45 Astronautics & Astrodynamics

Introduction[1]

Humankind has always strived to develop flight vehicles that can fly faster and higher. Eventually, there comes a height above the surface of the Earth where the normal atmosphere starts to thin out so that airplanes can no longer fly there. This height is the beginning of what is known as outer space, which is usually referred to as space. The fastest and highest aircraft ever flown, the SR-71 Blackbird, achieved a speed of 3,540 kph (2,200 mph) at an altitude of about 24 km (15 miles) before it ran out of atmosphere to fly as an aircraft.

Theodore von Kármán rationalized that space must begin where the atmosphere’s density becomes asymptotically thin to create aerodynamic lift. 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 (about 62 miles or 328,000 ft) and 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.

Why would humans, as 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, with no air to breathe and ultra-high and ultra-low temperature extremes. Distances are enormous, and even the Moon, the closest planet to the Earth, is very far away at about 384,400 km (239,000 miles), on average. Many 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?

Stephen Hawking, the renowned astrophysicist and the former Lucasian Professor of Mathematics at the University of Cambridge, spoke on the subject “Why we should go into space” for NASA’s 50th Anniversary Lecture Series. Professor Hawking, now since deceased, was severely disabled with ALS and spoke using a computer, but listen carefully as to why he advocates the need to go into space, return to the Moon, and eventually make an attempt to go to Mars.

It soon becomes clear that if it were not for the development of rockets, satellites, and other space systems, humankind would know much less about their planet, and 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 things 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!”

It is much 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 is still very much 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 is still being explored and tested, and new discoveries could challenge current understanding.

There are billions upon billions of planets; this is well known. But are there also other civilizations on those planets? It seems likely that humans are not the only living things in the universe. Might humans one day see intelligent living beings arrive from another planet? Will they look like us? Most likely not – the randomness of evolution suggests that 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 needed to answer these questions and better understand the possibilities of extraterrestrial life and other civilizations.

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 the Earth taken from the Orion spacecraft, as shown below, eminently makes the point. Despite the many technological advancements and understanding, the knowledge of the universe is still limited, and it remains an enigma in many ways. As shown below, the photograph of Earth taken from space is a humbling reminder of Earth’s small place in the cosmos. It certainly inspires humankind to continue exploring and learning more about the universe.

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

Learning Objectives

  • Learn about the history of astronomy and the planets.
  • Appreciate the challenges of humankind reaching into space.
  • Understand the basic 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.
  • Be aware of how to transfer spacecraft between orbits.
  • Understand the challenges of interplanetary travel, such as a mission to Mars.

What is Out There?

Humans have roamed the Earth for 300,000 years, but civilization has only existed for about 10,000 years. This time is a mere blink of an eye in the time scales of the universe, which is estimated to be about 14 billion years old, i.e., human civilization has been around for only about 0.0000001% of the established time. Advancing technology is one of the most characteristic of all human activities, the fascination with the stars and the exploration of our universe being just one motivation. However, only in the last 500 years has humankind started to make more sense of our place in the universe, thanks to significant advancements in science and engineering.

The earliest contributors to the science of the universe and space included observations of the planets by the Greek philosopher Aristotle. 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 was later to be disproved by Galileo and other scientists who used telescopes to observe the movements of the stars and the planets. The astronomer Claudius Ptolemaeus also believed the Sun, Moon, and other planets circled the Earth.

We now understand that the Earth is not at the center of the universe and that it orbits around the Sun as part of our solar system.  However, Aristotelian cosmology and so-called geocentrism or the belief that the Earth is at the center of the universe, dominated astronomy and science for over a 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 started to provide more evidence of heliocentrism, the idea that the Earth and other planets revolve around the Sun. Our understanding of the universe and its vast size and complexity has advanced even further, and the Earth is just a small planet in a spiral galaxy among billions of others.

In 1532, the Polish astronomer Nicolaus Copernicus was the first to propose that the Sun was the central focus around which the planets rotated in circular orbits, the Earth turning once daily on its axis. Copernicus’s heliocentric model of the universe was a major departure 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 new discoveries, theories, and insights into the nature of the universe, and helped lay 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 as a whole.

The Italian astronomer and scientist Galileo Galilei supported Copernicus’s theories that the Earth rotated on its axis once every 24 hours and simultaneously revolved around the Sun once a year. Starting about 1609, Galileo built on the concept of a spyglass or monocular to develop various astronomical telescopes with increasing magnifications. Galileo played a key role in the Scientific Revolution by using telescopes to observe the heavens and gather empirical evidence in support of heliocentrism. 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 important contributions to science and astronomy and is considered one of the greatest scientists in history. His observations of the Moon, Venus, Saturn, and other celestial objects were groundbreaking and challenged long-held beliefs about the nature of the universe. Galileo discovered that the surface of the Moon is not smooth and was made of many craters from the impacts of asteroids, nor was it translucent, a common belief at that 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” and his legacy continues to inspire scientists and astronomers.

The 17th-century German astronomer Johannes Kepler, a student of Tycho Brahe, made the profound observation that the planets moved around the Sun in elliptical orbits, not circular orbits, as shown in the schematic below. His major discovery that the planets move in elliptical orbits, not circular orbits, was a major departure from the previously held views and helped establish the idea that the universe was governed by scientific and mathematical laws. Kepler’s laws provided a basis for the work of later scientists, including Isaac Newton, who used these laws to develop his laws of gravitation and describe the behavior of all objects in the universe, not just the planets. It seems that the true significance of Kepler’s work remained unrecognized until Newton rediscovered it. Kepler’s work, however, was a significant step towards a more complete understanding of the solar system and the universe as a whole and continues to influence astronomical research.

 

Kepler was to conclude that the planets moved in elliptical orbits around the Sun, his observations and scientific reasoning becoming the basis for his three laws of planetary motion.

What is the closest planet to Earth besides the Moon?

Most people will say that the closest planet to Earth is Mars. But the closest planet to Earth, on average, is actually Venus. The average distance between Earth and Venus is about 41 million km (25.2 million miles), while the closest distance between the two planets is about 38 million km (23.6 million miles). Mercury is closer to the Sun than Earth and is, on average, 77 million km (48 million miles) away from the Earth. However, because of its elliptical orbit, the distance between Mercury and Earth can vary significantly.

Kepler’s laws, published between 1609 and 1619, were not immediately accepted. Indeed, Galileo himself questioned Kepler’s observations and disregarded his book Astronomia nova. In English, the full title of Kepler’s book is the 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. Kepler’s laws were later confirmed by other astronomical observations, particularly the transit time projections of Venus and Mercury across the face of the Sun, although these observations are now only of historical significance. Kepler’s laws provided the foundation for the laws of gravitation developed by Isaac Newton, and they are still used in modern celestial mechanics. Since then, Kepler’s laws have had a significant impact on our understanding of the solar system and have been instrumental in space exploration and satellite navigation. Kepler’s Astronomia nova soon became one of the most important books in the history of astronomy, and Kepler’s work remains relevant in modern astrophysics and planetary science.

