Rockets & Launch Vehicles

J. Gordon Leishman

59 Rockets & Launch Vehicles

Introduction

Rockets launch payloads, such as satellites and space probes, into Earth orbit. Launch vehicles are highly specialized and tailored to specific missions and payloads. For example, some launch vehicles (see Figure 1) are designed to place payloads, such as satellites, into low Earth orbit, whereas others are intended to send spacecraft into deep space. The choice of launch vehicle will depend on factors such as the desired orbit, payload mass, and size. In addition to military and civilian applications, there is also a growing interest in commercial and tourism-related payloads. Companies such as SpaceX and Blue Origin are developing launch vehicles for both commercial satellite launches and human spaceflight, enabling broader access to space and a wider range of applications.

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

Solid- and liquid-propellant rocket engines are commonly used in launch vehicles and can be combined to achieve specific performance characteristics. Solid-fuel rocket boosters can provide high initial thrust at liftoff. In contrast, liquid-propellant engines can provide more precise thrust control and greater efficiency once the vehicle is in flight. In addition, the number of stages in a launch vehicle can vary with mission requirements. Some launch vehicles have a single stage, whereas others have multiple stages that are separated sequentially during flight. The multi-stage approach enables the payload to achieve higher velocities and altitudes with less propellant than a single-stage rocket design.

Learning Objectives

Appreciate the various types of rockets and launch vehicles, as well as their applications in space missions.
Know how to derive and use the rocket equation to solve simple problems, such as determining burnout velocities.
Understand the staging process and its use for launch vehicles.
Know about missiles and the ballistics of projectiles.

Types of Rockets

There are various types of rockets designed for different purposes and applications. Launch vehicles transport payloads into space. Commercial rockets provide satellite launch services, with some models featuring partial reusability to reduce operational costs. Crewed spacecraft launch vehicles, such as SpaceX’s Falcon 9, can transport humans into orbit. Space probes explore celestial bodies, suborbital rockets facilitate space tourism, and military rockets and missiles serve in defensive and offensive roles. Experimental rockets, such as research and hobbyist rockets, are designed to support scientific experiments and personal projects. The diversity in rocket types reflects the multifaceted nature of space exploration, defense, and scientific research.

Launch Vehicles

A representative launch profile of a rocket designed to send a payload into space is shown in Figure 2. At launch, the thrust from the engines will exceed the rocket’s weight, allowing it to accelerate quickly away from the pad. The rocket’s weight decreases rapidly from the high propellant consumption, allowing it to continue accelerating as it gains altitude. As the rocket exits the lower denser atmosphere above approximately 60,000 ft (18,000 m), it will be flying at supersonic speed. It also begins to pitch into a more horizontal flight path, and the rocket gains translational velocity, allowing the payload to reach its initial equilibrium orbital velocity and altitude.

Representative launch profile for a two-stage rocket. The first stage may be recovered or break up and crash into the sea; first-stage recovery significantly reduces launch costs.

Several minutes into the ascent, staging occurs in which the first stage is jettisoned, and the second-stage rocket engine is ignited. The first stage then reenters the atmosphere and either burns up (depending on the staging altitude) or breaks apart and crashes into the ocean. In some launch systems, the first stage or solid rocket boosters may be recovered and reused. For example, the Space Shuttle solid rocket boosters were recovered by parachute, whereas many other solid rocket boosters are expended.

The upper stage (or stages) then continues to accelerate into space. The rocket engines will shut down when the payload reaches the required initial orbital velocity and altitude, known as a parking orbit. A second burn is then performed after the orbit has stabilized, placing the spacecraft into its final orbit. At this point, the payload, such as a satellite, is deployed. Rendezvousing a spacecraft with the International Space Station (ISS) involves a series of precise maneuvers to match the ISS’s orbit. The spacecraft gradually approaches the ISS, conducting proximity operations to ensure alignment. During the final approach, automated systems and minor thruster adjustments guide the spacecraft to a gentle docking with the ISS.

