example, the International Space Station is in LEO at an altitude of about 402 km (250 miles) above the Earth’s surface. It must travel at approximately 27,600 kph (7.67 km/s or 17,150 mph) to remain in orbit. Many satellites, however, orbit in geosynchronous Earth orbit (GEOS) at an altitude of more than 35,786 km (22,236 miles) above the surface and so only need to travel at approximately 10,783 km/h (3 km/s) or 6,700 mph to maintain their orbits.

Check Your Understanding #2 – Height of a geosynchronous orbit

How high above the surface of the Earth is a satellite in a geosynchronous orbit? Assume a circular orbit. Assume also that \mu_E = 398,600.5 km{^3} s^{-2}, the angular velocity of the Earth, \omega_E, is 7.29 \times 10^{-5} rad s^{-1}, and the radius of the Earth, R_E, is 6,378.1 km.

Show solution/hide solution.

A geosynchronous orbit is an orbit in which a satellite maintains an approximately fixed position above a point on Earth’s surface. To maintain such a position, the satellite’s angular velocity, \omega_s, about the Earth must equal the Earth’s angular velocity, \omega_E. We know that

    \[ V_s = \sqrt{ \frac{\mu_E}{R_s} } \]

where R_s is the radius of the satellite’s orbit relative to the Earth’s center. Also, from simple physics

    \[ V_s = \omega_s R_s \]

Equating these two latter equations gives

    \[ \omega_s R_s = \sqrt{ \frac{\mu}{R_s} } \]

Solving for the orbital radius, R_s, gives

    \[ R_s = \left( \frac{\mu}{{\omega_s}^2} \right)^{1/3} \]

We know that \mu_E = 398,600.5 km{^3} s^{-2} = 3.985 \times 10^{14} m{^3} s^{-2}. Also, based on the known angular rotational speed of the Earth and its average radius, then \omega_E = 7.29 \times 10^{-5} rad s^{-1}. Therefore, calculating R_s gives

    \[ R_s = \left( \frac{\mu_E}{{\omega_E}^2} \right)^{1/3} = \left( \frac{3.985 \times 10^{14}}{(7.29 \times 10^{-5})^2} \right)^{1/3} = 42,168.8~\mbox{km} \]

This result indicates that a satellite in a geosynchronous orbit is at a height of R_s - R_E above Earth’s surface, where R_E is Earth’s radius, 6,378.1 km. Therefore, satellites must be placed at about 35,790.7 km (22,236 miles) above the surface.

Orbit Equation

The orbit equation describes the orbital motion of a smaller mass around a larger, at-rest total mass. It is often referred to as a solution to the two-body problem. After a mass or payload, e.g., a satellite, is in orbit, the principles of orbital mechanics govern its motion and position. Orbital mechanics studies the motion of planets, satellites, and space vehicles under the influence of gravity, their own thrust, and other forces. The study of orbital mechanics enables engineers to plan and execute missions that explore the solar system and deploy and control satellites for various purposes, including communication, navigation, weather prediction, and Earth observation.

Energy Equation

After launch, a satellite will have some initial velocity, momentum, and total mechanical energy imparted by the launch vehicle up to the point of final burnout. This energy has two components: kinetic energy, associated with motion, and gravitational potential energy, associated with position in the gravitational field. The balance between these two forms of energy determines the satellite’s motion.

As the satellite moves along its orbit, changes in altitude result in corresponding changes in gravitational potential energy, which are exchanged with kinetic energy. This continuous exchange governs the orbital motion. Earth’s gravity provides the centripetal attraction, while the satellite’s velocity prevents it from falling directly toward Earth. Other effects, such as atmospheric drag, solar radiation pressure, and gravitational perturbations, can modify the orbit over time.

For a satellite of mass m_s orbiting a central mass M, the gravitational potential energy, U, is defined relative to zero at infinity, so that

(7)   \begin{equation*} U = -\frac{G \, M \, m_s}{r} \end{equation*}

This result can be obtained by evaluating the work required to move a mass from a reference location at R_i to a position at R_f, i.e.,

(8)   \begin{equation*} W = -\int_{R_i}^{R_f} \frac{G \, M \, m_s}{r^2} \, dr = - G \, M \, m_s \int_{R_i}^{R_f} \frac{1}{r^2} \, dr \end{equation*}

Evaluating the integral gives

(9)   \begin{equation*} W = - G \, M \, m_s \left( \frac{1}{R_i} - \frac{1}{R_f} \right) \end{equation*}

Taking the reference location at infinity, R_i \rightarrow \infty, then

(10)   \begin{equation*} U = W = -\frac{G \, M \, m_s}{R_f} \end{equation*}

or, in general form,

(11)   \begin{equation*} U = -\frac{G \, M \, m_s}{r} \end{equation*}

Because the reference point is at infinity, the potential energy is negative for a bound orbit. The closer the satellite is to the central body, the more negative the value of U, corresponding to a more strongly bound state.

The kinetic energy, K, is

(12)   \begin{equation*} K = \frac{1}{2} m_s V^2 \end{equation*}

In polar coordinates, the velocity may be written as

(13)   \begin{equation*} V^2 = \dot{r}^{\,2} + (r \, \dot{\theta})^2 \end{equation*}

so that

(14)   \begin{equation*} K = \frac{1}{2} m_s \left[ \dot{r}^{\,2} + (r \, \dot{\theta})^2 \right] \end{equation*}

Therefore, the total mechanical energy is

(15)   \begin{equation*} E = K + U \end{equation*}

or

(16)   \begin{equation*} E = \frac{1}{2} m_s \left[ \dot{r}^{\,2} + (r \, \dot{\theta})^2 \right] - \frac{G \, M \, m_s}{r} \end{equation*}

For a satellite orbiting the Earth, this expression is commonly written using the Earth’s gravitational parameter \mu_E = G M_E, giving

(17)   \begin{equation*} E = \frac{1}{2} m_s \left[ \dot{r}^{\,2} + (r \, \dot{\theta})^2 \right] - \frac{\mu_E \, m_s}{r} \end{equation*}

Equation of Motion of a Satellite (Two-Body Problem)

Recall that the total mechanical energy of a satellite is

(18)   \begin{equation*} E = \frac{1}{2} m_s \left[ (r \, \overbigdot{\theta})^2 + \overbigdot{r}^{\, 2} \right] - \frac{\mu \, m_s}{r} \end{equation*}

where, for the two-body problem,

(19)   \begin{equation*} \mu = G \left( M + m_s \right) \end{equation*}

The motion of the satellite is most conveniently described in polar coordinates (r,\theta). The radial and transverse components of acceleration are

(20)   \begin{equation*} a_r = \overbigdot{\overbigdot{r}} - r \, \overbigdot{\theta}^{\, 2} \end{equation*}

(21)   \begin{equation*} a_\theta = r \, \overbigdot{\overbigdot{\theta}} + 2 \, \overbigdot{r} \, \overbigdot{\theta} \end{equation*}

Applying Newton’s second law in the transverse direction, noting that no force acts in the transverse direction, gives

(22)   \begin{equation*} m_s \left( r \, \overbigdot{\overbigdot{\theta}} + 2 \, \overbigdot{r} \, \overbigdot{\theta} \right) = 0 \end{equation*}

which can be written as

(23)   \begin{equation*} \frac{d}{dt} \left( r^2 \, \overbigdot{\theta} \right) = 0 \end{equation*}

Therefore, angular momentum is conserved, i.e.,

(24)   \begin{equation*} m_s \, r^2 \, \overbigdot{\theta} = \mbox{constant} \end{equation*}

or, per unit mass,

(25)   \begin{equation*} r^2 \, \overbigdot{\theta} = h = \mbox{constant} \end{equation*}

Applying Newton’s second law in the radial direction gives

(26)   \begin{equation*} m_s \left( \overbigdot{\overbigdot{r}} - r \, \overbigdot{\theta}^{\, 2} \right) = -\frac{\mu \, m_s}{r^2} \end{equation*}

or

(27)   \begin{equation*} \overbigdot{\overbigdot{r}} - r \, \overbigdot{\theta}^{\, 2} = -\frac{\mu}{r^2} \end{equation*}

Using the conservation of angular momentum, r^2 \, \overbigdot{\theta} = h, so that

(28)   \begin{equation*} \overbigdot{\theta} = \frac{h}{r^2} \end{equation*}

Substituting into the radial equation gives

(29)   \begin{equation*} \overbigdot{\overbigdot{r}} - \frac{h^2}{r^3} = -\frac{\mu}{r^2} \end{equation*}

The solution of this differential equation leads to the orbit equation

(30)   \begin{equation*} r(\theta) = \frac{p}{1 + e \cos(\theta - \phi)} \end{equation*}

where

(31)   \begin{equation*} p = \frac{h^2}{\mu} \end{equation*}

and e is the eccentricity of the orbit, and \phi is a constant defining the orientation of the orbit. The angle \theta is the true anomaly measured from periapsis. This equation describes all possible conic-section orbits depending on the value of e.

Solutions to the Orbit Equation

In canonical form, the solutions to the orbit equation are written in polar coordinates as

(32)   \begin{equation*} r(\theta) = \frac{p}{1 + e \cos \theta} \end{equation*}

where

(33)   \begin{equation*} p = \frac{h^2}{\mu} \end{equation*}

is the semi-latus rectum, and e is the eccentricity of the orbit.

This equation is one of the most fundamental results in orbital mechanics and describes a family of curves known as conic sections. Depending on the value of e, four types of orbital trajectories are obtained:

  • e = 0: circular orbit
  • 0 < e < 1: elliptical orbit
  • e = 1: parabolic trajectory
  • e > 1: hyperbolic trajectory
Four orbital trajectories can be derived from the orbit equation depending on the eccentricity value.

Kepler’s Laws of Orbital Motion

Johannes Kepler concluded that the planets in the solar system moved around the Sun in elliptical orbits. Using the meticulous astronomical measurements compiled by his mentor, Tycho Brahe, Kepler made the following conclusions:

  1. All planets move in elliptical orbits with the Sun at one focus.
  2. A line joining any planet to the Sun sweeps out equal areas in equal times.
  3. The square of the period of any planet about the Sun is proportional to the cube of the planet’s mean distance from the Sun.

Kepler then developed his three foundational principles of orbital mechanics, which, in their general form, apply to any mass orbiting a larger, central mass. These are now known as Kepler’s Laws of Orbital Motion. At the time of their formulation, they were considered exact laws, and over the centuries, they have yielded very close approximations to actual planetary motion.

Kepler’s laws can also be deduced from Newton’s laws of motion and the law of universal gravitation, although the formal annunciation of Newton’s laws came about nearly a century later. Indeed, many scientists have pointed out that Isaac Newton referred to Johannes Kepler’s work in the traditional formulation of his gravitational theory in the publication of his Principia. Newton was to show that Kepler’s laws are a particular case of the gravitational motion of n bodies where all the bodies may be treated as point masses. Their motion is not affected by each other except for the gravitational attraction of a large central mass. Therefore, Kepler’s laws are solutions to the so-called two-body problem.

Kepler’s 1st Law

Kepler’s first principle (or law) was developed from his observations that the planets’ orbital trajectories are ellipses about the Sun with the Sun at the focus: “A satellite describes an elliptical orbit around its center of attraction.” Figure 10 shows an annotated elliptical orbit with semi-major length a and semi-minor length b.

An elliptical orbit is described in terms of its eccentricity and the major and minor lengths.

This first law permits two conclusions to be drawn about how it applies to planets or satellites in orbit about another planet or body:

  1. Because they move in a curvilinear path, an external force, i.e., the force of gravitational attraction, acts upon them.
  2. They conserve mechanical energy and angular momentum in the absence of external forces.

Notice that the value of \theta is known as the true anomaly, which is the angle between the perigee and the position vector to the orbiting mass or satellite. If the semi-major axis and eccentricity for an orbit are known, then at perigee or a true anomaly \theta = 0^{\circ}, then

(34)   \begin{equation*} R_p = a (1 - e) \end{equation*}

At apogee, or a true anomaly of \theta = 180^{\circ}, then

(35)   \begin{equation*} R_a = a (1 + e) \end{equation*}

Kepler’s 2nd Law

By studying the orbit of Mars, Kepler noticed that a line drawn between the Sun and Mars swept out equal areas in equal times, as illustrated in Figure 11. He reasoned that when the planet is closest to the Sun, or at perihelion, it must travel faster than when further from the Sun at its aphelion. This observation became the basis for Kepler’s second law or “law of areas,” i.e., “A satellite sweeps out equal areas in equal times around their center of attraction.”