Kepler’s next book, Epitome Astronomiae Copernicanae or Epitome of Copernican Astronomy, was read widely by astronomers during the 17th century. The publication of his book, however, was delayed until after Kepler’s death because of its “controversial” content. Nevertheless, many established astronomers and scientists of the era continued to challenge Kepler’s theories as a physical basis for explaining planetary and other celestial motions.

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

What is Space?

The aerodynamic environment in which aircraft fly has been well studied and standardized mathematical models of the atmosphere 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 at all, but this is untrue. Concepts such as pressure, temperature, and thermal transfer still have meaning in space, and concepts such as radiation now also need to be considered. Space is an almost perfect vacuum at extremely low pressure, primarily devoid of matter. Unlike air and other gases, space does not impede the motion of objects and they have no 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. In other regions, matter can accumulate to form planets, stars, and galaxies; the Earth is in one of these lucky regions.

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, such as cosmic rays, solar winds, and magnetic fields. These radiation and fields can affect the behavior of spacecraft and other objects in space, and must be taken into account when designing spacecraft and planning space missions. The study of the space environment, known as space physics, also involves understanding the interactions between these various forms of radiation and other matter in space, and how they influence the behavior of objects in the solar system. Space physics plays a critical role in enabling successful space missions, from communication and navigation to scientific exploration and human spaceflight.

Space and the universe as a whole are vast, and it extends to 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 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 in perspective, the speed of light is very fast and is about one million times the speed of sound in air on Earth.

The value of the speed of light can be used to confirm that the Earth is, on average, 384,400 km (39,000 miles) from the Moon, and the Earth is also 310.6 million km (193 million miles) from the Sun. That does not sound too far at all 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 reason is why scientists and engineers measure immense space distances in light-years, the distance that light can travel in one year, and it is equivalent to about 10 trillion km (6 trillion miles)! Yes, space is most definitely vast.

The Earth is part of a galaxy, which is part of an arm of a massive collection of stars that is swirling through space and throughout the universe. Indeed, our own galaxy, the Milky Way, is estimated to be just one of about two trillion (!) or more galaxies in the universe. Groups of galaxies are found to be arranged in clusters, and clusters then become part of superclusters. These superclusters stretch thousands of light-years across the universe, interspersed with dark voids and other features such as black holes.

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

Scientists have recently detected numerous exoplanets, or planets outside our solar system, and the number of known exoplanets has rapidly increased in recent years with the advent of new technologies and methods of detecting these distant worlds. The discovery of exoplanets has also sparked a new era of research into the potential for extraterrestrial life and the search for potentially habitable planets. The study of astronomy continues to be a fascinating and rapidly advancing field that has the potential to reveal new and exciting information about the universe and our place in it.

What is a “light-year”?

A light-year is the distance light travels in one year. Light travels through space at about 300,000 km per second (186,000 miles per second) and covers about 9.46 trillion km (5.88 trillion miles) per year. We use light-time 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 over 4 years even if we could travel at the speed of light, which is currently impossible according to our current understanding of physics. To put it into perspective, the fastest spacecraft ever launched by humanity, the Parker Solar Probe, takes about 90 years to travel 1 light-year at its maximum speed. These vast distances, combined with the long time it takes for light to travel from distant objects, make it a challenge for us to explore and understand the universe beyond our solar system.

What is Astrodynamics?

Astrodynamics, which is sometimes referred to as “orbital mechanics,” is a critical engineering field for the design and control of spacecraft missions. It involves the application of mathematical and physical principles to understand and predict the motion of satellites, as well as the calculation of spacecraft trajectories, orbit determination, and control. The field is extentensive, and the subject can only be introduced in this eTextbook.

The study of astrodynamics is also crucial for many aspects of space exploration, from designing interplanetary missions to maintaining satellites in orbit for communication, navigation, and other purposes. The principles of astrodynamics are used to determine the necessary propulsion systems, propellant requirements, and control strategies required to put a satellite into orbit, keep it there, and maneuver it to achieve its mission goals.

Getting into Space

Rockets or launch vehicles are used to send satellites, space probes, and human-carrying spacecraft into orbit around the Earth. A representative launch profile of a rocket is shown in the figure below. At the moment of the initial launch, the thrust produced by the rocket motors will be greater than the vehicle’s weight, so the rocket accelerates away quickly from the pad. The rocket’s weight then rapidly decreases because of the high consumption of propellant, so it continues to accelerate as it gains altitude. It also begins to pitch over to a more horizontal flight path under the effects of gravity, so the rocket quickly gains translational velocity. Finally, as it begins to exit the atmosphere at about 60,000 ft (about 20,000 m), it will by flying at supersonic speeds.

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

A few minutes into the ascent, staging will occur on the rocket booster where the first stage is jettisoned as it runs out of propellant, and the rocket motor for the second stage is ignited. The empty first stage then falls back toward the surface and either burns up in the atmosphere (depending on the staging altitude) or 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 at speeds of thousands of kilometers per hour as it orbits the Earth. The payload can then be configured to perform its intended mission, whether that be deploying satellites, conducting scientific experiments, or even transporting astronauts to the International Space Station. The complexity of the launch, ascent, and orbital insertion processes highlights the importance of having a highly skilled and knowledgeable engineering team, a thorough understanding of the environment, and the use of reliable and tested technologies to ensure the success of a rocket launch.

Gravitational Laws of Attraction

The payload (e.g., a satellite) stays in orbit because it has been given momentum and kinetic energy by the chemical energy liberated in the propellant by the rocket motors. In orbit, the balance of forces can be explained using the principles of statics. There will be an outward-directed force, called centrifugal force, which for equilibrium must be balanced by an inward-directed force caused by the effects of gravitational attraction toward the Earth. This balance between gravity and centrifugal forces keeps the satellite orbiting around Earth, and any change to this balance of forces will cause it to either fall to Earth or escape into space.

It is often said that Isaac Newton was responsible for discovering gravity. The legend is that he was sitting 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 fantasy if it were to give himself claim to the discovery of gravity. Although Isaac Newton is widely credited with formulating the laws of gravity, Robert Hooke had previously established the concept of a gravitational force and its attraction between bodies. However, it was Newton who provided the mathematical framework for understanding gravity and formulated the law of universal gravitation.

Universal Law of Gravitation

Hooke and Newton understood that the “force of gravity” must be related to the masses and distance between two bodies, i.e., the bigger they are and the 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 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 masses, 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 the figure below.

 

Hooke and Newton proposed that “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 36 is called the Universal Law of Gravitation. The minus sign denotes that this force is attractive, i.e., the masses are attracted toward each other. This fundamental work established Isaac Newton as one of the most influential scientists in history.

Gravitational Constant

Newton did not determine the value of the universal gravitational constant before his death. Its value was subsequently determined by Henry Cavendish who measured the average density and mass of the Earth, M_E, which is 5.97 \times 10^{24} kg. The accepted value of the universal gravitational constant is  G = 6.67428 \times 10^{-11} N m^2 kg^{-2}.

Consequently, Eq. 36 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 about 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*}

What is the Value of g ?