Spacecraft

Spacecraft are utilized in various applications. Some are designed for specialist missions, such as planetary exploration or Earth observation, while others are more general-purpose and can be used for multiple tasks. Spacecraft typically consist of numerous subsystems and components, including a payload (the primary equipment or instruments for the mission), a propulsion system (for maneuvering and trajectory control), communication systems (for transmitting and receiving data), and power systems (such as solar panels or batteries); see Figure 3.

One essential function of a spacecraft is orbit insertion, which involves placing the spacecraft into a specific orbit around a planet or other celestial body. This goal requires a carefully planned trajectory and precise firings of the rocket engines to achieve the desired orbit. However, spacecraft can be highly complex and require extensive testing and development on Earth to ensure reliability and safety in space. This work involves testing in simulated space environments and rigorous quality control procedures to ensure that all components meet strict performance standards. This approach minimizes the risk of malfunctioning in space, which can be disastrous and result in substantial financial loss.

Missiles

Missiles can be broadly categorized into two main types: ballistic and cruise. Ballistic missiles are used for long-range strikes against threats. They are launched high into the upper atmosphere or the fringes of space, following a parabolic trajectory before reentering the atmosphere and striking their target. Ballistic missiles are easier to intercept than cruise missiles because they follow a predetermined trajectory that is difficult to alter after launch; see Figure 4.

Comparison of the high-arc trajectory followed by a ballistic missile with the low-altitude, maneuverable flight path of a cruise missile.

Cruise missiles are designed to fly at low altitudes and to follow more maneuverable flight paths to evade threats and defenses. They can be launched from various platforms, including aircraft, ships, and ground-based launchers. Cruise missiles can be more challenging to detect and intercept because they fly at lower altitudes than ballistic missiles and are fast and highly maneuverable.

In addition to their propulsion system, targeting, guidance, and warhead systems, missiles require advanced sensors and communication systems to navigate accurately to their targets and avoid obstacles. Consequently, missiles are highly complex weapon systems that require extensive testing and development to ensure their reliability and effectiveness. In addition, they are subject to strict regulations and controls, and their use is governed by international law.

Miscellaneous Types of Rockets

Sounding rockets gather data on atmospheric conditions, such as temperature, pressure, and wind speed, at altitudes that are difficult to reach with aircraft or balloons. They are typically small, single-stage rockets launched into suborbital trajectories, carrying scientific instruments and sensors to collect various types of data.

Jet-Assisted Takeoff (JATO), also known as Rocket-Assisted Takeoff (RATO), is a technique that uses rockets to provide additional thrust during takeoff, particularly when the aircraft is heavily loaded or operating from a short runway. The rocket engines temporarily increase the aircraft’s acceleration, enabling it to take off and climb from short runways. The most famous RATO aircraft is a modified Lockheed Martin C-130 Hercules. This four-engine turboprop military transport airplane has served as a utility aircraft for the Blue Angels. The RATO system features solid rocket engines mounted on the sides of the aircraft’s fuselage.

Rockets are also used to provide emergency lifelines to ships that are in distress. In this application, a rocket-powered line is fired from shore or another ship to the stranded vessel, enabling rescuers to establish a connection and provide assistance. Rockets could also deliver relief materials to inaccessible areas during natural disasters or humanitarian crises. However, this approach would require developing reliable, cost-effective rocket delivery systems, as well as appropriate infrastructure and logistical support.

Rocket Equation

The rocket equation is widely used for sizing rockets and estimating propellant loads. The Russian scientist Konstantin Tsiolkovsky, who published it in 1903 in his work on spaceflight theory, is credited with deriving it. However, Robert Goddard and Hermann Oberth also derived the rocket equation during the 1920s, independently of Tsiolkovsky and of each other.

Derivation

Consider an accelerating rocket of mass $M$ propelled by the engine’s thrust, as shown in Figure 5. If gravity and aerodynamic drag are neglected, then Newton’s second law applied to the rocket gives

(1) $\begin{equation*} M \frac{dV}{dt} = T \end{equation*}$

where $dV/dt$ is the acceleration of the rocket. This equation is strictly valid for a rocket in space or for an idealized flight segment in which gravity and drag losses are neglected. These losses can be included later as corrections to the ideal rocket equation.

The rocket equation embodies the fundamental principle of conservation of momentum.