Kepler’s second law states that a line drawn between the central mass and a satellite sweeps out an equal area in an equal amount of time.

Notice that the areas of the small shaded areas can be written as

(36)   \begin{equation*} dA = \frac{1}{2} r^2 \, d \theta \end{equation*}

where d\theta is the angular distance. The angular momentum of a satellite is conserved, i.e.,

(37)   \begin{equation*} m \, r^2 \, \overbigdot{\theta} =\mbox{constant} \end{equation*}

so for a constant mass, m, then r^2 \, \overbigdot{\theta} = constant. The change in swept area with respect to time is

(38)   \begin{equation*} \frac{dA}{dt} = \frac{1}{2} r^2 \left( \frac{d \theta}{d t} \right) = \frac{1}{2} r^2 \, \overbigdot{\theta} = \mbox{constant} \end{equation*}

which confirms Kepler’s second law.

Kepler’s second law also permits two conclusions to be drawn about how it applies to planets or satellites in orbit about another planet or body:

  1. Planets do not move with constant speed along their orbits.
  2. A planet moves fastest at perihelion (or perigee) and slowest at its aphelion (or apogee).

Kepler’s 3rd Law

Kepler’s third law, established much later than the first two laws, states: “The square of the orbital period is directly proportional to the cube of the semi-major axis (i.e., the average orbital distance) of the orbit.” The third law appeared in what Kepler called his opus magnum, namely the book “Ioannis Keppleri Harmonices mundi libri V,” or in English “The Five Books of Johannes Kepler’s The Harmony of the World,” which was published in 1619. More precisely, his third law quantifies that the orbital period, T, is proportional to the cube of the semi-major length, a, of the elliptical orbit, i.e.,

(39)   \begin{equation*} T^2 \ \propto \ a^3 \quad \text{or} \quad T \ \propto \ a^{3/2} \end{equation*}

Kepler’s third law, often called the “harmonic law,” also verifies that the larger the orbit, the longer it takes for the satellite to traverse it.

Kepler’s third law can also be derived from Newton’s universal law of gravitation, i.e., by using

(40)   \begin{equation*} F_{\rm grav} = -\frac{G \, m_1 \, m_2}{R^2} \end{equation*}

Consider a satellite of mass m_s traveling at a velocity V in a circular orbit of radius R about a central mass M. The force equilibrium is such that the outward centrifugal force equals the gravitational attraction, i.e.,

(41)   \begin{equation*} \frac{m_s \, V^2}{R} = \frac{G \, M \, m_s}{R^2} \end{equation*}

The satellite will travel a distance of 2\pi \, R in one orbital period T, so V is

(42)   \begin{equation*} V = \frac{2\pi \, R}{T} \end{equation*}

Therefore,

(43)   \begin{equation*} V^2 = \frac{4 \pi^2 R^2}{T^2} \end{equation*}

so that

(44)   \begin{equation*} \frac{G M \, m_s}{R^2} = \frac{m_s \left(4 \pi^2 R^2\right)}{R \, T^2} \end{equation*}

which, on rearrangement, gives

(45)   \begin{equation*} T^2 = \left( \frac{4 \pi^2}{G \, M} \right) R^3 \end{equation*}

noting that this latter result does not depend on the satellite’s mass. This result applies strictly to a circular orbit, where R is constant. The conclusion is that

(46)   \begin{equation*} T^2 \ \propto \ R^3 \end{equation*}

which is Kepler’s third law for a circular orbit. For an elliptical orbit, the radius R is replaced by the semi-major axis a, giving

(47)   \begin{equation*} T^2 = \left( \frac{4\pi^2}{G \, M} \right) a^3 \end{equation*}

Kepler’s third law enables predictions of the orbits of any satellite or object in space orbiting a central body. Although it was previously derived under the assumption of a circular orbit, it also applies to an elliptical orbit. This relationship can be more formally quantified by

(48)   \begin{equation*} { T = 2 \pi \sqrt{ \frac{a^3}{\mu }} } \end{equation*}

where

(49)   \begin{equation*} \mu = G \left( M + m_s \right) \end{equation*}

and where a is the length of the semi-major axis of the satellite’s elliptical orbit.

Now consider the case of the orbital mass m_1 of one satellite about a large central mass M with period T_1 and semi-major axis length a_1, and let m_2 be the mass of another satellite about M with period T_2 and semi-major axis length a_2. Then the use of Eq. 48 gives

(50)   \begin{equation*} G \left( M + m_1 \right) = 4 \pi^2 \frac{{a_1}^{3}}{{T_1}^{2}} \end{equation*}

and

(51)   \begin{equation*} G \left( M + m_2 \right) = 4 \pi^2 \frac{{a_2}^{3}}{{T_2}^{2}} \end{equation*}

Therefore,

(52)   \begin{equation*} \frac{M + m_1}{M + m_2} = \left( \frac{a_1}{a_2} \right)^3 \left( \frac{T_2}{T_1} \right)^2 \end{equation*}

which is the correct form of Kepler’s third law.

However, because m_1 and m_2 are generally much smaller than M, even for the most massive planets that orbit the Sun, such as Jupiter, then the quantity on the left side of Eq. 52 is almost unity, i.e.,

(53)   \begin{equation*} { \left( \frac{a_1}{a_2} \right)^3 \left( \frac{T_2}{T_1} \right)^2 \approx 1 } \end{equation*}

so that

(54)   \begin{equation*} \frac{T_2}{T_1} = \left( \frac{a_2}{a_1} \right)^{3/2} \end{equation*}

which is the most commonly used form of Kepler’s third law.

Check Your Understanding #3 – Orbit time of Mars?

The Earth orbits around the Sun in one year, i.e., T_E =365.256 days. (Note: Ever wonder why we have a leap year every four years?) The length of the semi-major axis of the Earth’s orbit is a_E = 1.495 \times 10^8 km. Now, if the semi-major axis, a_M, of Mars is 2.278 \times 10^8 km, then what will the orbit time, T_M, of Mars be?

Show solution/hide solution.

Using Kepler’s third law, then

    \[ T = 2 \pi \sqrt{ \frac{a^3}{\mu}} \approx 2 \pi \sqrt{ \frac{a^3}{G \, M}} \]

Both the Earth and Mars orbit around the Sun, so the value of G \, M is the same for both planets, so that

    \[ \frac{T_M}{T_E} = \sqrt{ \frac{{a_M}^3}{{a_E}^3} } = \left( \frac{a_M}{a_E} \right)^{3/2} \]

Inserting the numerical values gives

    \[ T_M = T_E \left( \frac{a_M}{a_E} \right)^{3/2} = \left( \frac{2.278}{1.495} \right)^{3/2} = 696.96~\mbox{days} = 1.88~\mbox{years} \]

Mirror Theorem

The Mirror Theorem[4] describes a special symmetry in the Newtonian n-body problem. In a gravitational system of n bodies, a mirror configuration occurs when each body is moving perpendicular to its position vector from the system’s center of mass. At such a time, the system undergoes no net expansion or contraction, and all motion is tangential.

The Mirror Theorem states that if the system passes through a mirror configuration at time t_0, then the motion after that instant is the exact mirror image in time of the motion before. The paths of the bodies retrace themselves, and velocities reverse. Suppose the system passes through two such mirror configurations at times t_1 and t_2. In that case, the entire motion is periodic, i.e., all bodies return to their original positions and velocities after a time T = t_2 - t_1. This symmetry arises from the time-reversibility of Newton’s laws and provides an analytical tool for identifying periodic solutions in multi-body gravitational systems.

Determining the Mass of a Planet

One question people often ask is how the mass of a planet is determined. However, any planet with a satellite orbiting it, whether natural or artificial, can be used to measure the planet’s mass by studying the satellite’s orbital motion and applying Kepler’s third law.

Consider first the Earth’s orbit around the Sun, with the Sun having mass M_S and the Earth mass M_E, and an orbital semi-major axis a_E. Then using Eq. 50 gives

(55)   \begin{equation*} G \left( M_S + M_E\right) = 4 \pi^2 \frac{{a_E}^{3}}{{T_E}^{2}} \end{equation*}

Consider next a satellite of mass m_s in orbit around Earth with period T_s and orbital semi-major axis a_s. In this case, then

(56)   \begin{equation*} G \left( M_E + m_s\right) = 4 \pi^2 \frac{{a_s}^{3}}{{T_s}^{2}} \end{equation*}

Therefore, using both of these prior equations, then

(57)   \begin{equation*} \frac{M_E + m_s}{M_S + M_E} = \left( \frac{a_s}{a_E} \right)^3 \left( \frac{T_E}{T_s} \right)^2 \end{equation*}

The quantities a_s, a_E, T_E, and T_s, on the right-hand side of Eq. 57, can all be measured, so the Earth’s mass relative to the Sun can then be determined. It turns out that the Earth’s mass is approximately 333,000 times less than the Sun’s mass. As a reference, the Sun’s mass is estimated to be roughly 1.989 \times 10^{30} kg.

The uncertainty in the measurement of Earth’s mass is also related to the uncertainty in the gravitational constant, G. Modern measurements of G have been repetitions of the classic Cavendish experiment, reducing the uncertainty in measuring the Earth’s mass. It is generally accepted that the Earth’s mass, M_E, is 5.9722 \times 10^{24} kg. The Earth’s mass is also somewhat variable, depending on the relative balance between accretion of inward-falling material, including micrometeorites and cosmic dust, and loss of atmospheric gases.

Check Your Understanding #4 – Calculating the mass of Mars

The Viking Mars orbiter probe was placed in a circular orbit at an altitude of 17,000 km above the Martian surface. Doppler measurements of the transmitted signals from the probe indicated that it was at an orbital velocity, V_V, of 1.46 km/s. Calculate the mass of Mars, M_M, in terms of the mass of the Sun, M_S, given that the diameter of Mars, d_M, is 6,770 km and the radius of the Mars orbit about the Sun, a_M, is 1.524 AU. Note: 1 AU = {1.495 \times 10^8} km; 1 year = 3.156 \times 10^7 s.

Show solution/hide solution.

For the Sun and Mars, using Kepler’s third law gives

    \[ T_M = a_M^{3/2} = 1.524^{3/2} = 1.88~\mbox{years} \]

For Mars and the Viking probe, the orbital radius of the probe measured from the center of Mars is

    \[ a_V = 17,000 + \frac{d_M}{2} = 17,000 + \frac{6,770}{2} = 20,385~\mbox{km} = 1.3635 \times 10^{-4}~\mbox{AU} \]

Therefore, the time for Viking to orbit Mars, T_V, will be

    \[ T_V = \frac{2 \pi \, a_V}{V_V} = \frac{2 \pi \times 20,385}{1.46} = 87,727.9~\mbox{s} = 2.7797 \times 10^{-3}~\mbox{years} \]

Using Kepler’s third law for Mars about the Sun and for Viking about Mars gives

    \[ G \left( M_S + M_M \right) = 4 \pi^2 \frac{a_M^3}{T_M^2} \]

and

    \[ G \left( M_M + m_V \right) = 4 \pi^2 \frac{a_V^3}{T_V^2} \]

where m_V is the mass of the Viking probe. Because M_M \ll M_S and m_V \ll M_M, then

    \[ \frac{M_M}{M_S} \approx \left( \frac{T_M}{T_V} \right)^2 \left( \frac{a_V}{a_M} \right)^3 \]

Substituting the values gives

    \[ \frac{M_M}{M_S} = \left(\frac{1.88}{2.7797 \times 10^{-3} } \right)^2 \left( \frac{1.3635 \times 10^{-4}}{1.524} \right)^3 \]

Therefore,

    \[ \frac{M_M}{M_S} = 3.28 \times 10^{-7} \]

and

    \[ M_M = 3.28 \times 10^{-7} \, M_S \]

So, the mass of Mars is about 3,050,000 times less than that of the Sun.

How long does it take the Moon to go around Earth?

It takes 27 days, 7 hours, 43 minutes, and 11.461 seconds for the Moon to complete one full orbit around the Earth. This period is called the sidereal month. However, the Moon takes about 29.5 days to complete one cycle of phases, called the synodic month. The difference between the sidereal and synodic months is that, as the Moon orbits around the Earth, the Earth also orbits around the Sun. Therefore, the Moon must travel farther in its orbit to catch up and compensate for the added distance to complete the rotational phase cycle.