Everyone knows that the effects of gravity on the surface of the Earth is equivalent to an acceleration of 9.81 m/s2 (or 32.17 ft/s2), which is given the symbol g. The question is: Where does this value come from?  Its value is easy to determine using the universal law of gravitation, which states, in general, that

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

so for a mass m on the surface of the Earth and using Newton’s second law that F = m a, then

(6)   \begin{equation*} F_g = -\frac{G \, M_E \, m}{{R_E}^2} = - m g \end{equation*}

where M_E is the mass of the Earth and R_E is the radius of the Earth. The Earth is not exactly a sphere but an oblate spheroid with an equatorial bulge. However, the assumption that the Earth is a sphere is close enough for this analysis, as shown in the figure below.

The mass and radius of the Earth can be used to calculate the acceleration under gravity on the surface.

Recall that the mass of the Earth, M_E, is 5.97 \times 10^{24} kg, the radius of the Earth R_E  is 6.3781 \times 10^6 m, and  G = 6.67428 \times 10^{-11} N m^2 kg^{-2}. Therefore, rearranging the foregoing equation gives

(7)   \begin{equation*} g = \frac{G \, M_E }{{R_E}^2} \end{equation*}

Substituting the known values gives

(8)   \begin{equation*} g = \frac{ (6.67428 \times 10^{-11}) \times (5.97 \times 10^{24} )}{(6.3781 \times 10^6)^2} = 9.81~\mbox{m/s$^2$} \end{equation*}

As might be expected, the familiar value of g comes from the direct application of the universal law of gravitation to a mass sitting on the surface of the Earth. Because of the use of g in a variety of problems in orbital mechanics, it is often given the symbol g_0.

Worked Example #1

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

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

On planet X, the contribution of the mass and radius of the planet is going to change. 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 experienced by the astronaut on planet X would be approximately 1.78 times greater than the force of gravity experienced on Earth. This means that you would weigh approximately 1.78 times more on planet X than you do on Earth.

Circular Orbits

Consider now a satellite in a circular orbit around the Earth. Circular orbits arise whenever the gravitational force on a satellite balances the centrifugal force needed to move it with uniform circular motion, as shown in the figure below.

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

The force equilibrium of a satellite of mass m_s is

(9)   \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

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

so that the orbital speed is inversely proportional to the square-root of its orbital radius. Notice that this foregoing result does not depend on the mass of the satellite.

It can be concluded from Eq. 10 that satellites that orbit closer to Earth (i.e., the value of R_s is less) experience stronger effects of gravity, as shown in the figure below. Therefore, to stay in orbit, they must travel faster than a satellite that is orbiting farther away from the Earth.  In a geosynchronous orbit (GEOS), the satellite takes 24-hours to complete one orbit around the Earth, so it appears to be stationary in the sky relative to an observer on the Earth. This behavior is useful for communication satellites that need to remain over a specific location on the Earth’s surface. The altitude of GEOS is high enough that the satellite is not affected by the drag from the Earth’s atmosphere, which would cause its orbit to decay.

 

Low Earth orbit (LEO) is only about 402 km (250 miles)  above the surface of the Earth. Many satellites, however, are in a geosynchronous orbit (GEOS) at a height of more than 35,406 km (22,000 miles).

Low Earth orbit (LEO) is much lower and closer to the Earth’s surface and so satellites are subject to drag from the Earth’s atmosphere, which requires more energy to maintain the orbit and also limits the lifespan of such satellites. For example, the International Space Station is in LEO at only about 402 km (250 miles) above the surface of the Earth. It must travel at a speed of about 27,600 kph (7.67 km/sec or 17,150 mph) to stay in orbit! Many satellites, however, orbit in GEOS at a height of more than 35,406 km (22,000 miles) above the surface, and so only (!) have to travel at about  10,783 kph, 3 km/sec or 6,700 mph) to maintain their orbits.

Worked Example #2

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.

A geosynchronous orbit is one in which the satellite maintains its position (approximately) over a point on the surface of the Earth. To maintain such a position, the angular velocity of  the satellite, \omega_s, about the Earth must match to the angular velocity of the Earth, \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 center of the Earth. 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 means that a satellite in geosynchronous orbit will be at a height of R_s - R_E above the surface of the Earth, where R_E for the Earth is 6,378.1 km. This means that such satellites will be at a height of about 35,790.7 km (22,236 miles) above the surface.

Orbit Equation

The orbit equation explains the orbital motion of a lesser mass about a higher total mass at rest. It is often referred to as a solution of the two-body problem. After a mass or payload, e.g., a satellite, is in orbit, its motion and position are governed by the principles of orbital mechanics. Orbital mechanics is the study of the motions of planets as well as satellites and space vehicles moving under the influence of gravity, their own thrust, etc. The study of orbital mechanics helps engineers to plan and execute missions to explore the solar system and to 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 and momentum as well as total energy. This energy has been imparted by the launch vehicle up through the point of final burnout. It has two parts, namely kinetic energy, which associated with its motion, and its gravitational potential energy, which is associated with its altitude or height. The motion of the satellite is determined by the balance between these two forms of energy. As the satellite moves along its orbit, the gravitational potential energy changes because of the decrease in altitude, and this energy is transformed into kinetic energy.

This balance between gravitational potential energy and kinetic energy allows the satellite to remain in orbit. The gravitational attraction from the Earth acts to keep the satellite moving in its path, while its velocity and momentum prevents it from falling back towards the Earth. The stability of the satellite’s orbit is also affected by other factors such as atmospheric drag, solar radiation pressure, and gravitational perturbations from other celestial bodies.

For circular orbits, the gravitational potential energy, which is usually given the symbol U, of a satellite of mass m_s that orbits around a central mass M, is

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

which is obtained by using the universal law of gravitation.

This latter result is easily proved by determining the work needed to raise a mass m_s from one altitude, R_i, to some other altitude, R_f. The total work to do this can be written as

(12)   \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}  \left( \frac{1}{r^2} \right)\, dr \end{equation*}

Evaluating the integral gives

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

In orbital mechanics, the reference point for the evaluation of the potential energy is assumed to be at infinity, i.e., R_i = \infty. In this case then U = W (energy is conserved) so

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

or in general

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

Notice that because the reference for the potential energy is set to infinity then the potential energy is negative if the satellite is bound in an orbit around the Earth or another planet. Therefore, the closer the two masses are then the stronger the gravitational potential energy of the satellite and the more negative the value of U.

The kinetic energy, which given the symbol K, is best expressed in terms of polar coordinates, i.e.,

(16)   \begin{equation*} K = \frac{1}{2} m_s V^2 = \frac{1}{2} m_s V_{\theta}^2 + \frac{1}{2} m_s \dot{r}^2 \end{equation*}

which can be rationalized from the schematic shown below.

The total kinetic energy of a satellite about a central mass may have angular and radial components.