The thrust produced by the rocket motor is

(2) $\begin{equation*} T = \overbigdot{m} \, V_{\rm eq} \end{equation*}$

where $\overbigdot{m}$ is the mass flow rate of the propellant during the burn, and $V_{\rm eq}$ is the equivalent exhaust velocity from the rocket motor’s nozzle. Using Eq. 2, then

(3) $\begin{equation*} M \left( \frac{dV}{dt} \right) = \overbigdot{m} \, V_{\rm eq} \end{equation*}$

which gives the acceleration of the rocket during the burn as

(4) $\begin{equation*} \frac{dV}{dt} = \left( \frac{\overbigdot{m}}{M} \right) V_{\rm eq} \end{equation*}$

The change in mass of the rocket is $-dM$ (from the use of propellant), so the time rate of decrease of mass is equal to the mass flow rate, i.e.,

(5) $\begin{equation*} -\frac{dM}{dt} = \overbigdot{m} \end{equation*}$

so that

(6) $\begin{equation*} dV = -\left( \frac{dM}{M} \right) V_{\rm eq} \end{equation*}$

Separating the variables and integrating them over the burn time gives

(7) $\begin{equation*} \int_{V_1}^{V_2} dV = -V_{\rm eq} \int_{M_0}^{M_b} \left( \frac{dM}{M} \right) \end{equation*}$

where ${M_0}$ and ${V_1}$ are the initial mass and velocity of the rocket at $t = t_1$ , respectively, and $M_b$ and ${V_2}$ are the final mass and velocity, respectively, at time ${t = t_2}$ .

After integration of the previous equation, the change in the velocity of the rocket is

(8) $\begin{equation*} \Delta V = V_{\rm eq} \ln \left( \frac{M_0}{M_b} \right) \end{equation*}$

This latter equation, known as the rocket equation, is highly useful in mission performance analysis and vehicle sizing. In some ways, it is analogous to the Breguet equations used in aircraft performance analysis. Notice that $M_b$ is given by

(9) $\begin{equation*} M_b = M_0 - M_P \end{equation*}$

where $M_P$ is the mass of propellant used.

The rocket equation is often written in terms of the initial mass, ${M_0}$ , and the burnout mass, ${M_b}$ , i.e., the mass of the rocket after the propellant is fully expended, as

(10) $\begin{equation*} V_b - V_0 = \Delta V = V_{\rm eq} \ln \left( \frac{M_0}{M_b} \right) \end{equation*}$

where $V_0$ is the initial velocity, and the burnout velocity is $V_b$ . For a launch vehicle, its initial velocity on the pad is zero ( $V_0 = 0$ ), so the burnout velocity for the rocket (or the first stage) will be

(11) $\begin{equation*} V_b = V_{\rm eq} \ln \left( \frac{M_0}{M_b} \right) = V_{\rm eq} \ln \left( \frac{M_0}{M_0 - M_P} \right) \end{equation*}$

Effects of Gravity

If gravity losses are included (but no aerodynamic drag losses) in the case of a pure vertical launch, then the rocket equation in Eq. 10 is modified to

(12) $\begin{equation*} \Delta V = V_{\rm eq} \ln \left( \frac{M_0}{M_b} \right) - \Delta V_g \end{equation*}$

The second term, ${\Delta V_g}$ , is referred to as the gravity loss. Notice that for a rocket going up vertically, the gravity loss term reduces the attainable velocity of the rocket at the burnout time, i.e.,

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

where ${g_0}$ is standard gravity, which is equal to 9.81 m/s ${^{2}}$ or 32.17 ft/s ${^{2}}$ . Therefore, the rocket equation with the gravity loss term becomes

(14) $\begin{equation*} \Delta V = V_{\rm eq} \ln \left( \frac{M_0}{M_b} \right) - g_0 \, t_b = V_{\rm eq} \ln \left( \frac{M_0}{M_0 - M_P} \right) - g_0 \, t_b \end{equation*}$

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 the gravity loss term is

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

where $\gamma = 90^{\circ}$ when the rocket flies vertically and $\gamma = 0$ when it flies horizontally. However, properly evaluating the latter term requires more specific information about the launch profile, i.e., $\gamma(t)$ . Neglecting the gravity loss term is generally not justified for launch vehicles, because it can represent a significant fraction of the total required $\Delta V$ , often on the order of 1–2 km/s depending on the ascent profile.