Orbital Energy

Orbital mechanics is primarily governed by the satellite’s total energy, which consists of kinetic and potential energy. The total mechanical energy of an object in orbit around another mass must remain constant as a result of the conservation of energy. In addition to energy, the laws of orbital mechanics are also influenced by the law of universal gravitation, which describes the attractive force between two masses, and the laws of motion, which relate how an object moves in response to forces acting on it. These fundamental laws help explain the motion and other characteristics of satellites orbiting Earth or other celestial bodies.

Circular Orbits

The kinetic energy of the satellite, K, is

(58)   \begin{equation*} K = \frac{1}{2} m_s V^2 \end{equation*}

and the associated potential energy, U, is

(59)   \begin{equation*} U = -\frac{G \, M \, m_s}{R_s} \end{equation*}

Therefore, the total energy is

(60)   \begin{equation*} E = K + U = \frac{1}{2} m_s V^2 -\frac{G \, M \, m_s}{R_s} \end{equation*}

The force equilibrium for a satellite of mass m_s in a circular orbit is

(61)   \begin{equation*} \frac{m_s \, V_s^2}{R_s} = \frac{G \, M \, m_s}{R_s^2} \end{equation*}

where V_s is the tangential or orbital speed of the satellite, and R_s is the radius of its orbit as measured from the center of mass, M. Rearranging the preceding equation gives

(62)   \begin{equation*} \frac{1}{2} m_s \, V_s^2  = \frac{1}{2} \left( \frac{G \, M \, m_s}{R_s} \right) \end{equation*}

and so the energy equation becomes

(63)   \begin{equation*} E = K + U = \frac{1}{2} m_s V^2 - \frac{G \, M \, m_s}{R_s } = \frac{1}{2} \left( \frac{G \, M \, m_s}{R_s} \right) -\frac{G \, M \, m_s}{R_s} \end{equation*}

After rearrangement then

(64)   \begin{equation*} E = K + U = -\frac{1}{2}\left( \frac{G \, M \, m_s}{R_s} \right) \end{equation*}

which states that the satellite’s energy in orbit around mass M is negative. This result also holds for an elliptical orbit.

Negative Potential Energy?

What does it mean when the energy given by Eq. 64 is negative? This situation is called a bound orbit, in that the satellite is bound to a mass or planet, i.e., it is gravitationally bound in the same manner that the Earth is bound to the Sun and the Moon is bound to the Earth. A bound orbit is a closed orbit, either circular or elliptical, and the body or satellite will remain around the source of the gravitational attraction. Circular orbits have the minimum energy. Bound circular and elliptical orbits always have E < 0. Unbound trajectories, such as parabolic orbits and hyperbolic orbits, will have E > 0, and the object will escape from the source of the gravitational attraction.

Elliptical Orbits

It will be apparent that when a satellite follows an elliptical rather than a circular orbit, its total energy E is also conserved. However, in a non-circular orbit, there must always be an interchange between a satellite’s kinetic energy, K, and its potential energy, U. This latter result is a direct consequence of Kepler’s second law: a satellite moves faster at perigee and slower at apogee, as shown in Figure 12.

Because of momentum and energy conservation, a satellite will speed up at its perigee and slow down at its apogee.

The energy equation for a satellite of mass m in an elliptical orbit with a semi-major axis length a about a central mass M can be written as the sum of its kinetic energy and the potential energy, i.e.,

(65)   \begin{equation*} E = K + U = \frac{1}{2} m V^2 - \frac{G \, M \, m}{r} \end{equation*}

The total kinetic energy can be written as

(66)   \begin{equation*} K = \frac{1}{2} m_s V_s^2 \end{equation*}

and the total mechanical energy becomes

(67)   \begin{equation*} E = \frac{1}{2} m_s V_s^2 - \frac{G \, M \, m}{r} \end{equation*}

The energy can also be written as

(68)   \begin{equation*} E = -\left(\frac{G \, M \, m_s }{r_1 + r_2 } \right) = -\left(\frac{\mu m_s }{a } \right) \end{equation*}

where

(69)   \begin{equation*} \mu = G \left( M + m_s \right) \approx G \, M \end{equation*}

The energy is usually written per unit mass of the satellite, i.e.,

(70)   \begin{equation*} E = -\frac{\mu}{2 a } \end{equation*}

Because the orbital energy is constant, the orbit’s overall size remains constant. Therefore, a conclusion is that the satellite will move the fastest at its closest point to the central mass M, i.e., at its perigee (or perihelion) where r = r_1 = a (1 - e) and slowest at its farthest point, i.e., at its apogee (or aphelion) where r = r_2 = a (1 + e). It can be shown that the velocity, V, of a satellite in an elliptical orbit, where r is the distance between the satellite and the central mass, is given by

(71)   \begin{equation*} V = \sqrt{ \mu \left( \frac{2}{r} - \frac{1}{a} \right)} \end{equation*}

which is historically called the vis-viva equation. Therefore, at perihelion, its velocity is

(72)   \begin{equation*} V_p = \sqrt{ \frac{\mu}{a} \left( \frac{1 + e}{1 - e} \right)} \end{equation*}

and at aphelion, its velocity is

(73)   \begin{equation*} V_a = \sqrt{ \frac{\mu}{a} \left( \frac{1 - e}{1 + e} \right) } \end{equation*}

The eccentric anomaly, E, is another angular parameter that defines the position of a body moving along an elliptic orbit. Indeed, the eccentric anomaly is one of three angular parameters (“anomalies”) that define a position along an orbit, the other two being the true anomaly, as previously mentioned, and the mean anomaly, M. The eccentric anomaly E is related to the mean anomaly M by Kepler’s equation, i.e.,

(74)   \begin{equation*} M = E - e \sin E \end{equation*}

which is a transcendental equation. The symbol conflict with the mass should be noted. Given E and {e} in Kepler’s equation, it is trivial to find M. However, if M and {e} are known, then E must be determined numerically using an interactive approach.

Lagrange Points

In the restricted three-body problem, a small object of negligible mass can remain in a fixed position relative to two more massive orbiting bodies if placed at specific locations where the gravitational and centrifugal forces balance. These locations are known as Lagrange points. This effect is named after the mathematician Joseph-Louis Lagrange. There are five such points, denoted L_1 through L_5, and they represent solutions to the equations of motion in a rotating frame of reference.

Consider a system of two massive celestial bodies, such as the Sun and Earth, orbiting their common center of mass. In the rotating frame, the positions of the two primary bodies remain fixed. A third body, with negligible mass compared to the other two, moves under their gravitational influence, as shown in Figure 13. In this frame, the equations of motion can be written to include a centrifugal potential, thereby yielding an effective potential. Lagrange points are found by solving for positions where the net force, including both gravitational and centrifugal contributions, vanishes.

Lagrange points are positions in space where objects sent there tend to stay put. At Lagrange points, the gravitational pull of two large masses precisely equals the centripetal force required for a small object to move with them.

The five Lagrange points are defined as follows:

  1. The point L_1 lies along the line connecting the two massive bodies, between them. It is the position where the gravitational pull of both bodies and the centrifugal force balance such that the third body orbits with the same angular velocity. The L_1 point is commonly used for solar observatories, such as the SOHO spacecraft, because it provides an uninterrupted view of the Sun.
  2. The point L_2 lies along the line connecting the two bodies, but beyond the smaller of the two. It also allows the third body to orbit with the same angular velocity as the primary bodies. This point is ideal for deep space telescopes, such as the James Webb Space Telescope, because it offers a stable thermal environment and minimal interference from the Sun or Earth.
  3. The point L_3 lies on the opposite side of the larger body, directly opposite the smaller one. It is the least accessible of the colinear points and has no current practical use in mission design.
  4. The points L_4 and L_5 form equilateral triangles with the two massive bodies and lie ahead of and behind the smaller body in its orbit, respectively. These points are stable for systems in which the mass ratio between the primary bodies exceeds approximately 24.96, as in the Sun-Earth and Sun-Jupiter systems. Small perturbations near these points result in bounded oscillatory motion, making them dynamically stable. Natural objects such as Jupiter’s Trojan asteroids are found near its L_4 and L_5 points.
  5. In contrast, the colinear points L_1, L_2, and L_3 are dynamically unstable. A spacecraft placed at these points requires continuous station-keeping or may instead occupy a stable periodic or quasi-periodic orbit around the point, such as a halo orbit or Lissajous orbit.

Lagrange points are of significant importance in mission planning, as they enable spacecraft to remain in stable or quasi-stable positions relative to the Earth and Sun with minimal propellant usage, thereby supporting continuous observation, communication, and deep-space exploration objectives.

Check Your Understanding #5 – Orbital motion of a comet

A comet moves in an elliptical orbit about the Sun, and its orbit is coplanar with Earth’s orbit. The comet is observed to cross the Earth’s orbit at a time where its heliocentric velocity is 31.0 km/s, and its true anomaly is 140^{\circ}. Taking the Earth’s orbit to be circular, calculate the semi-major axis length and eccentricity of the comet’s orbit and the time before the comet next crosses the Earth’s orbit.

Show solution/hide solution.

At time t then r = 1.0 AU, {\scriptstyle{f}} = 140^{\circ}, and V = 31.0 km/s = 6.544 AU/year. Using the vis-viva equation for a body in an elliptical orbit, then

    \[ V = \sqrt{ \mu \left( \frac{2}{r} - \frac{1}{a} \right)} \]

In this case, r = 1 AU, so

    \[ \left( \frac{2}{r} - \frac{1}{a} \right) = \frac{V^2}{\mu} = \frac{6.544^2}{4 \pi^2} = 1.0848 \]

Therefore,

    \[ \frac{1}{a} = 2.0 - 1.0848 = 0.915 \]

So, the semi-major length of the orbital axis is a = 1.093~\mbox{AU}. We also know that for an elliptical orbit that

    \[ r = \frac{a(1 - e^2}{1 + e \cos f} \]

which on rearrangement leads to a quadratic in {e}, i.e.,

    \[ a e^2 + \cos f e + (1 - a) = 0 \]

The solution for the eccentricity is {e} = 0.80625, which is a very stretched elliptical orbit. Because

    \[ { r = a ( 1 - e \cos E) } \]

where E is the eccentric anomaly, then

    \[ \left( 1 - \frac{r}{a} \right) \frac{1}{e} = \cos E = \left( 1 - \frac{1}{1.093} \right) \frac{1}{0.8063} = 0.1052 \]

so that E = 83.96^{\circ} at orbit crossing. The period of the comet’s orbit (paying attention to the units) is

    \[ \tau = 2 \pi \sqrt{ \frac{a^3}{\mu} } = \sqrt{ a^3} = \sqrt{ 1.093^3} = 1.142~\mbox{years} \]

At {\scriptstyle{f}} = 140^{\circ} then E is 83.96^{\circ} = 1.465 radians, which using Kepler’s equation means that

    \[ M = E - e \sin E \]

where M is the mean anomaly, and so M = 0.6636 in this case. Also, the mean angular velocity (orbit frequency) is

    \[ n = \frac{2 \pi}{T} = \frac{2 \pi}{1.142} = 5.501~\mbox{rads/year} \]

Therefore, the orbit time is

    \[ t_1 = \frac{M}{n} = \frac{0.6636}{5.501} = 0.1206~\mbox{years} \]

At {\scriptstyle{f}} = 220^{\circ} and E = 276.04^{\circ}, then now

    \[ m = E - e \sin E = 4.818 - 0.80625 \sin 4.818 = 5.6196 \]

Therefore,

    \[ t_2 = \frac{M}{n} = \frac{5.6196}{5.501} = 1.0216~\mbox{years} \]

and so the time to wait for the next orbit crossing of the comet will be

    \[ t_2 - t_1 = 1.0216 - 0.1207 = 0.901~\mbox{years} \]

Energy Considerations of Changing Orbits

It is often required that a satellite or other spacecraft be able to change its orbit from a lower to a higher orbit or vice versa, as shown in Figure 14. After launch, a satellite may be placed in an initial (lower) parking orbit until the orbit stabilizes, after which it is boosted to its final altitude. Raising the orbital altitude requires energy and propellant. A rocket engine must be fired to create work, W, to increase the radius of the orbit from R_1, where it has energy E_1, to a radius R_2, where it has energy E_2. Often, the orbital altitude relative to the surface of the Earth, {h}, is used such that R_1 = R_E + h_1 and R_2 = R_E + h_2, where R_E is the radius of the Earth.

It is often required that a spacecraft be able to change its orbit, which requires a burn from the rocket engine.