Because V_{\theta} = r \omega_{\theta} = r \dot{\theta}} (i.e., solid body rotation) then this latter equation can be written as

(17)   \begin{equation*} K = \frac{1}{2} m_s \left( (r \dot{\theta})^2 + \dot{r}^2 \right) - \frac{G M m_s}{r} \end{equation*}

Therefore, the total energy is

(18)   \begin{equation*} E = K + U = \frac{1}{2} m_s \left( (r \dot{\theta})^2 + \dot{r}^2 \right) - \frac{G M m_s}{r} \end{equation*}

In the case of a satellite of mass m_s orbiting the planet Earth, then

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

Equation of Motion of a Satellite

Recall that the energy equation is

(20)   \begin{equation*} E = \frac{1}{2} m_s \left( (r \dot{\theta})^2 + \dot{r}^2 \right) + \frac{\mu \, m_s}{r} \end{equation*}

where the satellite has a mass m_s. Expanding out, then

(21)   \begin{equation*} E = \frac{1}{2} m_s (r \dot{\theta})^2 + \frac{1}{2} m_s \dot{r}^2 + \frac{\mu \, m_s}{r} \end{equation*}

noting that this latter equation involves both r and \theta, i.e., it has two degrees of freedom.

Start first with the \theta coordinate. In this case

(22)   \begin{equation*} \frac{d}{dt} \left( \frac{\partial B}{\partial \dot{\theta}} \right) = 0 = \frac{d(r^2 \dot{\theta})}{dt} \end{equation*}

implying that r^2 \dot{\theta} is constant, which is the angular momentum of the satellite about the central mass. This means that angular momentum is conserved, i.e.,

(23)   \begin{equation*} m_s r^2 \dot{\theta} =\mbox{constant} \end{equation*}

assuming, of course, no thrust is applied to the satellite. This latter equation is often written in terms of “per unit mass” so the momentum per unit mass is

(24)   \begin{equation*} r^2 \dot{\theta} = h = \mbox{constant} \end{equation*}

Leaving out the all the detailed steps of the derivation, which can be found in any textbook on orbital mechanics, it can be shown that the equation of motion in the r direction is

(25)   \begin{equation*} \frac{h^2}{r^3} + \frac{\mu}{r^2} = 0 \end{equation*}

where

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

It can be further shown that the solution to this equation for a given value of m_s is

(27)   \begin{equation*} r (\theta) = \frac{ \displaystyle{\frac{h^2}{\mu}}}{1 + \left(\displaystyle{\frac{c \, h^2}{\mu}} \right) \cos (\theta - \phi)} \end{equation*}

where h, c, and \phi are constants. The angle \theta is measured between the vector r and the axis of periapsis, which is also called the true anomaly. This foregoing algebraic equation describes the orbital trajectory of the satellite of mass m_s around another larger mass M and is usually referred to as the orbit equation.

Solutions to the Orbit Equation

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

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

where p is called the parameter of the orbit and e is called the eccentricity of the orbit.

This foregoing solution is one of the most fundamental equations in all of orbital mechanics, and gives several different orbital trajectories, called conic sections or conics. As shown in the figure below, there are four orbital trajectories that can be derived from the orbit equation, depending on the value of e, these being:

          1. e = 0, the orbit is circular.
          2. e < 1, the orbit is elliptical.
          3. e = 1, the orbit is parabolic.
          4. e > 1, the orbit is hyperbolic.
Four orbital trajectories can be derived from the orbit equation, depending on the value of the ecentricity.

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 that is orbiting about another, larger, central mass, are now known as Kepler’s Laws of Orbital Motion. At the time of their formulation they were considered to be exact laws, and over the centuries they have been found to give very close approximations to actual planetary motions.

Kepler’s laws can also be deduced from Newton’s laws of motion and 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 formal formulation of his gravitational theory in the publication of his Principea. Newton was to show that Kepler’s laws are a special case of the gravitational motion of n bodies where all the bodies may be treated as point masses and their motion is not affected by each other except for the gravitational attraction of 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 orbital trajectories of the planets are ellipses about the Sun with the Sun at the focus, i.e., “A satellite describes an elliptical orbit around its center of attraction.” The figure below shows an annotated elliptical orbit with semi-major length a and semi-minor length b.

An elliptical orbit can be described in terms of its eccentricity as well as 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 are moving in a curvilinear path, an external force is acting upon them, which is the force of gravitational attraction.
  2. Once they move in an orbit, they will have constant momentum and energy.

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

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

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

(30)   \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 center of the Earth and that of Mars swept out an equal area in an equal amount of time, as illustrated in the figure below. He reasoned that when the planet is closest to the Sun, or at perihelion, it must be traveling faster than when it is farther from the Sun at its aphelion. This observation became the basis for Kepler’s second law, 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 a 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

(31)   \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.,

(32)   \begin{equation*} m r^2 \dot{\theta} =\mbox{constant} \end{equation*}

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

(33)   \begin{equation*} \frac{dA}{dt} = \frac{1}{2} r^2 \frac{d \theta}{d t} = \frac{1}{2} r^2 \dot{\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 is moving fastest when it is at perihelion (or perigee) and slowest at its aphelion (or apogee).

Kepler’s 3rd Law

Kepler’s third law, which was established much later than the first two laws, states: “The square of the orbital period is directly proportional to the cube of the average distance between the satellite and the center of attraction.” 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.,

(34)   \begin{equation*} T^2 \, \propto  \, a^3 \end{equation*}

or

(35)   \begin{equation*} T \, \propto \, a^{3/2} \end{equation*}

Kepler’s third law, unsurprisingly, also verifies that the larger the orbit then 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

(36)   \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 a central mass M. The force equilibrium is that the outward inward centrifugal force equals the gravitational attraction, i.e.,

(37)   \begin{equation*} \frac{m_s \, V^2}{R_s} = \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

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

Therefore,

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

which on rearrangement gives

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

noticing that this latter result does not depend on the mass of the satellite. The conclusion is that

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

which is Kepler’s third law.

Kepler’s third law allows predictions of the orbit of any satellite or object in space travelling around a central body. While it has been previously derived by assuming a circular orbit, it also applies to an elliptical orbit. This relationship can be more formally quantified by

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

where

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

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

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. 42 gives

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

and

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

Therefore,

(46)   \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 more 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. 46 is almost unity, i.e.,

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

so that

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

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

Worked Example #3

The Earth orbits around the Sun in one year, i.e., T_E =365.256 days. (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 10^9 km. Now, if the semi-major axis, a_M, of Mars, is 2.278 \times 10^9 km, then what will the orbit time, T_M, of Mars be?

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

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

Measuring the Mass of a Planet

One of the questions students often ask is how we know the masses of the planets. However, any planet that has a satellite orbiting about it, either natural or artificial, can be used to measure the planet’s mass by studying the orbital motion of the satellite and using Kepler’s third law.

Consider first the orbit of the Earth around the Sun, the Sun having mass M_S and the Earth having mass M_E and orbital semi-major axis length a_E. Then using Eq. 44 gives

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

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

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

Therefore, using both of these prior equations, then

(51)   \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. 51 can all be measured, so the mass of the Earth relative to the Sun can then be determined. It turns out that the mass of the Earth is approximately 333,000 times less than the mass of the Sun. As a reference, the Sun is estimated to have a mass of approximately 1.989 \times 10^{30} kg.

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

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, it takes the Moon about 29.5 days to complete one cycle of phases, which is called the synodic month. The difference between the sidereal and synodic months is because as the Moon orbits around the Earth, the Earth also orbits around the Sun. Therefore, the Moon must travel a little farther in its orbit to compensate for the added distance to complete the phase cycle.