Burnout Height

For a single-stage rocket, the vertical height achieved at the burnout time, $t_b$ , is

(16) $\begin{equation*} H_b = \int_0^{t_b} V \, dt \end{equation*}$

If the propellant mass flow rate is assumed constant, then the mass of the rocket varies with time as

(17) $\begin{equation*} M(t) = M_0 - \overbigdot{m}\, t \end{equation*}$

so that the flight velocity as a function of time is

(18) $\begin{equation*} V(t) = V_{\rm eq} \ln \left( \frac{M_0}{M(t)} \right) - g_0 \, t \end{equation*}$

then the height achieved at $t_b$ is

(19) $\begin{equation*} H_b = \int_0^{t_b} V(t)\, dt \end{equation*}$

After integration, then

(20) $\begin{equation*} H_b = V_{\rm eq} t_b \left( \frac{R-1-\ln R}{R-1} \right) - \frac{1}{2} g_0 t_b^2 \end{equation*}$

where

(21) $\begin{equation*} R=\frac{M_0}{M_b} \end{equation*}$

or equivalently,

(22) $\begin{equation*} H_b = V_{\rm eq} t_b \left( \frac{\dfrac{M_0}{M_b}-1-\ln\!\left(\dfrac{M_0}{M_b}\right)} {\dfrac{M_0}{M_b}-1} \right) - \frac{1}{2} g_0 t_b^2 \end{equation*}$

Rocket Mass Breakdown

The initial mass ${M_0}$ of the rocket vehicle can be written as the sum

(23) $\begin{equation*} M_0 = M_P + M_S + M_L \end{equation*}$

where $M_P$ is the mass of the propellant, $M_S$ is the structural mass of the rocket, and $M_L$ is the mass of the payload. The burnout mass is reached when all of the propellant is exhausted, which is given by

(24) $\begin{equation*} M_b = M_0 - M_P = M_S + M_L \end{equation*}$

As in most engineering fields, it is convenient to use dimensionless quantities. For rockets, the initial mass-to-burnout mass ratio, $R$ , is defined as

(25) $\begin{equation*} R = \frac{M_0}{M_b} \end{equation*}$

Likewise, the payload ratio, $\lambda$ , is defined by

(26) $\begin{equation*} \lambda = \frac{M_L}{M_0 - M_L} = \frac{M_L}{M_P + M_S} \end{equation*}$

Finally, the structural mass coefficient, $\epsilon$ , is defined by

(27) $\begin{equation*} \epsilon = \frac{M_S}{M_P + M_S} = \frac{M_S}{M_0 - M_L} \end{equation*}$

Using these definitions, the mass ratio $R$ can be written as

(28) $\begin{equation*} R = \frac{M_0}{M_b} = \frac{1 + \lambda}{\epsilon + \lambda} \end{equation*}$

Therefore, in terms of the mass and payload ratios and the structural mass coefficient, then

(29) $\begin{equation*} \Delta V = V_{\rm eq} \ln R = V_{\rm eq} \ln \left( \frac{1 + \lambda}{\epsilon + \lambda} \right) \end{equation*}$

Payload mass ratios can vary significantly depending on the mission objectives, destination, and launch vehicle configuration. However, historical data show that structural mass coefficients are relatively consistent across various rockets and launch vehicles. The structural mass coefficient, denoted by $\epsilon$ , represents the ratio of structural mass to the combined structural and propellant mass of the stage. Its approximate constancy among different vehicle designs is convenient for preliminary design studies, where an estimated value of $\epsilon$ can be assumed without requiring a detailed structural analysis. Typical values of $\epsilon$ for various launch vehicles are listed in the table below.