Consider a satellite or spacecraft of mass m_s in an initially circular orbit of radius R_1 about the Earth, which has mass M_E. Its orbital energy is

(75)   \begin{equation*} E_1 = -\frac{1}{2}\left( \frac{G \, M_E \, m_s}{R_1} \right) \end{equation*}

To get to the higher orbit, the conservation of energy requires

(76)   \begin{equation*} E_2 = E_1 + W \end{equation*}

where W is the work required to raise the orbital altitude. Inserting the values of E_1 and E_2 gives

(77)   \begin{equation*} -\frac{1}{2}\left( \frac{G \, M_E \, m_s}{R_2} \right) = -\frac{1}{2} \left( \frac{G \, M_E \, m_s}{R_1} \right) + W \end{equation*}

Therefore, the work (or energy) needed to raise the orbit is

(78)   \begin{equation*} W = \frac{1}{2} \left( \frac{G \, M_E \, m_s}{R_1} \right) - \frac{1}{2}\left(\frac{G \, M_E \, m_s}{R_2} \right) \end{equation*}

and further rearranging gives

(79)   \begin{equation*} W =\frac{1}{2} G \, M_E \, m_s \left( \frac{1}{R_1} - \frac{1}{R_2} \right) =\frac{1}{2} G \, M_E \, m_s \left( \frac{R_2 - R_1}{R_1 \, R_2} \right) \end{equation*}

Finally, in terms of orbital altitude, then

(80)   \begin{equation*} W = \frac{1}{2} G \, M_E \, m_s \left( \frac{ h_2 - h_1}{(h_1 + R_E)  (h_2 + R_E)} \right) \end{equation*}

Similar principles apply to elliptical orbits.

Hohmann Transfer Orbit

The Hohmann cotangential elliptic transfer orbit is a minimum-energy orbital maneuver used to transfer a spacecraft between two orbits at different altitudes, as shown in Figure 15. The maneuver is accomplished by placing the spacecraft into an elliptical transfer orbit, called a Hohmann transfer orbit, in which the perigee and apogee of the trajectory are tangential to the initial and target orbits. The procedure is named after Walter Hohmann, whose early work on space flight and rocket science helped lay down the foundations of astrodynamics.

A Hohmann cotangential elliptic transfer orbit can be used to transition a spacecraft between two orbits.

The transfer maneuver needs excess energy, so the spacecraft uses two impulsive engine burns. The first burn is required to increase {V_1}, break out of the initial circular orbit, and establish the elliptical transfer orbit. The second burn provides the spacecraft with the velocity required at the altitude of the final (target) orbit, {V_2}.

Notice that a second burn is needed because the spacecraft loses its kinetic energy as it reaches the aphelion of the transfer orbit, so it will be too slow to stay at the final orbital altitude without the {\Delta V} boost obtained using the second burn to circularize the orbit. Depending on the spacecraft’s orientation upon reaching the target orbit, it may need to be reoriented so that the engine(s) can fire in the orbital direction to increase its speed.

Assuming a transfer orbit around the Earth, for the lower (initial) orbit, solving for {V_1} gives

(81)   \begin{equation*} V_1 = \sqrt{ \frac{G \, M_E}{R_1} } \end{equation*}

and for the higher (final) orbit, then

(82)   \begin{equation*} V_2 = \sqrt{ \frac{G \, M_E }{R_2} } \end{equation*}

Remember that {V_2} in the higher, final orbit will be lower than {V_1} in the lower, initial orbit. However, at the end of the transfer orbit, the spacecraft’s velocity will still be less than the needed {V_2} without a burn.

The total energy of the spacecraft in the initial orbit, E_1, is

(83)   \begin{equation*} E_1 = -\frac{1}{2}\left( \frac{G \, M_E \, m_s}{R_1} \right) \end{equation*}

and in the final orbit, it must be

(84)   \begin{equation*} E_2 = -\frac{1}{2}\left( \frac{G \, M_E \, m_s}{R_2} \right) \end{equation*}

In the Hohmann elliptical transfer orbit, the energy E_T will be

(85)   \begin{equation*} E_T = -\frac{1}{2} \left( \frac{G \, M_E \, m_s}{a_T }\right) = -\left( \frac{G \, M_E \, m_s}{(R_1 + R_2)} \right) \end{equation*}

noticing that the length of the semi-major axis of the elliptical transfer orbit is

(86)   \begin{equation*} a_T = \frac{R_1 + R_2}{2} \end{equation*}

Consider the point in time just after the first burn to leave the initial orbit, in which a change in velocity, \Delta V_1, is required. In this case, the total energy just after the burn can be written as

(87)   \begin{equation*} -\frac{G \, M_E \, m_s}{(R_1 + R_2) } + \frac{1}{2} m_s {V_p}^2 = -\frac{G \, M_E \, m_s}{R_1} \end{equation*}

where V_p is the velocity immediately after the burn, i.e., at the perigee of the elliptical transfer orbit. Notice that the mass of the spacecraft, m_s, cancels out in this latter equation so that

(88)   \begin{equation*} \frac{1}{2} V_p^2 = \frac{G \, M_E}{R_1} -\left( \frac{G \, M_E }{R_1 + R_2} \right) \end{equation*}

Solving for V_p gives

(89)   \begin{equation*} V_p = \sqrt{ 2 G \, M_E \left( \frac{1}{R_1} - \frac{1}{R_1 + R_2} \right) } \end{equation*}

or in terms of \Delta V_1, then

(90)   \begin{equation*} V_p = V_1 + \Delta V_1 = \sqrt{ 2 G \, M_E \left( \frac{1}{R_1} - \frac{1}{R_1 + R_2} \right)} \end{equation*}

After the spacecraft is established in the transfer orbit, it will continue under gravity until it reaches the height of the final orbit. At this point, a second burn will be needed. The velocity, V_a, which is at the aphelion of the transfer orbit and just before the second burn, will be

(91)   \begin{equation*} V_a = \sqrt{ 2 G \, M_E \left( \frac{1}{R_2} - \frac{1}{R_1 + R_2} \right) } \end{equation*}

where it will be noticed that R_1 in the first term inside the parentheses in Eq. 90 has been replaced by R_2 in Eq. 91.

However, at the end of the transfer orbit, this velocity will be too low to maintain the final orbit. Without a second burn, the spacecraft will return to a lower orbit and remain in the transfer orbit indefinitely. The velocity in the final circular orbit (Eq. 82) must be

(92)   \begin{equation*} V_2 = \sqrt{ \frac{G \, M_E }{R_2} } \end{equation*}

so the needed \Delta V_2 can then be calculated, i.e.

(93)   \begin{equation*} \Delta V_2 = V_2 - V_a \end{equation*}

The corresponding transfer time, \tau will be half of the elliptical transfer orbit time, T, i.e.,

(94)   \begin{equation*} \tau = \frac{T}{2} = \frac{2 \pi}{2} \sqrt{ \frac{a^3}{G \, M_E} } = \pi \sqrt{ \frac{a^3}{G \, M_E} } \end{equation*}

The values of {\Delta V} at each burn can be calculated using the rocket equation, i.e.,

(95)   \begin{equation*} \Delta V = I_{\rm sp} \, g_0 \ln \left( \frac{M_0}{M_f} \right) \end{equation*}

where {M_0} is the initial mass of the spacecraft at the beginning of the burn, and {M_f} is the final (burnout) mass after the burn during which a propellant mass of {M_P} is consumed, i.e., M_f = M_0 - M_P.

Using the rocket equation gives

(96)   \begin{equation*} \frac{M_0}{M_f} = \exp \left( \frac{\Delta V}{I_{\rm sp} \, g_0} \right) = \frac{M_0}{M_0 - M_P} \end{equation*}

and so

(97)   \begin{equation*} \frac{M_P}{M_0} = 1 - \exp \left( -\frac{\Delta V}{I_{\rm sp} \, g_0} \right) \end{equation*}

Of course, the reverse process described can be used to decrease a spacecraft’s orbital altitude. A retro-burn will be required at the higher orbit to reduce velocity and initiate the orbital transfer. At the end of this transfer, as gravitational potential energy increases (it becomes less negative), a retro-burn will be needed to break out of the elliptical orbit and circularize it at the lower orbital altitude.

Check Your Understanding #6 – Raising an orbit using a Hohmann transfer

Calculate the two velocity increments required to change the orbit of the Viking probe from a circular orbit at 17,000 km above the Martian surface to another circular orbit of a height of 30,000 km above the surface. Assume a Hohmann cotangential elliptic transfer orbit. Also, determine the semi-major axis, eccentricity, and transfer time of the transfer orbit. The diameter of Mars, d_M, is 6,770 km. Note: 1 AU = {1.495 \times 10^8} km; 1 year = 3.156 \times 10^7 s. Bonus: Why could the probe not be placed at an orbital altitude of 800,000 km?

Show solution/hide solution.

Notice that in this case, R_1 is 20,385 km or 1.36354 \times 10^{-4} AU and R_2 is 33,385 km or 2.2331 \times 10^{-4} AU. Therefore, for the elliptical transfer orbit, the semi-major axis, a, will be

    \[ a = \frac{R_1 + R_2}{2} = \frac{20,385 + 33,385}{2} = 26,885~\mbox{km} \]

The corresponding eccentricity of the transfer orbit will be

    \[ e = \frac{R_2 - R_1}{R_2 + R_1} = \frac{33,385 - 20,385}{33,385 + 20,385} = 0.2418 \]

The orbital transfer time, \tau, will be

    \[ \tau = \pi \sqrt{ \frac{a^3}{G \, M_M} } \]

where G \, M_M = \mu in this case is for Mars. From Example #2, it was shown that

    \[ M_M = 3.2811 \times 10^{-7} \, M_S \]

so

    \[ G \, M_M = 1.2453 \times 10^{-5} \]

which will be in units of AU years. Therefore, the transfer time, \tau, will be

    \[ \tau = \pi \sqrt{ \frac{a^3}{G \, M_M} } = \pi \sqrt{ \frac{(1.2453 \times 10^{-5})^3}{1.2453 \times 10^{-5}} } = 2.105 \times 10^{-3}~\mbox{years} = 18.45~\mbox{hours} \]

The {\Delta V} increments to initiate and end the transfer will be

    \[ \Delta V_1 =\sqrt{ \frac{G \, M_M}{R_1} } = \sqrt{ \frac{1.2453 \times 10^{-5}}{1.36354\times 10^{-4}} } = 0.09133~\mbox{AU/year}  = 0.167~\mbox{km/s} \]

and

    \[ \Delta V_2 =\sqrt{ \frac{G \, M_M}{R_2} } = \sqrt{ \frac{1.2453 \times 10^{-5}}{2.2331 \times 10^{-4}} } = 0.05577~\mbox{AU/year}  = 0.147~\mbox{km/s} \]

Bonus: At this orbital altitude, it is easily shown that the spacecraft would need a {\Delta V} to escape the Martian gravity and head off into space.

There is no dark side of the moon, really!

“There is no dark side of the moon, really. Matter of fact, it’s all dark. The only thing that makes it look light is the Sun.” The Moon rotates on its own axis and experiences phases of daylight and darkness, just like on Earth. However, because of the Moon’s non-uniform mass distribution, which resembles a dumbbell, it behaves like a pendulum under gravity as it orbits the Earth. Therefore, it takes the Moon the same time to turn once on its own axis as it takes to do one orbit around the Earth. This means that only one side of the Moon (the near side) is always visible from Earth. We never see the far side of the Moon, but it is not always dark there for the same reason the near side is not always dark: all sides are dark once a month.

Escape Velocity

How much velocity does it take to escape the Earth’s gravitational field? Escape velocity is the speed at which a spacecraft’s energy is sufficient that it never achieves orbit. In other words, how fast does a spacecraft need to travel such that it continues into space and never returns to Earth under the effects of its gravitational field? The answer lies in the energy and speed given to a spacecraft.

Consider a spacecraft of mass m_s in orbit around the Earth, as shown in Figure 16. The rocket performs a burn to impart additional energy, E_i. Conservation of energy requires that the final energy, E_f, is

(98)   \begin{equation*} E_f = E_i \end{equation*}

Therefore, the initial energy is

(99)   \begin{equation*} E_i = \frac{1}{2} m_s V_i^2 -\frac{G \, M \, m_s}{R_i} \end{equation*}

where R_i is the radius of the initial starting orbit.

To “escape” the gravity of planet Earth and head out into deeper space, a spacecraft needs a velocity of over 11 km/s.