Orbital Energy

Orbital mechanics is mostly governed by the energy of the orbiting satellite, which consists of kinetic energy 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, as well as the laws of motion, which describe how an object moves in response to forces acting on it. These fundamental laws help explain the motion and stability of satellites in orbit around Earth and other celestial bodies.

Circular Orbit

The kinetic energy of the satellite, K, is

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

and the associated potential energy, U, is

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

Therefore, the total energy is

(54)   \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

(55)   \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 foregoing equation gives

(56)   \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

(57)   \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

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

which says that the energy of the satellite in an orbit around the mass M is negative. This result can also be proven for an elliptical orbit.

Negative Potential Energy?

What does it mean when the energy given by Eq. 58 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 orbits that are circular and elliptical 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 orbit, 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 result is a direct outcome of Kepler’s second law in that a satellite will speed up when it is closer to the central mass at its perigee and slow down when it is further away at its apogee, as shown in the figure below.

Because of conservation of momentum and energy, 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.,

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

The kinetic energy can also be written as

(60)   \begin{equation*} K = \frac{1}{2} m V^2  = \frac{1}{2} \left( \frac{G \, M \, m}{a} \right) \end{equation*}

so the total mechanical energy becomes

(61)   \begin{equation*} E = -\left( \frac{G \, M \, m}{2a} \right) \end{equation*}

Because the orbital energy is constant, the overall size of the orbit 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, and slowest at its farthest point, i.e. at its apogee.

Changing Orbits

It is often required that a satellite or other spacecraft be able to change its orbit from a lower orbit to a higher orbit, or vice-versa, as shown in the figure below. For example, after launch a satellite may be parked in an initial (lower) orbit until the orbit stabilizes, then it is boosted to its final altitude.

To raise the orbit requires energy (and propellant), in that a rocket motor must be fired to create work, W, to raise 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 motor.

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

(62)   \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 then conservation of energy requires

(63)   \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

(64)   \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

(65)   \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

(66)   \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

(67)   \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*}

Worked Example #4

Consider the energy required to raise a spacecraft from an orbital altitude of 250 km to 300 km. The mass of the spacecraft, m_s, is 1,500 kg. Assume that the mass of the Earth, M_E, is 5.97 \times 10^{24} kg, the radius of the Earth R_E is 6.3781 \times 10^6 m, and  G = 6.67428 \times 10^{-11} N m^2 kg^{-2}.

The work (and energy) required will be

    \[ W = \frac{1}{2} G \, M_E \, m_s \left( \frac{h_2 - h_1}{(h_1 + R_E) (h_2 + R_E)} \right) \]

The value of \frac{1}{2} G \, M_E \, m_s is

    \[ \frac{1}{2} G \, M_E \, m_s = \frac{1}{2} (6.67428 \times 10^{-11}) \times (5.97 \times 10^{24}) \times (1,500.0) = 2.9885\times 10^{17} \]

Inserting the other values gives

    \[ W = 2.9885\times 10^{17} \left( \frac{300.0 - 250.0}{(250.0 + 6,378.1)(300.0 + 6,378.1)} \right) \]

and so

    \[ W = 3.376 \times 10^{6}~\mbox{MJ} \]

Orbital Launch

Rocket launches are very exciting events to watch! The synchronized launch process, the flaming exhaust and clouds of smoke, followed by the rising spacecraft into the sky can be awe-inspiring. The technology and engineering behind the launches, the science and exploration that they enable, and the thrill of human endeavor all contribute to the excitement. Additionally, watching a rocket launch live or on TV can be an educational and inspiring experience, especially for students and younger people who are interested in science and space. Even those of us who live in Florida never get tired of watching rocket launches!

NASA’s SLS rocket carrying the Orion spacecraft launches on the Artemis 1 flight on December 16, 2022, from Launch Complex 39B at NASA’s Kennedy Space Center in Florida.

Drama and excitement aside, the energy and propellant requirements for a rocket (booster) and its payload with the \Delta V needed to reach a specific orbital altitude h_s can be estimated using the principles of energy conservation in conjunction with Tsiolkovsky’s rocket equation. The rocket equation is

(68)   \begin{equation*} \Delta V = V_{\rm ex} \ln \left( \frac{M_f}{M_0} \right) \end{equation*}

where V_{\rm ex} is the equivalent exit velocity from the particular rocket motor, 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 at burnout time, t_b, during which a propellant mass M_p is consumed. In this case it is assumed, for simplicity, that it is a single-stage launch vehicle (i.e., no staging).

The equivalent exit velocity can be written in terms of the specific impulse, I_{\rm sp}, as

(69)   \begin{equation*} V_{\rm ex} = I_{\rm sp} \, g_0 \end{equation*}

where g_0 is the reference value of acceleration under gravity at mean sea-level on Earth, i.e., g_0 = 9.81 m/s. Therefore, the rocket equation can also be written as

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

where the value of I_{\rm sp} depends on the type of rocket motor and its propellant. If the value of I_{\rm sp} is known, then the rocket equation can be used to determine the propellent mass needed to give a certain \Delta V for the satellite or spacecraft to reach an orbit at the required altitude, h_s, as shown in the figure below.

Orbital mechanics can be used to estimate the propellent mass needed to lift a payload into orbit.

For a launch, several factors will influence the \Delta V required, including the needed orbital altitude above the surface of the Earth, h_s, the orbital inclination, and launch latitude (this affecting the initial energy,) the effects to overcome gravity when the rocket is going vertically, and the aerodynamic drag on the rocket in the lower atmosphere.

For the satellite or spacecraft to reach the required orbital altitude, h_s, then the \Delta V required to give the needed kinetic energy is

(71)   \begin{equation*} \Delta V = \sqrt{ \frac{G \, M_E}{R_E + h_s} } = \Delta V_K \end{equation*}

noting that this result does not depend on the mass of the satellite or spacecraft. This latter component is the dominant energy for high orbits.

The \Delta V associated with creating the needed potential energy for the same orbit is

(72)   \begin{equation*} \Delta V = \sqrt{ \frac{ G \, M_E }{R_E} } \sqrt{ 2 - \frac{R_E}{R_E + h_s} } - \sqrt{ \frac{ G \, M_E}{R_E + h_s} } = \Delta V_U \end{equation*}

For low orbital altitudes with h \ll R_E then

(73)   \begin{equation*} \Delta V_U = \frac{h_s}{R_E} \sqrt{ \frac{G \, M_E}{R_E} } \end{equation*}

which is the dominant term in this case compared to the potential energy.

There is a gravitational effect that needs to be added to the total needed \Delta V, which is usually referred to as the gravity loss. For a rocket going up vertically, then

(74)   \begin{equation*} \Delta V_g = \int_0^{t_b} g_0 \, dt = g_0 \, t_b \end{equation*}

where t_b is the burnout time. If the rocket follows a curved trajectory and pitches over at a local trajectory angle \gamma (with respect to the horizon) as it increases altitude, then

(75)   \begin{equation*} \Delta V_g = \int_0^{t_b} \sin \gamma \, g_0 \, dt \end{equation*}

where \gamma = 90^{\circ} when the rocket flies vertically and \gamma = 0 when it flies horizontally. The proper evaluation of this latter term, therefore, requires information about the launch profile.