Typical structural mass coefficients for launch vehicles.
Vehicle Type	Typical Structural Mass Coefficient (ε)
Solid rocket stages	0.09–0.12
Liquid rocket stages (expendable)	0.07–0.10
Liquid rocket stages (reusable)	0.10–0.15
Upper stages (lightweight)	0.04–0.08
Heavy-lift first stages	0.08–0.12
Reusable boosters (e.g., Falcon 9 first stage)	0.12–0.18

Structural Mass Breakdown

The structural mass of a rocket comprises all components necessary to support and operate the propulsion and payload systems, excluding the propellant and payload. Using the nomenclature introduced previously, the initial mass is

(30) $\begin{equation*} M_0 = M_P + M_S + M_L \end{equation*}$

where $M_P$ is the propellant mass, $M_S$ is the dry structural and systems mass, and $M_L$ is the payload mass. The corresponding structural mass fraction is

(31) $\begin{equation*} \zeta_S = \frac{M_S}{M_0} \end{equation*}$

and the component mass fractions satisfy

(32) $\begin{equation*} \zeta_P + \zeta_S + \zeta_L = 1 \end{equation*}$

Structural mass includes the propellant tanks, interstage structures and bulkheads, engine mounts and thrust structure, aerodynamic fairings, feed and pressurization systems, avionics and wiring, thermal protection systems, and, for reusable vehicles, recovery and landing hardware. Therefore, the dry structural and systems mass may be written schematically as

(33) $\begin{equation*} M_S = M_t + M_i + M_e + M_f + M_d + M_a + M_{\rm TPS} + M_r + M_o \end{equation*}$

where $M_t$ is the tank mass, $M_i$ is the interstage and bulkhead mass, $M_e$ is the engine-mount and thrust-structure mass, $M_f$ is the fairing mass, $M_d$ is the feed and pressurization-system mass, $M_a$ is the avionics and wiring mass, $M_{\rm TPS}$ is the thermal-protection-system mass, $M_r$ is the recovery and landing-system mass, and $M_o$ represents other structural and systems mass. The recovery term applies only to reusable vehicles.

Dividing by the initial mass gives

(34) $\begin{equation*} \zeta_S = \frac{M_t+M_i+M_e+M_f+M_d+M_a+M_{\rm TPS}+M_r+M_o}{M_0} \end{equation*}$

or, equivalently,

(35) $\begin{equation*} \zeta_S = \zeta_t + \zeta_i + \zeta_e + \zeta_f + \zeta_d + \zeta_a + \zeta_{\rm TPS} + \zeta_r + \zeta_o \end{equation*}$

where each term is the corresponding component mass divided by $M_0$ .

The propellant tanks must withstand internal pressures and dynamic loads during flight, often requiring significant reinforcement, particularly in cryogenic applications where thermal insulation is also needed. Interstage adapters and bulkheads transfer thrust and aerodynamic loads between stages while maintaining structural integrity under changing load conditions.

The thrust structure mounts the engines and transmits their loads to the rocket body, often accounting for a substantial portion of the total structural mass. Aerodynamic fairings protect the payload during ascent and must remain lightweight while surviving aerodynamic and acoustic loading. Feed and pressurization systems, such as helium tanks used to maintain propellant-tank pressure during depletion, add further mass. Avionics and wiring, although individually lightweight, collectively form a non-negligible contribution when distributed throughout the vehicle. Thermal protection systems increase structural mass by incorporating heat-resistant materials. In reusable rockets, additional mass is incurred by descent-control and landing systems, such as grid fins, reaction-control thrusters, and deployable landing legs.

For a given initial mass, an increase in structural mass fraction reduces the fraction available for propellant and payload, i.e.,

(36) $\begin{equation*} \zeta_P + \zeta_L = 1-\zeta_S \end{equation*}$

The structural mass must therefore be minimized to maximize payload capability while still meeting requirements for strength, stiffness, thermal protection, durability, and reusability. For comparisons between stages, the structural mass coefficient is

(37) $\begin{equation*} \epsilon = \frac{M_S}{M_S+M_P} \end{equation*}$

which measures the structural fraction of the stage mass excluding payload. The relative contribution of each component to $M_S$ depends strongly on the stage type and its mission. Representative conceptual-design values are summarized in the following table. These values are illustrative rather than universal and should not replace a detailed structural mass analysis.