The final energy of the spacecraft is

(100)   \begin{equation*} E_f = \frac{1}{2} m_s V_f^2 -\frac{G \, M \, m_s}{R_f} \end{equation*}

where R_f is the final radius or distance from the Earth. Because E_f = E_i then

(101)   \begin{equation*} \frac{1}{2} m_s V_i^2 -\frac{G \, M \, m_s}{R_E} = \frac{1}{2} m_s V_f^2 -\frac{G \, M \, m_s}{R_f} \end{equation*}

To escape the Earth’s gravity, the spacecraft must fly so far away that R_f \rightarrow\infty. Also, suppose that no excess kinetic energy is assumed at this distance. In that case, this assumption will allow an estimate of the minimum initial velocity required to escape Earth’s gravity. The two terms on the right-hand side of the preceding equation are zero, so

(102)   \begin{equation*} \frac{1}{2} m_s V_i^2 -\frac{G \, M \, m_s}{R_i} = 0 \end{equation*}

which, on rearrangement, gives

(103)   \begin{equation*} \frac{1}{2} V_i^2 =\frac{G \, M }{R_i} \end{equation*}

so the minimum “escape” velocity, V_{\rm esc}, is given by

(104)   \begin{equation*} V_{\rm esc} = \sqrt{ \frac{2 G \, M_E }{R_i} } \end{equation*}

Remember that the rocket engines must perform a burn to leave the initial orbit and achieve this velocity.

Now, armed with these results, it is possible to estimate Earth’s escape velocity. Recall that the Earth’s mass, M_E, is 5.97 \times 10^{24} kg, the radius of the initial orbit will be the radius of the Earth R_E, which is 6.3781 \times 10^6 m, plus the orbital height, {h}. Also, G = 6.67428 \times 10^{-11} N m{^{2}} kg^{-2}. Assuming that the orbital height relative to the radius of the Earth is small, then

(105)   \begin{equation*} V_{\rm esc} = \sqrt{ \frac{2 \, G \, M_E }{R_E} } = \sqrt{ \frac{2 \times 6.67428 \times 10^{-11} \times  5.97 \times 10^{24} }{6.3781 \times 10^6 } }= 11.18~\mbox{km/s} \end{equation*}

which will be the minimum velocity to escape the Earth’s gravitational field.

All spacecraft designed to travel into deep space must have a velocity larger than 11.2 km/s and will ultimately follow hyperbolic trajectories away from the Earth.

Gravity Assist & Velocity Gain

A gravity assist, or gravitational slingshot, is a maneuver in which a spacecraft gains or loses heliocentric velocity by passing close to a moving planetary body, i.e., a flyby maneuver. This interaction allows the spacecraft to adjust its speed and trajectory without additional propellant, relying instead on momentum transfer from the planet orbiting a central body, such as the Sun.

The analysis of a gravity assist maneuver involves two reference frames: the heliocentric (inertial) frame and the planet-centric (non-inertial) frame. In the planet-centric frame, the planet is stationary, and the spacecraft approaches with velocity {\vec{v}_{\text{in}}} and exits with velocity \vec{v}_{\text{out}}, such that the magnitudes are equal because of conservation of energy, i.e.,

(106)   \begin{equation*} \| \vec{V}_{\text{in}} | = \| \vec{V}_{\text{out}} \| = V \end{equation*}

The spacecraft’s trajectory is hyperbolic relative to the planet, with the angle between \vec{V}_{\text{in}} and \vec{V}_{\text{out}} denoted by \theta, known as the deflection angle, as shown in Figure 17.

Gravity assist maneuver in the planet’s frame of reference.

The planet moves with velocity \vec{V}_p in the heliocentric frame. The spacecraft’s heliocentric velocities before and after the flyby are given by vector addition, i.e.,

(107)   \begin{equation*} \vec{V}_{\text{in}} = \vec{V}_p + \vec{V}_{\text{in}} \quad \text{and} \quad \vec{V}_{\text{out}} = \vec{V}_p + \vec{V}_{\text{out}} \end{equation*}

It is important to note that the change in heliocentric velocity results from the change in the direction of the spacecraft’s velocity in the planet’s frame. Because {\vec{v}_{\text{in}}} and \vec{v}_{\text{out}} have equal magnitudes but different directions, the heliocentric velocity changes as

(108)   \begin{equation*} \vec{V}_{\text{out}} - \vec{V}_{\text{in}} = \vec{v}_{\text{out}} - \vec{v}_{\text{in}} \end{equation*}

This result emphasizes that the spacecraft’s change in heliocentric velocity is determined by the vector difference between the incoming and outgoing velocities in the planet’s frame of reference. Therefore, the expression for the heliocentric velocity after the encounter is

(109)   \begin{equation*} { \vec{V}_{\text{out}} = \vec{V}_{\text{in}} + (\vec{v}_{\text{out}} - \vec{v}_{\text{in}}) } \end{equation*}

Care must be taken to distinguish velocity vectors in different frames to avoid misinterpreting the maneuver’s dynamics.

Although the spacecraft’s speed relative to the planet does not change, the change in direction of \vec{v} leads to a change in the heliocentric speed. The gain in heliocentric speed is calculated by evaluating the difference in magnitude between \vec{V}_{\text{out}} and \vec{V}_{\text{in}}. Assuming a symmetric flyby and deflection angle \theta, the heliocentric speed gain is

(110)   \begin{equation*} \Delta V_H = 2 V_p \sin\left( \frac{\theta}{2} \right) \end{equation*}

This latter expression arises from the geometric relationship between the incoming and outgoing velocity vectors and the planetary velocity. The direction of deflection relative to the planet’s motion determines whether the spacecraft gains or loses speed. The maximum possible gain occurs when the deflection angle is largest and oriented such that the spacecraft reverses direction in the planet’s frame of reference. To clarify, the gravity-assist maneuver does not violate conservation of energy or momentum. In the planet’s frame of reference, the spacecraft’s energy is conserved. In the heliocentric frame, the spacecraft gains energy at the expense of the planet’s orbital momentum. However, this loss is negligible because of the planet’s large mass relative to the spacecraft.

Gravity assists are essential in interplanetary mission planning, allowing spacecraft to reach distant destinations using far less propellant than would otherwise be required. By precisely targeting the flyby geometry and timing, mission designers can exploit planetary motion to achieve substantial changes in heliocentric velocity and trajectory. This technique has been successfully implemented in numerous high-profile missions. The Voyager 1 and 2 spacecraft used a series of gravity assists from Jupiter and Saturn to accelerate out of the solar system. The Cassini-Huygens mission to Saturn utilized flybys of Venus, Earth, and Jupiter to achieve the required velocity for orbital insertion around Saturn. Similarly, the Juno spacecraft performed an Earth gravity assist to gain sufficient speed for its transfer to Jupiter.

Mission to Mars?

Humans landed on the Moon over half a century ago but have yet to return. It is likely, however, that they will return to the Moon within this decade. However, the Moon is a long way away, about 570 km (275,000 miles), and it takes about three days to get there in a spacecraft. At least, it can be reached using conventional propulsion systems.

But what about Mars? Mars is one of Earth’s two closest planetary neighbors, Venus being the other. Mars, often called the red planet, is close enough to be seen with the naked eye on a dark night without cloud cover, looking like a bright red point of light in the sky. So if it can be seen, then surely humans should be able to go there? However, the average distance between Earth and Mars is 225,000,000 km (140,000,000 miles), about 500 times further than the Moon! Mars is also 1.5237 AU from the Sun, where 1 AU (Astronomical Unit) is the average distance from the Sun to Earth, which is {1.495 \times 10^8} km.

How does a spacecraft reach Mars? How long will it take? And how much propellant will be needed? Can astronauts even survive the long journey? How long will it take them to get back? Humans have sent probes to Mars and other planets, but can humans visit Mars or colonize it? It should be noted that humans have yet to colonize the Moon or Antarctica; Mars is much more distant and hostile to humans than either.

Trans-Martian Injection

Some engineering answers to these questions can be estimated using the principles previously established for a Hohmann cotangential elliptic transfer orbit, as shown in Figure 18 for trans-Martian injection and transit to Mars. It is assumed that the orbits of Earth and Mars are circular rather than elliptical; this simplification considerably reduces the mathematics and provides reasonable estimates of the problem parameters.

A Hohmann cotangential elliptic transfer orbit can be used to transfer a spacecraft between Earth and Mars.

One assumption is that the spacecraft is already in Earth orbit, fully fueled, and ready to travel to Mars. Of course, getting the spacecraft into orbit with the needed propellant is also a significant challenge. The timing of the spacecraft’s departure for Mars on the transfer orbit is critical because it must arrive at the exact orbital location at the same time. Otherwise, the spacecraft will get there too early or too late to rendezvous with Mars. Likewise, the spacecraft’s return from Mars to Earth must be timed to permit an Earth rendezvous, re-entry, and landing.

The time it takes to get to or from Mars can be estimated using Kepler’s third law, i.e.,

(111)   \begin{equation*} T^2 = a^3 \end{equation*}

If the period T about the Sun is measured in units of years and a is measured in AU units, then for Earth T_E = 1 and a_E = 1, and for Mars a_M = 1.5237. The semi-major axis length of the transfer orbit will be

(112)   \begin{equation*} { a_T = \frac{a_E + a_M}{2} = \frac{1.0 + 1.5237}{2} = 1.26185~\mbox{AU} } \end{equation*}

Therefore, for the transfer orbit to Mars, the orbital period is

(113)   \begin{equation*} T_T = \sqrt{ {a_T}^3 } = \left( 1.26185 \right)^{3/2} = 1.41746~\mbox{years} \end{equation*}

However, this is the time for a complete orbit, so the transfer time to Mars will be half of this value, yielding an average flight time of 0.71 years, or approximately 8.5 months.

The exact transfer time for a mission to Mars depends on the relative positions of Mars and Earth in their elliptical orbits around the Sun at any given time, as well as the gravitational influence of other planets along the trajectory, such as Jupiter. However, it will take much longer to reach Mars than to reach the Moon. Even in the best scenario, a human mission to and from Mars, with just a few weeks or months spent on the surface, will likely take a couple of years. Nevertheless, this might not occur with conventional (chemical) propulsion systems.

Orbital Velocities

It is first necessary to calculate the orbital velocity of the Earth and Mars about the Sun. Again, circular orbits are assumed. For the Earth, its orbital velocity, V_E, is

(114)   \begin{equation*} V_E = \sqrt{ \frac{G \, M_S}{R_E} } \end{equation*}

where M_S is the mass of the Sun. For Mars, its orbital velocity is V_M, i.e.,

(115)   \begin{equation*} V_M = \sqrt{ \frac{G \, M_S }{R_M} } \end{equation*}

At the perihelion of the transfer orbit, then

(116)   \begin{equation*} V_p = V_E + \Delta V_1 \end{equation*}

and at the aphelion of the transfer orbit

(117)   \begin{equation*} V_M = V_a + \Delta V_2 \end{equation*}

Conservation of angular momentum of the spacecraft at the perihelion and aphelion of the transfer orbit gives

(118)   \begin{equation*} m_s V_p R_E = m_s V_a R_M \end{equation*}

or

(119)   \begin{equation*} \frac{V_a}{V_p} = \frac{R_E}{R_M} \end{equation*}

Conservation of energy for the movement about the Sun requires

(120)   \begin{equation*} \frac{1}{2} m_s {V_p}^2 - \frac{G \, m_s \, M_S}{R_E} = \frac{1}{2} m_s {V_a}^2 - \frac{G \, m_s \, M_S}{R_M} \end{equation*}

noting that m_s cancels out on both sides of the preceding equation to give

(121)   \begin{equation*} \frac{1}{2}{V_p}^2 - \frac{G \, M_S}{R_E} = \frac{1}{2} {V_a}^2 - \frac{G \, M_S}{R_M} \end{equation*}

Using Eqs. 119 and 121 and after some algebra gives

(122)   \begin{equation*} V_p = \sqrt{ \frac{2 G \, M_S \, R_M}{R_E (R_E + R_M) }} \end{equation*}

The {\Delta V} for the first burn to exit the Earth’s orbit into the Hohmann transfer orbit will be

(123)   \begin{equation*} \Delta V_1 = V_p - V_E = \sqrt{\frac{2 G \, M_S \, R_M}{R_E (R_E + R_M)}} - \sqrt{\frac{G \, M_S}{R_E}} \end{equation*}

which, after rearrangement, becomes

(124)   \begin{equation*} \Delta V_1 = \sqrt{\frac{G \, M_S}{R_E}} \left(\sqrt{\frac{2 R_M}{R_E + R_M}} - 1\right) \end{equation*}

For the second burn, then

(125)   \begin{equation*} \Delta V_2 = V_M - V_a = \sqrt{\frac{G \, M_S}{R_M}} - \sqrt{\frac{2 G \, M_S \, R_E}{R_M (R_E + R_M)}} \end{equation*}

which gives

(126)   \begin{equation*} \Delta V_2 = \sqrt{\frac{G \, M_S}{R_M}} \left( 1 - \sqrt{\frac{2 R_E}{R_E + R_M}} \right) \end{equation*}

Inserting the known values of G, M_S, etc., shows that \Delta V_1 \approx 3.0 km/s and \Delta V_2 \approx 2.7 km/s.