Finally, there is an aerodynamic drag on the rocket as it flies through the lower atmosphere. This drag force, D,  can be expressed as

(76)   \begin{equation*} D = \frac{1}{2} \varrho V_{\infty}^2 \, C_D \, A_{\rm ref} = q \, C_D \, A_{\rm ref} \end{equation*}

where q is the dynamic pressure, i.e., q = \frac{1}{2} \varrho V_{\infty}^2 where \varrho is the local density of the air, and V_{\infty} is the true airspeed, the latter values also requiring information about the launch profile. The density of the air can be represented by an ISA model for both the low and extended atmospheres. The reference area, A_{\rm ref}, in Eq. 76 is usually taken as the projected frontal area of the rocket. The drag coefficient C_D will also depend on the exact shape of the rocket and its flight Mach number.

It is apparent that drag of the rocket given by Eq. 76 is proportional to the density of the air and the square of the airspeed. This result show that to minimize the net aerodynamic effects the rocket should ascend vertically and as slowly as possible, but this is contrary to the need to accelerate the rocket as quickly as possible to minimize gravitational effects. Nevertheless, the dynamic pressure during the launch needs particular emphasis because this quantity affects the aerodynamic-induced structural loads on the rocket. As shown in the typical launch profile below, the value of q increases rapidly after launch, reaching a maximum then decreasing quickly as the altitude increases and the density of the air diminishes. In many cases, the value of the “maximum q” on the rocket will be limited, requiring the rocket motors to be throttled down temporarily to prevent excessive aerodynamic loads.

A representative launch profile for a rocket. The criticality of the “max. q” in the flight profile requires that the rocket motors be throttled.

The drag coefficient, C_D, for a rocket is often represented as

(77)   \begin{equation*} C_D =\left \{\begin{array}{ll} 0.2 \quad \mbox{ for $0 < M_{\infty} \leq 0.85$} \\[12pt] 0.11 + \displaystyle{\frac{0.82}{M_{\infty}^2}} - \displaystyle{\frac{0.55}{M_{\infty}^4}} \quad \mbox{for $M_{\infty} > 0.85$} \end{array} \right. \end{equation*}

where M_{\infty} is the Mach number of the free-stream aerodynamic flow. The \Delta V increment required to overcome aerodynamic drag (again, using the rocket equation) will be

(78)   \begin{equation*} \Delta V_d = \sqrt{ \int_0^{t_a} \frac{D \, V_{\infty}}{M} \, dt } \end{equation*}

where M is the instantaneous mass of the rocket and t_a is the launch time over which aerodynamic forces are considered.

Therefore, the total \Delta V_T is the sum

(79)   \begin{equation*} \Delta V_T = \Delta V_K + \Delta V_U + \Delta V_g +\Delta V_d \end{equation*}

The required propellant mass, M_p, can then be estimated from the rocket equation, i.e.,

(80)   \begin{equation*} \Delta V_T = I_{\rm sp} \, g_0 \ln \left( \frac{M_f}{M_0} \right) = I_{\rm sp} \, g_0 \ln \left( \frac{M_0 - M_p}{M_0} \right) \end{equation*}

again assuming no staging. Therefore, using the principles of logarithms, then

(81)   \begin{equation*} \frac{M_p}{M_0} = 1 - \exp \left( -\frac{\Delta V_T}{I_{\rm sp} \, g_0} \right) \end{equation*}

If staging is used, then the \Delta V for each stage would be calculated as each stage is depleted of propellant and the empty stage is then discarded.

For a satellite or other spacecraft to reach an orbital altitude of 300–400 km, the needed \Delta V is about 10 km/s. The aerodynamic contribution, \Delta V_d, is the lowest effect compared to the other three. The needed \Delta V also depends on the latitude of the launch. The initial energy given to the rocket from the Earth’s rotation is higher for launch sites closer to the equator. The needed \Delta V is lower if the rocket is launched in the direction of the Earth’s rotation (toward the east). For launches from Cape Canaveral, for example, the \Delta V boost is about 0.3 km/s, and is certainly not insignificant in terms of propellant requirements.

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 the diagram below. The maneuver is accomplished by placing the spacecraft into an elliptical transfer orbit, called a Hohmann transfer orbit, the perigee and the apogee of this trajectory being 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 is used to fly a spacecraft between two orbits at different altitudes, in this case about the Earth.

The transfer maneuver needs excess energy, so the spacecraft uses two impulsive engine burns. The first burn is required to increase V_1 and break out of the initial circular orbit and establish the elliptical transfer orbit. The second burn is used to obtain the velocity of the spacecraft needed 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 when it reaches the target orbit, it may need to be reoriented so that the engine(s) can fire in the orbital direction to speed it up.

For the lower (initial) orbit, solving for V_1 gives

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

and for the higher (final) orbit, then

(83)   \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 velocity of the spacecraft 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

(84)   \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

(85)   \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 the energy E_T will be

(86)   \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

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

Consider the point in time just after the first burn used 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

(88)   \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

(89)   \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

(90)   \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

(91)   \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, then it will continue under the action of gravity until it just 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

(92)   \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. 91 has been replaced by R_2 in Eq. 92.

However, at the end of the transfer orbit, this velocity will be too low to maintain the final orbit. Without a second burn, therefore, the spacecraft will simply return to the lower orbit and continue indefinitely in the transfer orbit. The velocity in the final circular orbit (Eq. 83)
must be

(93)   \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.

(94)   \begin{equation*} \Delta V_2 = V_2 - V_a \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_f}{M_0} \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_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

(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 to that described can be used to decrease the orbital altitude of a spacecraft. In this case, a retro-burn will be needed at the higher orbit to reduce the velocity and start the orbital transfer. At the end of this transfer, as gravitational potential energy increases (it becomes more negative), a retro-burn will be needed to break out of the elliptical orbit and so circularize it at the lower orbital altitude.

There is no dark side of the moon!

“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 non-uniform mass distribution inside the Moon, which is like a dumbell, it acts like a pendulum as it rotates around 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 only one side of the Moon (called the nearside) is always visible from the Earth. We never see the far side of the Moon, but it is not always dark there for the same reason as the nearside is not always dark.

Escape Velocity

How much velocity does it take to escape the Earth’s gravitational field? The escape velocity is a particular case where the spacecraft’s energy is high enough that an orbit is never accomplished. In other words, how fast does a spacecraft need to travel such that it continues into space and never comes back to Earth under the effects of the Earth’s gravitational field? The answer lies in the amount of energy and speed given to a spacecraft.

Consider a spacecraft of mass m_s in an orbit around the Earth, as shown in the figure below. The rocket executes a burn to give it some initial 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 very high velocity of over 11 km/second.

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 needs to fly far away that R_f \rightarrow\infty. Also, if is assumed that there is no excess kinetic energy at this distance, this assumption will allow an estimate of the minimum initial velocity to escape from the Earth’s gravity. The two terms on the right hand side of the foregoing 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 to obtain this velocity, the rocket motors need to execute a burn to leave the initial orbit.

Now it is possible to estimate the escape velocity for the Earth. Recall that the mass of the Earth, 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 } }= 11.178~\mbox{km/s} \end{equation*}

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

All spacecraft that are designed to head out into deep space must be given a velocity that is larger than 11 km/s, and will ultimately follow a hyperbolic trajectories away from the Earth.