Representative breakdown of launch-vehicle structural mass.
Structural Component	Expendable Liquid Stage (% of $M_S$ )	Reusable Liquid Booster (% of $M_S$ )
Propellant tanks, $M_t$	42	32
Interstage structures and bulkheads, $M_i$	8	6
Engine mounts and thrust structure, $M_e$	28	24
Aerodynamic fairings, $M_f$	5	0
Feed and pressurization systems, $M_d$	7	6
Avionics and wiring, $M_a$	4	4
Thermal protection system, $M_{\rm TPS}$	1	8
Recovery and landing systems, $M_r$	0	15
Other structure and systems	5	5

The table shows why reusable boosters generally have higher structural mass coefficients than expendable stages. Mass that would otherwise be available for propellant or payload must be allocated to thermal protection, grid fins or other descent-control devices, landing gear, additional avionics, and the structural reinforcement required for repeated flight.

Impulse

The total impulse produced by a rocket engine is defined as the integral of thrust over the burn time, $t_b$ , i.e.,

(38) $\begin{equation*} I = \int_0^{t_b} T \, dt \end{equation*}$

Therefore, total impulse is the area under the thrust-versus-time curve.

For an ideal rocket engine, the thrust may be written as

(39) $\begin{equation*} T = \overbigdot{m} \, V_{\rm eq} \end{equation*}$

where $\overbigdot{m}$ is the positive propellant mass flow rate leaving the rocket and $V_{\rm eq}$ is the equivalent exhaust velocity. The total impulse is therefore

(40) $\begin{equation*} I = \int_0^{t_b} \overbigdot{m} \, V_{\rm eq} \, dt \end{equation*}$

The total impulse is the integral of thrust over the burn time and is represented by the area under the thrust-versus-time curve.

If $V_{\rm eq}$ is constant, then

(41) $\begin{equation*} I = V_{\rm eq} \int_0^{t_b} \overbigdot{m} \, dt \end{equation*}$

The total propellant mass expelled during the burn is

(42) $\begin{equation*} M_P = \int_0^{t_b} \overbigdot{m} \, dt \end{equation*}$

so

(43) $\begin{equation*} I = M_P \, V_{\rm eq} \end{equation*}$

If thrust is also constant, then

(44) $\begin{equation*} I = T \, t_b = M_P \, V_{\rm eq} \end{equation*}$

This result represents the total thrust impulse generated through the exhaust momentum flux. However, a rocket is a variable-mass system because propellant continuously leaves the vehicle. Consequently, the thrust impulse cannot generally be equated directly to the difference between the initial and final momentum of the remaining rocket, i.e.,

(45) $\begin{equation*} I \ne M_b \, V_b - M_0 \, V_0 \end{equation*}$

To see why, consider the momentum of the remaining rocket

(46) $\begin{equation*} p = M \, V \end{equation*}$

Its differential is

(47) $\begin{equation*} d(MV) = M \, dV + V \, dM \end{equation*}$

Therefore,

(48) $\begin{equation*} \frac{d(MV)}{dt} = M \, \frac{dV}{dt} + V \, \frac{dM}{dt} \end{equation*}$

The second term is nonzero because the rocket mass decreases continuously during the burn. The momentum carried away by the expelled propellant must therefore be included in the momentum balance.

For a small amount of propellant mass $-dM$ expelled backward with equivalent velocity $V_{\rm eq}$ relative to the rocket, conservation of momentum gives

(49) $\begin{equation*} M \, dV = - V_{\rm eq} \, dM \end{equation*}$

or

(50) $\begin{equation*} dV = - V_{\rm eq} \frac{dM}{M} \end{equation*}$

Because $dM<0$ during propellant consumption, the resulting velocity increment $dV$ is positive.