These values will require a significant amount of propellant, all of which must be carried on board the spacecraft. More propellant is required to place the spacecraft into a Mars orbit and to descend to the surface. Staging will be required to progressively reduce the spacecraft’s mass at each phase of flight, including the Mars landing. However, no matter how reductions in propellant mass can be argued, getting to Mars can still be expected to require an enormous amount of propellant.

Propellant Requirements

Based on the preceding analysis, estimating the propellant mass needed to fly to Mars is possible. The mass needed depends on the type of rocket engine(s), but can be estimated from the rocket equation, i.e.,

(127)   \begin{equation*} \Delta V = I_{\rm sp} g_0 \ln \left( \frac{M_0}{M_f} \right) \end{equation*}

where I_{\rm sp} is the specific impulse for the rocket engine, {M_0} is the initial mass of the spacecraft at the beginning of the burn, and M_f is the final (burnout) mass during which a propellant mass, M_P, is consumed, i.e., M_f = M_0 - M_P.

Using the rocket equation gives

(128)   \begin{equation*} \frac{M_f}{M_0} = \exp \left( -\frac{\Delta V}{I_{\rm sp} \, g_0} \right) = \frac{M_0 - M_P}{M_0} \end{equation*}

and so

(129)   \begin{equation*} \frac{M_P}{M_0} = 1 - \exp \left( -\frac{\Delta V}{I_{\rm sp} \, g_0} \right) \end{equation*}

A value for I_{\rm sp} \, g_0 can be assumed to make sense of this result for the Mars mission. For example, for a rocket engine using liquid hydrogen-oxygen as a propellant, {I_{\rm sp} \, g_0 \approx 4,400} m/s or 4.4 km/s. The hydrogen-oxygen reaction is the most energetic chemical reaction. However, there are limits to the amount of energy that can be extracted from chemical processes, which constrain what is possible with current technology and what is not.

For the first impulsive burn on a mission to Mars, then

(130)   \begin{equation*} \frac{M_P}{M_1} = 1 - \exp \left( -\frac{3.0}{4.4} \right) = 0.494 \end{equation*}

This means the spacecraft will consume nearly half of its initial mass, M_1, as propellant to reach the transfer orbit.

Consider the second burn at the end of the transfer orbit, recognizing that the initial mass M_1 has now been reduced significantly to M_2 = M_1 - 0.494 M_1 = 0.506 M_1. In this case, the additional mass of propellant needed to circularize the orbit of the spacecraft relative to its initial mass will be

(131)   \begin{equation*} \frac{M_P}{M_1} = 0.561 \left[ 1 - \exp \left( -\frac{2.7}{4.4} \right) \right] = 0.257 \end{equation*}

which is significantly less than the first burn because the spacecraft now has much less mass.

However, the preceding analysis indicates that reaching Mars will require a substantial amount of propellant, accounting for at least 75% of the spacecraft’s initial mass at launch from Earth orbit. There are significant engineering implications in designing a spacecraft capable of reaching Mars, including integrating payloads such as life support systems and supplies. For example, if the spacecraft has a total mass of 50,000 kg, approximately 37,500 kg of that will be required as propellant. For a 100,000 kg spacecraft, approximately 75,000 kg will be propellant, among other components. A further amount of propellant will be required for any mid-course corrections, a retro-burn to lower the spacecraft (or part of it) into a Mars orbit, and, finally, for descent to the surface.

Getting Back to Earth

The astronauts have finally arrived on Mars and set up a Martian base. They collect data and do experiments. What about getting them back from Mars to Earth? Again, the required propellant can also be estimated using reasonable assumptions. Returning from Mars requires that the spacecraft carry the propellant needed for the return journey, which must also be included in the spacecraft’s initial mass when it leaves Earth. It requires propellant to transport propellant to Mars for the return mission to Earth. Now, the engineering problems are becoming more complex.

Assume that the net values of {\Delta V} needed on the return flight are the same as on the outbound flight. On a mission to Mars, when carrying enough propellant to come back to Earth, to enter and leave the Mars orbit, then

(132)   \begin{equation*} { \frac{M_P}{M_1} = 1 - \exp \left( -\frac{2 \times 2.7}{4.4} \right) = 0.707 } \end{equation*}

and to leave and re-enter Earth’s orbit

(133)   \begin{equation*} \frac{M_P}{M_1} = 1 - \exp \left( -\frac{2 \times 3.0}{4.4} \right) = 0.744 \end{equation*}

If these velocity increments are treated as a single equivalent requirement, then the total return-mission propellant fraction is

(134)   \begin{equation*} \frac{M_P}{M_1} = 1 - \exp \left( -\frac{2(2.7) + 2(3.0)}{4.4} \right) = 0.925 \end{equation*}

Therefore, under these simplified assumptions, about 92.5% of the initial spacecraft mass would have to be propellant, leaving only about 7.5% for structure, payload, crew systems, and reserves.

What does this mean? How can it take more propellant mass than a spacecraft to reach Mars and return to Earth? In this case, the answer is that it is impossible, at least with current propulsive technology and the assumptions used in this analysis, which include the possibility of accomplishing the Mars mission with a single spacecraft. If this is difficult to believe, watch the video “Tyranny of the Rocket Equation” by accomplished NASA astronaut Donald Pettit.

However, not all is lost. Such a mission is feasible with two or three other spacecraft: one carrying the astronauts, and the others carrying supplies and the required propellant. Remember that even a mission to the Moon and return requires approximately 95% of the total spacecraft mass as propellant, leaving 5% as payload, including the astronauts and all their supplies. Achieving the initial mass for a Mars mission to a parking orbit around Earth will also require a substantial additional propellant load.

So, is it Possible?

Even under the best circumstances, the engineering challenges in reaching Mars are now apparent. In addition to orbital mechanics, the astronauts sent to Mars must be protected from radiation exposure and have sufficient life support and a suitable habitat to maintain human performance and health. These are all enormous engineering challenges that require substantial financial resources. Perhaps the conclusion is that Mars is beyond our current reach, at least if the astronauts sent there plan to return to Earth without sending out tremendous quantities of supplies and propellant in advance, nor following up with resupply missions. In the meantime, it does not stop humankind from sending probes to study Mars and beyond; sending out human explorers and bringing them back safely to Earth is another matter entirely. Moreover, even if a mission to Mars were technically feasible, the enormous costs are unlikely to be justified by any single organization or country.

Aerobraking

Aerobraking is an orbital maneuver in which a spacecraft uses aerodynamic drag from a planet’s atmosphere to reduce its orbital energy, rather than relying entirely on onboard propulsion. This technique is particularly valuable for interplanetary missions because it enables significant propellant mass savings during orbit insertion and circularization.

The spacecraft begins in a highly elliptical orbit, typically after a capture maneuver that places it into orbit around a planet. The spacecraft enters the upper atmosphere at each periapsis passage, experiencing drag that reduces its kinetic energy and gradually lowers its apoapsis. By performing multiple controlled passes through the atmosphere, the orbit can be transitioned into a desired lower-energy orbit, as shown in Figure 19.

An aerobraking maneuver must be done in small steps to prevent thermal overheating.

The aerodynamic drag force acting on the spacecraft is modeled using the standard equation for bluff body drag, i.e.,

(135)   \begin{equation*} F_D = \frac{1}{2} \varrho_{\infty} \, V^2 \, C_D \, A_{\rm ref} \end{equation*}

where \varrho_{\infty} is the atmospheric density, V is the spacecraft’s velocity relative to the atmosphere, C_D is a drag coefficient that depends on the vehicle’s shape and orientation, and A_{\rm ref} is the effective cross-sectional area on which C_D is based. Because aerobraking usually requires some form of heatshield, an initial estimate for C_D can be based on a closed hemispherical body, i.e., C_D \approx 0.42 based on the maximum projected cross-sectional area.

This drag force opposes the spacecraft’s velocity vector and performs negative work. The change in the spacecraft’s mechanical energy over a single atmospheric pass can be expressed as

(136)   \begin{equation*} \Delta E = -\int_{\rm entry}^{\rm exit} F_D \, ds \end{equation*}

where {s} describes the path traveled through the atmosphere between the entry and exit points. The orbital energy per unit mass in a two-body system is given by

(137)   \begin{equation*} E = \frac{V^2}{2} - \frac{GM}{r} \end{equation*}

so a loss of energy leads to a change in the orbit’s shape, primarily lowering the apoapsis while the periapsis remains close to the atmospheric boundary.

Aerobraking is most effective when performed at periapsis, where orbital velocity and atmospheric density are maximized. However, the maneuver must be carefully controlled to prevent overheating or excessive structural loads. The heat load during an aerobraking pass can build up rapidly because

(138)   \begin{equation*} q \, \propto \, \varrho_{\infty} \, V^3 \end{equation*}

which shows that small increases in velocity or atmospheric density can lead to significant increases in thermal loads. Therefore, spacecraft using aerobraking must be designed with appropriate thermal protection systems.

Atmospheric density is not constant and varies with solar activity, time of day, and altitude. Accurate prediction of atmospheric conditions is essential to maintain a safe periapsis altitude. In practice, spacecraft perform periapsis raises or lowers using thrusters to adjust drag exposure between passes. The process typically takes weeks to months, with frequent ground-based updates because of the high uncertainty in estimating atmospheric drag.

Aerobraking has been successfully employed in several missions. Mars Global Surveyor performed over 900 aerobraking passes to reduce its apoapsis from 56,000 km to 450 km. Mars Odyssey and Mars Reconnaissance Orbiter also used aerobraking for orbit insertion and circularization. The Venus Express mission employed aerobraking to study the structure of the upper atmosphere, lowering the spacecraft’s altitude for atmospheric sampling. While aerobraking offers substantial mass savings, it comes with trade-offs in complexity, duration, and thermal issues. It is most advantageous for missions with tight mass constraints and tolerant mission timelines.

Check Your Understanding #7 – Mission to Mars

Human colonization of space remains a hugely ambitious goal. NASA’s return to the Moon has begun with its Artemis program. Make a list of some of the technical and other challenges in sending humans back to the moon, Mars, and perhaps even deep space. Remember that Mars is over 250 times further away than the Moon.

Show solution/hide solution.

Here is a list of some of the technical and other challenges in sending humans back to the moon, Mars, and into deep space:

  1. Radiation exposure: Space travel exposes astronauts to high radiation levels, which can cause various health problems, including increased cancer risk, cognitive decline, and cardiovascular disease.
  2. Life support systems: Providing astronauts with a sustainable, life-supporting environment is a significant challenge, especially during long-duration missions to the moon and Mars.
  3. Propulsion systems: A significant challenge is to develop reliable, efficient, and safe propulsion systems capable of transporting astronauts and their equipment over large distances.
  4. Habitat and resource utilization: Providing a suitable habitat for astronauts and utilizing local resources, such as water and minerals, will be critical for long-duration missions.
  5. Human performance and health: Ensuring astronauts’ physical and psychological well-being during long-duration missions is a significant challenge.
  6. Cost and funding: Sending humans back to the moon, Mars, and deep space is costly and resource-intensive, requiring significant funding and political support.
  7. Technological advancements: Significant technological advancements, including propulsion, life support, habitat design, and materials science, will be required to make human space exploration possible.
  8. Mission design and operations: Planning, designing, and executing complex missions to the moon, Mars, and beyond requires significant expertise and experience in mission operations and logistics.

In conclusion, sending humans back to the moon, Mars, and into deep space is a highly ambitious goal that requires overcoming a range of technical and other challenges, including radiation exposure, life support, propulsion systems, habitat and resource utilization, human performance and health, cost and funding, technological advancements, and mission design and operations.

Orbital Decay

For satellites in LEO, the aerodynamic drag, no matter how small, can be sufficient to diminish the satellite’s energy, causing it to slowly spiral toward the surface of the Earth, as shown in Figure 20. The drag force acts as a perturbation to the orbital trajectory. The satellite’s shape is now more important than it was when it was farther out in space, in a higher orbit, and the density of the upper atmosphere is also more important. Asymmetric drag forces may also cause the satellite to tumble, a process that can be irreversible if the orbit becomes too low. The decay of satellites in Earth orbit is a significant problem that requires close attention in mission planning and longevity.