What is a Black Hole?

Everyone has heard of black holes, right? They dot the universe and are considered a region where the effects of gravity are so strong that nothing, including light, has enough energy to escape. The effect of gravity is so strong in a black hole because so enough 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 from the signatures of high-energy radiation, and by the manner in which the stars behave differently to other stars under the extremely powerful gravitational forces. An interesting image of a black hole is shown below.

 

he 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
This image of black hole was taken by the Event Horizon Telescope. The image shows a bright ring formed as light bends in the intense gravity around a black hole that is 6.5 billion times the mass of the Sun.

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

(106)   \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

(107)   \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 calculated this exact solution for the first time.

Because the value of G = 6.67428 \times 10^{-11} N m^2 kg^{-2} and the speed of light is c = 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 was the same mass 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 much more massive than the Sun.

Theory of General Relativity

In 1915, Einstein published his theory of general relativity, which showed that massive objects cause a distortion in spacetime, leading to the curvature of nearby objects’ paths, as well as the phenomenon of gravitational lensing. The theory of general relativity was confirmed by the 1919 solar eclipse observation, which showed that light from stars near the Sun was deflected by the Sun’s gravity. Since then, the theory of general relativity has been extensively tested and verified, and it has become a cornerstone of modern physics and cosmology. It has also provided the foundation for understanding black holes, the expanding universe, and the formation of structures in the universe.

Einstein’s famously quoted result is

(108)   \begin{equation*} E = mc^2 \end{equation*}

although in Einstein’s original paper in 1905 he writes the relationship only as

(109)   \begin{equation*} m = \frac{E}{c^2} \end{equation*}

which describes the relationship between relativistic mass, m, and energy, E, where c is the speed of light. It is a statement that mass is composed of energy. Relativistic mass is related to rest mass, m_0, using

(110)   \begin{equation*} m = \frac{m_0}{\sqrt{ 1 - \displaystyle{\frac{u^2}{c^2}}}} \end{equation*}

where u is the relative speed of the mass. In the more general case, the equation of special relativity can be written as

(111)   \begin{equation*} E^2 = ( p c)^2 + (mc^2)^2 \end{equation*}

where p is momentum. Therefore, Eq. 111 can be used to relate and unify the concepts of energy, momentum, and invariant mass. The values of E and p depend on the relative movement of the observer relative. In the special case where the body is at rest, it becomes Eq. 108. Einstein’s theory of special relativity has developed into a fundamental scientific paradigm that connects mass, energy, momentum, space, and time, and it is accepted and used in modern astrophysics.

What happens when an object approaches you at the speed of light?

If an object approaches you, as a stationary observer, at just below the speed of light, the wavelength of the light emitted from it becomes longer, effectively shifting the light towards the red end of the spectrum. This effects is known as relativistic redshift, and is a consequence of the fact that the observer and the light are moving relative to one another. This redshift effect is predicted by Einstein’s theory of special relativity and is consistent with observations of high-speed astronomical objects, such as quasars and gamma-ray bursts. As the object passes you by, then its light will becomes blue shifted. If the object is traveling toward you at the speed of light, the light would appear infinitely redshifted and would no longer be visible to you.

Mission to Mars?

Humans landed on the Moon over a half a century ago but have not yet returned. It is likely, however, that they will return to the Moon within this decade. The Moon is a long way away, about 442.,570 km (275,000 miles), and it takes about three days to get there in a spacecraft, but 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, 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), so about 500 times further than the Moon! Mars is also 1.5237 AUs away from the Sun, one AU (Astronomical Unit) being the average distance from the Sun to the Earth.

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? Yes, humans have certainly sent probes to Mars and other planets, but can humans visit Mars or maybe even colonize it? It should be remembered that humans have yet to colonize the Moon or even Antarctica on our own planet; Mars is much more distant and hostile to humans than Antarctica or the Moon.

Some of the engineering answers to these questions can be estimated with the help of the principles previously established for a Hohmann cotangential elliptic transfer orbit, as shown in the figure below for transit to Mars. It is assumed that the orbits of Earth and Mars are circular rather than elliptical; this assumption simplifies the mathematics considerably and still gives 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 made is that the spacecraft is already in orbit around the Earth, and is fully fueled with propellant and ready to make the journey to Mars. Of course, getting the spacecraft into orbit with the needed propellant is also a significant challenge. The timing of when the spacecraft sets out for Mars on the transfer orbit is critically important, because they must both reach the exact orbital location simultaneously. 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 allow for an Earth rendezvous, reentry, and landing.

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

(112)   \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

(113)   \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

(114)   \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 one orbit, so to the transfer time to Mars will be half of this value giving an average flight time of 0.71 years or about 8.5 months.

The exact transfer time on a mission to Mars will also depend on how close Mars and Earth are in their elliptical orbits around the Sun at any one time, as well as the gravitational influence of other planets along the way, such as Jupiter. However, it is going to take a lot longer to get to Mars than 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, is likely going to take a couple of years. Nevertheless, this might not happen at all when using our conventional (chemical) propulsion systems.

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

(115)   \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.,

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

At the perigee of the transfer orbit then

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

and at the apogee of the transfer orbit

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

Conservation of angular momentum of the spacecraft at the perigee and apogee of the transfer orbit gives

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

or

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

Conservation of energy for the movement about the Sun requires

(121)   \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 foregoing equation to give

(122)   \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. 120 and 122 and after some algebra gives

(123)   \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

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

After some more algebra, then

(125)   \begin{equation*} \Delta V_2 = V_M - V_a = \sqrt{ \frac{2 G \, M_S \, R_M}{R_E (R_E + R_M)}} - \sqrt{ \frac{G \, M_S}{R_E} } = \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 has to be carried onboard the spacecraft. More propellant is also needed to get the spacecraft into Earth orbit, Mars orbit, and descend onto the surface. Staging will inevitably be required to progressively reduce the spacecraft’s mass at each phase of the flight, including a landing on Mars. 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.

Based on the preceding analysis, it is possible to estimate the propellant mass needed for the flight to Mars. The mass needed depends on the type of rocket motor(s) but can be estimated from the rocket equation, i.e.,

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

where I_{\rm sp} is the specific impulse for the rocket motor, 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

(127)   \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

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

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

For the first impulsive burn on a mission to Mars then

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

which means that the spacecraft will consume nearly half of its initial mass, M_1, in propellant in getting into the transfer orbit.

Consider now 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

(130)   \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, it becomes clear from the preceding analysis that getting to Mars will require a lot of propellant, totaling at least about 75% of the spacecraft’s original mass as it sets off from the Earth. Obviously, there are massive engineering implications in terms of designing a spacecraft capable of reaching Mars that is comprised of 75% propellant and only 25% spacecraft, including payload such as life-support and supplies.

For example, if the spacecraft has a total mass of 50,000 kg, then about 37,500 kg of that will be the needed propellant. For a 100,000 kg spacecraft, about 75,000 kg will need to be propellant, and so on. A further amount of propellant will be needed for any mid-course corrections, for a retro-burn to lower the spacecraft (or part of it) into a Mars orbit, and finally descend onto the surface. Even under the best circumstances, the engineering challenges in reaching Mars have now become more apparent. Besides the orbital mechanics, the astronauts that are sent to Mars will need to be protected from radiation exposure, and also have sufficient life support and a suitable habitat to maintain human performance and health.