Aerodynamic drag on satellites in LEO will cause their orbits to decay and slowly spiral toward the Earth.

Analysis

It is difficult to predict orbital decay with the desired certainty because of variations in the density of the upper atmosphere caused by tides, seasonal changes, solar activity, and other factors. However, reasonable estimates can still be made by using certain assumptions. The process involves an estimation of the aerodynamic drag force, D_s, on the satellite, which can be expressed conventionally as

(139)   \begin{equation*} D_s = \frac{1}{2} \varrho_{\infty} \, V_s^2 \, A_{\rm ref} \, C_D \end{equation*}

where V_s is its orbital velocity. In the absence of additional specific information, an ISA model for the extended atmosphere can be used to calculate the density, \varrho_{\infty}. The reference area, A_{\rm ref}, and the drag coefficient, C_D, in Eq. 139 will depend on the exact shape of the satellite. It is often assumed that in the absence of specific information, then C_D = 1 in the very outer regions of the atmosphere in the low-density or rarefied flow, increasing to C_D = 2 when approaching an orbital altitude of 150 km. It is found that as a result of the eccentricity of most satellite orbits, decay is more rapid at the apogee, which dips further into the atmosphere, than at the perigee, which, on average, tends to circularize the orbit.

For a satellite in an assumed circular orbit, the orbital velocity V_s can be approximated as

(140)   \begin{equation*} V_s \approx \sqrt{\frac{G \, M_E}{R_s}} \end{equation*}

where M_E is the Earth’s mass and R_s is the average radius of the satellite’s orbit. Aerodynamic drag acts opposite to the satellite’s velocity, causing a loss of energy and a gradual reduction in orbital radius.

The energy E of the satellite in a circular orbit is

(141)   \begin{equation*} E = -\frac{G \,  M_E \, m_s}{2 \,  R_s} \end{equation*}

where m_s is the mass of the satellite. The time rate of change of energy from the creation of aerodynamic drag will be

(142)   \begin{equation*} \frac{dE}{dt} = -D_s \, V_s \end{equation*}

Substituting the foregoing expressions for D_s and V_s, i.e., Eqs. 139 and 140, respectively, gives

(143)   \begin{equation*} \frac{dE}{dt} = -\frac{1}{2} \varrho_{\infty} \, V_s^3 \, A_{\rm ref} \, C_D \end{equation*}

Using Eq. 141, the time rate of change of energy is also given by

(144)   \begin{equation*} \frac{dE}{dt} = \frac{G \,  M_E \, m_s}{2 \, R_s^2} \left( \frac{dR_s}{dt} \right) \end{equation*}

Equating Eqs. 143 and 144 gives

(145)   \begin{equation*} -\frac{1}{2} \varrho_{\infty} \, V_s^3 \, A_{\rm ref}  \, C_D = \frac{G \, M_E \, m_s}{2 \, R_s^2} \left( \frac{dR_s}{dt} \right) \end{equation*}

and rearranging gives

(146)   \begin{equation*} \frac{dR_s}{dt} = -\frac{\varrho_{\infty} \, V_s^3 \, A_{\rm ref}  \, C_D \, R_s^2}{G \, M_E \, m_s} \end{equation*}

the minus sign confirming an orbital decay. The orbital speed is given by Eq. 140, so that

(147)   \begin{equation*} V_s^3 = \left( \sqrt{\frac{G \, M_E}{R_s}} \right)^3 = \frac{(G \, M_E)^{3/2}}{R_s^{3/2}} \end{equation*}

Therefore, substituting and rearranging gives

(148)   \begin{equation*} \frac{dR_s}{dt} = \overbigdot{R_s} = -\dfrac{\varrho_{\infty} \, A_{\rm ref} \, C_D}{m_s} \sqrt{\dfrac{G \, M_E}{R_s}} = -\varrho_{\infty} \left( \dfrac{ A_{\rm ref} \, C_D}{m_s} \right) \sqrt{\dfrac{G \, M_E}{R_s}} = -\dfrac{\varrho_{\infty}}{\beta} V_s \end{equation*}

where the grouping \dfrac{m_s}{A_{\rm ref} \, C_D} is classically known as the ballistic coefficient[5] and is given the symbol \beta. Notice that the units of \beta are mass per unit area.

The significance of Eq. 148 shows that the rate of decay of the orbital radius, R_s, is proportional to the aerodynamic drag coefficient, C_D, the cross-sectional reference area, A_{\rm ref} (on which the drag coefficient will be based), the atmospheric density, \varrho, and inversely proportional to the mass of the satellite, m_s. The decay rate also depends on the satellite’s speed, V_s, and orbital radius, R_s, with faster, closer satellites experiencing greater orbital decay, as might be expected.

Notice also that the higher the value of the ballistics coefficient, \beta, the lower the rate of orbital decay, an effect analogous to maximizing the range and velocity of a projectile. Because of their high values of C_D, satellites also tend to have very high values of \beta, typically between 100 for a small satellite[6] to over 5,000 for something like the ISS.

Dimensionless Form

The orbital decay can also be expressed in dimensionless form. If it is assumed that

(149)   \begin{equation*} \overbigdot{R_s} = \Psi_1 \left( \varrho_{\infty}, \, A_{\rm ref}, \,D, \,m_s, \,g, \,R_s, \,V_s \right) \end{equation*}

where \Psi_1 is some function to be determined, i.e., it depends on seven parameters, then a dimensional analysis will give that

(150)   \begin{equation*} \dfrac{\overbigdot{R_s}}{V_s} = \Psi_2 \left( C_D, \, \dfrac{V_s^2}{g \, R_s}, \, \dfrac{\varrho_{\infty} \, R_s^2}{m_s} \right) \end{equation*}

where \Psi_2 is some other function. The latter grouping in Eq. 150 is a buoyancy term, arising in hydrodynamics and aerodynamics, but of no consequence in space. Therefore, it can be concluded for the orbital decay problem that

(151)   \begin{equation*} \dfrac{\overbigdot{R_s}}{V_s} = \Psi_3 \left( C_D, \, \dfrac{V_s^2}{g \, R_s} \right) \end{equation*}

It will be seen that the dimensionless decay rate, \overbigdot{R_s}/V_s, is a function of the aerodynamic drag coefficient, C_D, and a second dimensionless parameter that is the ratio of potential to kinetic energy, i.e.,

(152)   \begin{equation*} \dfrac{g \, R_s}{V_s^2} \equiv  \dfrac{g \, R_s}{ V_s^2} \left( \dfrac{m_s}{m_s} \right) \equiv \frac{\text{Potential energy}}{\text{Kinetic energy}} \end{equation*}

This parameter indicates the ratio of gravitational energy to kinetic energy[7] for the satellite’s orbit. It also helps assess the relative dominance of gravitational forces over the centrifugal forces that keep the satellite in orbit, as summarized in Figure 21.

Orbital decay rate versus the ratio of gravitational energy to kinetic energy, which must be maintained at a value close to 1.0 to prevent excessive decay.

If the value of \dfrac{g \, R_s}{V_s^2} \ge 1, then gravitational forces dominate, and the satellite remains in a stable orbit (assuming no other perturbations). If the value of \dfrac{g \, R_s}{V_s^2} < 1, then orbital decay begins. The satellite will gradually lose altitude and eventually re-enter the atmosphere.

Consequences of Orbital Decay

Atmospheric drag significantly affects the LEO altitudes of crewed Earth-orbit spacecraft and satellites, even those at relatively high LEO, such as the Hubble Space Telescope, which has R_s \approx 540 km. Therefore, when the orbital decay reaches a critical level, a satellite must be boosted back into its initial orbit. Sufficient propellant must be carried on board the satellite to ensure this is achievable. The propellant required depends on the satellite’s mission requirements, projected life, and orbital decay rate. The ISS, for example, requires regular altitude boosts to counteract orbital decay. About once a month, the ISS fires its thrusters to recover its orbital altitude, a maneuver called a reboost. Accurately estimating the propellant required for a satellite in LEO is crucial, as having too little can cause mission failure, whereas carrying too much is unnecessary weight and cost.

Typically, satellites in LEO exceeding a radial decay rate of approximately 10^{-4} km/s are considered to be in a critical state of orbital decay. For a satellite with V_s \approx 7.5 km/s (similar to the ISS in LEO), a radial decay rate exceeding that threshold could see a significant orbital decay over a relatively short period. Typically, if a spacecraft comes within 160 km of the Earth’s surface, it will likely be in an irrecoverable orbit because of aerodynamic drag. It is unlikely that sufficient propellant remains to correct the orbit, unless the intention is to de-orbit the satellite and take it out of commission. Therefore, the spacecraft will enter a “graveyard spiral,” with its burn-up and disintegration typically beginning at an altitude of approximately 80 km.

Atmospheric Re-entry

Sending astronauts into orbit or beyond is challenging enough, but getting them back safely to Earth brings up a whole new set of technical challenges. As shown in Figure 22, this artist’s vision of a flaming spacecraft hurtling toward the Earth closely resembles the physical truth. High re-entry velocities, which can exceed Mach 12, generate substantial aerodynamic friction on the spacecraft, producing heat. So much heat is generated during re-entry that the spacecraft must be equipped with a thermal protection system (TPS), such as a heat shield, to prevent it from burning up or vaporizing.

An artist’s impression of the Apollo command module flying with the heat shield’s blunt end at a non-zero attack angle to control the re-entry profile.

NACA experiments showed that sharp objects traveling at hypersonic speeds become so hot from kinetic heating that they melt. Therefore, the preferred re-entry shape is a blunt body, which causes the primary shock waves to stand off from the spacecraft’s surface, as shown in Figure 23. This reduces kinetic heating and increases drag, thereby slowing it down. Although the process still generates significant heat, blunt body shapes are always preferred because they provide the necessary high ballistic coefficient. Some shapes, such as the Space Shuttle and SpaceX’s Starship, can also generate lift, thereby allowing the descent speed and re-entry trajectory to be controlled within certain bounds.

Diagram of different shapes of objects flying at hypersonic speeds at re-entry and how flow velocity if affected.
A blunt body is preferred for a re-entry shape, which causes the shock wave to stand off from the spacecraft, reducing kinetic heating and creating high drag.

During re-entry, the spacecraft experiences gravitational acceleration, pulling it toward Earth while simultaneously encountering aerodynamic drag opposing its motion, as described previously. The decelerations as the spacecraft enters the denser lower atmosphere may be significant, ranging up to 5 g. The balance between gravitational and aerodynamic forces governs the spacecraft’s deceleration and descent profile as it passes through varying atmospheric conditions. Understanding these forces is crucial for predicting and optimizing the trajectory to ensure a safe and controlled descent.

Most spacecraft designed to re-enter the Earth’s atmosphere for a landing will follow a ballistic or semi-ballistic trajectory, as shown in Figure 24. The key to a safe landing is careful energy management and the complete depletion of the initial orbital energy. The re-entry flight path has a very steep angle but is relatively short in duration. The steep re-entry angle creates atmospheric drag, slowing the spacecraft and progressively dissipating its potential and kinetic energy. In this regard, the spacecraft free-falls through the upper atmosphere at hypersonic Mach numbers, gradually slowing as the atmosphere becomes denser. The lower atmosphere produces significant aerodynamic drag, causing the spacecraft to become subsonic. The final stages of a landing require deploying parachutes and, if necessary, a final retro-burn, to ensure the spacecraft lands softly in the sea or on land.

A typical re-entry profile for a spacecraft returning to Earth.

The re-entry trajectory analysis involves solving the differential equations of motion that account for gravitational and aerodynamic drag. Given the problem’s complexity, numerical methods are employed to compute the spacecraft’s velocity and altitude as it descends through the atmosphere. The International Standard Atmosphere (ISA) model is used to represent variations in air density, temperature, and pressure with altitude, all of which influence the spacecraft’s dynamics and heat-transfer characteristics during re-entry. Many simulations are performed under varying assumptions, and the final predicted trajectory is obtained with reasonable confidence.

The basic equations of motion for a descending spacecraft can be written as

(153)   \begin{equation*} \frac{dV}{dt} = -g(h) + \frac{D}{m_s} \end{equation*}

(154)   \begin{equation*} \frac{dh}{dt} = -V \sin \gamma \end{equation*}

(155)   \begin{equation*} \frac{d\gamma}{dt} = \frac{L}{m_s \, V} - \frac{g(h)}{V} \cos \gamma \end{equation*}

where V is the spacecraft’s velocity, {h} is the altitude, \gamma is the flight path angle relative to the local horizontal, m_s is the mass of the spacecraft, g(h) is the gravitational acceleration as a function of altitude, D is the aerodynamic drag, and L is the aerodynamic lift. The gravitational acceleration g(h) decreases with altitude and is given by

(156)   \begin{equation*} g(h) = g_0 \left( \frac{R_E}{R_E + h} \right)^2 \end{equation*}

where {g_0} is the standard gravitational acceleration at Earth’s surface and R_E is the radius of the Earth. The aerodynamic lift can be neglected, but the drag is given by

(157)   \begin{equation*} D = \frac{1}{2} \, \varrho_{\infty} (h) \, V^2 \, C_D \, A \end{equation*}

where \varrho_{\infty} (h) is the atmospheric density as a function of altitude, C_D is the drag coefficient, and A is the reference area of the spacecraft on which C_D is based.

Managing the thermal environment during re-entry is critical, as the spacecraft’s kinetic energy must be dissipated safely. The convective heating rate at the surface of the vehicle can be approximated by

(158)   \begin{equation*} \overbigdot{q} \ \  \propto \ \ \sqrt{ \varrho_{\infty} (h)} \, V^3 \end{equation*}

where \overbigdot{q} is the surface heat flux. This relationship indicates that heating depends strongly on both atmospheric density and the cube of the velocity, making high-speed re-entry phases particularly severe for thermal loads.

Engineers optimize re-entry trajectories to achieve specific landing locations, such as in the Gulf of Mexico or the Atlantic Ocean. This optimization involves adjusting entry angles, velocities, and descent profiles to balance spacecraft performance with required safety margins. A thermal protection system (TPS), often incorporating heat shields, is employed to shield the spacecraft from intense heating during atmospheric re-entry. In some cases, TPS materials are designed to ablate and shed material, reducing temperatures on the spacecraft surface. Advancements in computational fluid dynamics (CFD) and heat-transfer simulations have enabled more detailed analyses of TPS performance, which remain crucial for ensuring vehicle survivability during re-entry.

Deep Space & Beyond

Ventures into deep space and beyond our solar system may start to infringe on the final frontier, whatever that is. For example, it has become clear that our current technology makes sending astronauts into deep space impossible. It is unthinkable that they could ever be returned safely to Earth, even after doing so.

The distances in deep space are enormous, and then some. Alpha Centauri, the Earth’s nearest Sun-like star system, is located (only!) 4.37 light-years away from the Earth, so about 25,000,000,000,000 miles (40 trillion km) away. Figure 25 gives some idea of the gigantic distances involved. Interstellar probes and robotic explorers such as Voyager 1 and 2 continue to send back data. However, it has taken several decades for them to reach their current position just beyond our solar system, which is less than 1 light day, and it will take them another 400,000 years to reach the nearest stars. However, these robotic probes carry the vision and inspiration of humankind, with the hope that one day humans may be able to venture there as well.

Image of a gigantic, gaseous cavity in deep space as captured by Webb space telescope. The scale is labeled in light-years, highlighting the enormity of even this tiny slice of the galaxy.

What is a Black Hole?

Everyone has heard of black holes, right? They dot the universe and are regions where gravity is so strong that nothing, including light, has sufficient energy to escape. The effect of gravity is so strong in a black hole because so much matter has been compressed into a small volume. It is not possible to see a black hole with a conventional telescope. However, the locations of black holes in the universe can be identified by the signatures of high-energy radiation and the way stars behave differently from other stars under extremely powerful gravitational forces. An interesting image of a black hole, shown in Figure 26, was taken by the Event Horizon Telescope.

The first picture of a black hole was made using observations of the center of galaxy M87 taken by the Event Horizon Telescope. The image shows a bright ring formed as light bends in the intense gravity around a black hole 6.5 billion times the Sun’s mass. Credits: Event Horizon Telescope Collaboration
The Event Horizon Telescope took this image of a black hole. The photo shows a bright ring formed by light bending in the intense gravity around a black hole with a mass 6.5 billion times that of the Sun.

By using Eq. 104 and replacing the escape velocity with the speed of light, {c}, then

(159)   \begin{equation*} c = \sqrt{ \frac{2 G \, M }{R_{\rm bh}} } \end{equation*}

Solving for R_{\rm bh}, which will be the radius of a black hole, gives

(160)   \begin{equation*} R_{\rm bh} = \frac{2 G \, M }{c^2} \end{equation*}

which is known as the Schwarzschild equation and is named after the German astronomer Karl Schwarzschild, who first derived this exact solution.

Because the value of G = 6.67428 \times 10^{-11} N m{^{2}} kg^{-2} and the speed of light is 299,792,458 m s^{-2}, then the radius of a black hole is typically small by planetary standards. For example, if the mass of a black hole were the same as the Sun (= 2 \times 10^{30} kg), then its radius would be about 3 km, according to the Schwarzschild equation. Most black holes, however, are significantly more massive than the Sun.

Theory of Relativity

In the early twentieth century, Albert Einstein introduced the theories of special relativity (1905) and general relativity (1915), which refined the classical descriptions of space, time, mass, and energy. These theories do not replace Newtonian mechanics in ordinary engineering applications; instead, they define the conditions under which Newtonian assumptions cease to be exact.

Special relativity addresses the behavior of matter and electromagnetic radiation in inertial reference frames when relative speeds approach the speed of light. One of its central results is the connection between inertia and energy. For a body at rest, the total energy is

(161)   \begin{equation*} E_0 = m_0 \, c^2 \end{equation*}

where m_0 is the invariant (rest) mass and c is the speed of light in vacuum. This relation states that a body with inertia possesses an intrinsic energy even when at rest; it does not imply a physical conversion of mass into energy.

For a moving body, special relativity relates energy and momentum through

(162)   \begin{equation*} E^2 = (p\,c)^2 + (m_0\,c^2)^2 \end{equation*}

where p is the linear momentum. Energy and momentum depend on the observer’s inertial reference frame, while the rest mass m_0 remains invariant. Equation 162 provides a compact statement of energy–momentum consistency across inertial frames and reduces to Eq. 161 when p = 0.

General relativity extends these ideas to gravitation and non-inertial reference frames. In this theory, gravitational effects are described in terms of spacetime geometry rather than as instantaneous forces acting at a distance. One of the earliest experimental confirmations occurred during the 1919 solar eclipse, when starlight was observed to be deflected as it passed near the Sun, consistent with Einstein’s predictions. General relativity forms the theoretical basis for gravitational lensing, black holes, and cosmology.

For most aerospace engineering applications, including atmospheric flight and orbital mechanics, velocities are small compared with the speed of light, and gravitational fields are weak. Under these conditions, Newtonian mechanics and classical gravitation provide an excellent approximation for forces, trajectories, and orbital motion. Relativistic effects become important primarily in applications involving high-precision timekeeping, such as satellite navigation, or in astrophysical and cosmological contexts. Accordingly, relativity serves here mainly to define the limits of validity of classical mechanics rather than as a routine analytical tool.

Relativistic corrections are essential for the accuracy of the Global Positioning System (GPS), where both gravitational and velocity-dependent timing effects must be accounted for. Without these corrections, GPS would accumulate position errors of several kilometers per day, rendering the system unusable. Two primary relativistic effects influence GPS satellite clocks. First, according to general relativity, clocks located in weaker gravitational fields run faster. Because GPS satellites orbit at an altitude of approximately 20,200 km, their onboard clocks advance by about 45 \mus per day relative to clocks on the Earth’s surface. Second, special relativity predicts that clocks moving relative to an inertial frame run more slowly. Because GPS satellites travel at about 3.9 km/s, this kinematic effect causes their clocks to lag by about 7 \mus per day. The combined result is a net clock advance of approximately 38 \mus per day relative to ground-based clocks. Even a timing error of 1 \mus corresponds to a position error of roughly 300 m. Without relativistic corrections, GPS positioning errors would accumulate at a rate of about 10 km per day.

Summary & Closure

The ability to successfully engineer rockets and spacecraft has enabled humankind to reach space. They have been used to study other planets and Earth. Planetary exploration using probes will continue in the coming decades, perhaps at an accelerated pace. Humans will likely return to the Moon before the end of this decade. However, the challenges in sending humans to even our nearest planets, including Mars, remain a much more ambitious technical goal. Nevertheless, humankind will likely continue to justify the means of exploring space, partly because commercial space endeavors have revitalized public interest in space. In addition, new space technologies, including reusable launch vehicles and more efficient rocket engines, will continue to significantly reduce payload launch costs.

Sending humans back to the Moon, Mars, and into deep space will be a highly ambitious goal that requires overcoming a range of technical and other challenges, including radiation exposure, life support, propulsion systems, habitat and resource utilization, human performance and health, cost and funding, technological advancements, and mission design and operations. Space exploration continues to push the boundaries of science and technology, helping us address some of the most fundamental questions about our universe. The development of the International Space Station and partnerships between countries worldwide have enabled long-duration human spaceflights, leading to significant scientific discoveries and technological advancements that benefit both space exploration and life on Earth.

5-Question Self-Assessment Quickquiz

For Further Thought or Discussion

  • Can you hit a golf ball hard enough to send it into orbit around the Earth? What about if you were on the Moon? Could you hit a golf ball into a lunar orbit?
  • Assume that someone dug a hole from one point on the surface of the Earth all the way to its core and then out the other side. What would happen if you then jumped into the hole? Would you pop out on the other side of the Earth?
  • Is there any gravity at the center of the Earth? Hint: Consider the effects of gravity on one mass located inside a larger mass.
  • Prove that the potential U at a point outside a spherical shell of mass M is G \, M /r, where r is the distance to the point from the center of the shell.
  • As a spacecraft leaves the Earth and heads to the Moon, it will progressively overcome the Earth’s gravity and be gradually attracted by the Moon’s gravity. At what point between the Earth and the Moon will the gravitational effects be in perfect balance?
  • As a pressure or sound source moves in the air, a Doppler effect comes into play relative to a fixed observer. What might happen to a light source as it nears the speed of light?
  • What was the Hafele–Keating experiment, and why was it important?

Other Useful Online Resources

Visit the following websites to learn more about space and orbital mechanics:

  • A lecture by Dr. Steven Hawking: “Why we should go into space.”
  • Carl Sagan’s 1994 “Lost” Lecture: “The Age of Exploration.”
  • Carl Sagan Lecture: Cosmos – Galaxies
  • The entire series of Carl Sagan video lectures.
  • A rare interview with Neil Armstrong, the first person to walk on the Moon.
  • A NASA video on: “Space Flight: The Application of Orbital Mechanics.”
  • NASA talk: “Escaping Earth’s Gravity: Space Launch System.”
  • NASA talk: “Path to Mars and Asteroid Mission: The First Step.”
  • NASA video: ” Space Flight: The Application of Orbital Mechanics.”
  • 25 Mind-Blowing Facts About the Apollo Space Missions – Smithsonian Channel.
  • Deriving Einstein’s most famous equation: Why does energy = mass x speed of light squared?

  1. The author is indebted to his teacher, Professor Archibald "Archie" Roy (1924–2012), who inspired many generations of engineering students about astronomy, space, and the future possibilities of space flight. Professor Roy was a Fellow of the Royal Society of Edinburgh, the Royal Astronomical Society, and the British Interplanetary Society. Professor Roy published 20 books and 70 scientific papers. He is also known for the "The Mirror Theorem" in celestial mechanics.
  2. The full significance of Kepler's work became clear after Newton established its theoretical foundation.
  3. In English, the full title of Kepler's book is "New Astronomy, Based upon Causes, or Celestial Physics, Treated by Means of Commentaries on the Motions of the Star Mars, from the Observations of Tycho Brahe."
  4. Roy, A. E. and Ovenden, M. W., "On a Class of Symmetric Solutions of the Problem of n Bodies,'' Monthly Notices of the Royal Astronomical Society, Vol. 115, No. 1, pp. 23–30, 1955.
  5. In the conventional ballistics of projectiles, the ballistic coefficient, \beta, of a bullet is a measure of its ability to overcome air resistance in flight. A higher value of \beta means it will be faster when it reaches the target.
  6. As a reference, a rifle bullet will have a value of \beta \approx = 0.1.
  7. It will be recognized that the classic Froude number used in hydrodynamics is simply the square root of this parameter. However, in the context of orbital mechanics, the parameter is most meaningful when expressed as an energy-level ratio.

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Introduction to Aerospace Flight Vehicles Copyright © 2022–2026 by J. Gordon Leishman is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License, except where otherwise noted.

Digital Object Identifier (DOI)

https://doi.org/https://doi.org/10.15394/eaglepub.2022.1066.n44

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