So, now the astronauts finally get to Mars and set up a Martian base. What about getting them back from Mars to Earth? Again, the propellant needed can also be estimated with some assumptions but reasonable ones. Getting back to Earth from Mars requires that the spacecraft carries the needed propellant for the return journey, which must also be part of the initial mass of the spacecraft when it leaves Earth. So it takes propellant to transport propellant to Mars for the return mission to Earth. Now the engineering problem is getting more difficult.

Assume that the 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

(131)   \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 reenter Earth orbit

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

Therefore, the needed propellant mass for an out-and-return mission to Mars will be 1.45 times the mass of the spacecraft.

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

However, not all is lost. With two or three or more other spacecraft, one carrying the astronauts and the others carrying supplies and the needed propellant, then such a mission may be, at least technically, possible. It should also be remembered that even a mission to the Moon and return will cost about 95% in propellant and 5% as spacecraft, including the astronauts and all of their supplies. Getting the initial mass for a Mars mission into a parking orbit around the Earth will also require an enormous amount of additional propellant.

It becomes clear that Mars is probably beyond our current reach, at least if the astronauts sent there plan to return to Earth without sending out tremendous quantities of supplies and lots of propellant in advance, as well as following up with resupply missions. Meantime, it does not stop humankind from sending probes out there to study Mars and beyond, but sending out human explorers and bringing them back safely to Earth is another matter entirely. Moreover, even if a mission to Mars was technically possible, the monumental costs involved are unlikely to be justified by any single country.

Deep Space & Beyond

Ventures into deep space and beyond our solar system may start to encroach on the final frontier, whatever that really is. For example, it has become clear that with our current technology, it is impossible to send astronauts into deep space, and unthinkable that they could ever be returned safely to Earth 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. Interstellar probes and robotic explorers, such as Voyager 1 and Voyager 2, are still sending back data. However, it has taken several decades for them to get where they are now, which is just beyond our solar system, and they will take another 40,000 years to get to 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 head out there too.

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

Orbital Decay

For satellites in LEO, the aerodynamic drag, no matter how small, is usually sufficient to diminish the satellite’s energy, causing it to slowly spiral toward the surface of the Earth, as shown in the figure below. The drag force acts as a perturbation to the orbital trajectory. Regarding aerodynamic drag, the shape of the satellite is now important (compared to when it was in space) as well as the density of the upper atmosphere. Any asymmetric drag forces may also cause the satellite to tumble.

 

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

Because of the variations in density in the upper atmosphere from seasonal variations and solar activity, it is not possible to predict orbital decay to a high degree of certainty. However, some estimates can still be made with the use of certain assumptions. The drag force, D, on the satellite can be expressed conventionally as

(133)   \begin{equation*} D = q \, C_D \, A_{\rm ref} \end{equation*}

where q is the dynamic pressure. The density of the air can be represented by an ISA model for the extended atmosphere. The reference area, A_{\rm ref}, and the drag coefficient, C_D, in Eq. 133 will depend on the exact shape of the satellite. It is often assumed that C_D = 1 in the very outer regions of the atmosphere in the low density or rarefied flow, increasing to a value of C_D = 2 when approaching an orbital altitude of 150 km. It is found that because of the eccentricity of most orbits, the decay is more rapid at the apogee than the perigee.

Atmospheric drag exerts a significant effects at the LEO altitudes of crewed Earth-orbit spacecraft and satellites, even those with relatively high LEO such as the Hubble Space Telescope. When the orbital decay reaches a critical level, the satellite must be boosted back into its initial orbit. To ensure that this can be done, sufficient propellant must be carried onboard the satellite.  The amount of propellant required depends on the satellite’s mission requirements, its projected life, and the rate of orbital decay. The ISS, for example, requires a regular altitude boost to counteract the effects of orbital decay. About once a month, the ISS fires its thrusters to recover its altitude, which is a maneuver called a reboost. Accurately estimating the amount of propellant required for a satellite in LEO is crucial, as having too little can cause a mission failure, while carrying too much propellant is an unnecessary weight and cost.

Atmospheric Reentry

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. The artistic vision of a flaming spacecraft hurtling toward the Earth is not so far from the physical truth. The high reentry velocities, which can reach Mach 12 and more, create tremendous amounts of friction on the spacecraft, creating heat. So much heat is produced during the reentry that the spacecraft must be designed with thermal protection so that it does not burn up and vaporize.

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

Work by the NACA showed that sharp objects flying at hypersonic speeds get too hot from kinetic heating and would melt. Therefore, the preference for a reentry shape is a bluff or blunt body, which causes the shock wave to stand off from the spacecraft, as shown in the figure below, reducing the kinetic heating and also creating high drag to slow it down. While the process still creates significant heat, blunt body shapes are always preferred. Some shapes, such as the Space Shuttle, can also create lift, allowing the descent to be controlled within certain bounds.

The preference for a reentry shape is a blunt body, which causes the shock wave to stand off from the spacecraft, thereby reducing the kinetic heating and creating high drag.

Most spacecraft designed to reenter the Earth’s atmosphere for a landing will follow a ballistic or semi-ballistic trajectory, as shown in the figure below. This type of reentry has a very steep angle and is of relatively short duration. The steep reentry angle creates the atmospheric drag needed to slow down the spacecraft. In this regard, the spacecraft free-falls through the atmosphere at very high Mach numbers, and it finally slows down as the atmosphere becomes denser. Thermal protection for the spacecraft is provided by a heat shield that also ablates and so sheds some material to reduce the reentry temperatures on the spacecraft. The final stages of a landing require the deployment of parachutes, and the spacecraft lands softly into the sea.

A typical reentry profile for a spacecraft returning to Earth.

Summary & Closure

The ability to successfully engineer rockets and spacecraft has already made it possible for humankind to reach space. They have been used study other planets as well as our own planet. Planetary exploration using probes will continue in the decades to come, perhaps at an accelerated rate. Humans will likely return to the Moon before the end of this decade. However, the technical and other 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 means to explore space, partly because commercial space endeavors have revitalized the public’s interest in space. In addition, new space technologies, including reusable launch vehicles and more efficient rocket engines, will continue to significantly reduce the costs of launching payloads into space.

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 and helps us answer some of the biggest questions about our universe. The development of the International Space Station and partnerships between countries around the world has enabled long-duration human spaceflights, leading to even greater scientific discoveries and technological advancements that benefit not only space exploration but also 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 diametric 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 in the hole? Would you pop out on the other side of the Earth?
  • Is there any gravity at the center of the Earth? Hint: Think about the effects of gravity on one mass that is 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 progressively 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 air, then a Doppler effect comes into play relative to a fixed observer. What might happen to a light source as its nears the speed of light?

Other Useful Online Resources

Visit the following web sites 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.”
  • 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 Archie Roy, who inspired many generations of engineering students about astronomy, space, orbital mechanics, and the future possibilities of space flight.