Integrating from the initial rocket mass $M_0$ and velocity $V_0$ to the burnout mass $M_b$ and velocity $V_b$ gives

(51) $\begin{equation*} \int_{V_0}^{V_b} dV = - V_{\rm eq} \int_{M_0}^{M_b} \frac{dM}{M} \end{equation*}$

Therefore,

(52) $\begin{equation*} V_b - V_0 = V_{\rm eq} \ln \left( \frac{M_0}{M_b} \right) \end{equation*}$

or

(53) $\begin{equation*} \Delta V = V_{\rm eq} \ln \left( \frac{M_0}{M_b} \right) \end{equation*}$

The distinction between impulse and velocity increment is important. The engine impulse depends on the total expelled propellant mass and equivalent exhaust velocity

(54) $\begin{equation*} I = M_P \, V_{\rm eq} \end{equation*}$

whereas the rocket velocity increment depends logarithmically on the ratio of initial mass to burnout mass

(55) $\begin{equation*} \Delta V = V_{\rm eq} \ln \left( \frac{M_0}{M_b} \right) \end{equation*}$

The logarithmic dependence occurs because each successive increment of thrust accelerates a rocket whose mass is continuously decreasing.

Specific Impulse

The measure of efficiency used in most rocket performance calculations is the specific impulse, which is the total impulse produced divided by the propellant used in terms of its weight. Specific impulse is a property of the propellant’s chemical composition and can be evaluated through a thermodynamic analysis. In general, one seeks to carry as little propellant as possible; therefore, the objective is to use a rocket with a high specific impulse.

The specific impulse, $I_{\rm sp}$ , is defined as

(56) $\begin{equation*} I_{\rm sp} = \frac{\mbox{\small Total~impulse}}{\mbox{\small Weight~of~propellant~burned}} = \frac{I}{M_P \, g_0} \end{equation*}$

Therefore, the higher the value of $I_{\rm sp}$ , the more efficient the rocket motor will be in producing thrust. It will be noted that the specific impulse for the rocket motor is analogous to the specific fuel consumption for an air-breathing engine. It is further apparent using Eq. 2 that if the thrust is approximately constant, then

(57) $\begin{equation*} I_{\rm sp} = \frac{V_{\rm eq}}{g_0} = \frac{T}{\overbigdot{m} \, g_0} = \frac{T}{\overbigdot{W}_P} \end{equation*}$

where it will be noticed that $I_{\rm sp}$ has units of time (seconds).

Therefore, the specific impulse is the total impulse, or change in momentum delivered, per unit weight of propellant consumed, which is a measure of the efficiency in producing thrust. The equivalent exhaust velocity, $V_{\rm eq}$ , depends to a large extent on the chemistry and efficiency of the combustion of propellant. High combustion temperatures, achieved with low-molecular-weight propellants, are essential for maximizing the specific impulse. To this end, the most widely used propellants are liquid hydrogen (LH₂) and liquid oxygen (LO₂ or LOX), as summarized in the table below.

Specific impulse values by propellant type and rocket type.
Fuel / Propellant Combination	Rocket Type	Specific Impulse (s)	Conditions
Ammonium perchlorate composite propellant (APCP)	Solid rocket motor	240–290	Sea level
RP-1 (kerosene) / Liquid oxygen (LOX)	Liquid bipropellant	280–310	Sea level
Liquid hydrogen (LH₂) / Liquid oxygen (LOX)	Liquid bipropellant	370–465	Sea level / Vacuum
UDMH / Nitrogen tetroxide (N₂O₄)	Hypergolic bipropellant	290–320	Vacuum
Hydrazine (monopropellant)	Monopropellant thruster	220–240	Vacuum
Xenon (electric ion propulsion)	Ion thruster	2,000–5,000	Vacuum
Liquid methane (CH₄) / Liquid oxygen (LOX)	Liquid bipropellant	360–380	Vacuum
Hydrogen / Nuclear thermal propulsion	Nuclear rocket engine	800–900	Vacuum

Check Your Understanding #1 – Burnout velocity of a single-stage rocket

Use the rocket equation to determine the burnout velocity and the maximum achievable height of a simple rocket launched vertically. Neglect the aerodynamic drag forces. Solve for the burnout velocity and maximum altitude given a burnout time of 60 seconds. The specific impulse is 250 seconds, the initial mass is 12,700 kg, and the propellant mass is 8,610 kg. Assume the launch as vertical, ignoring curvature and drag, and solve for the required velocity, assuming it will be redirected to tangential velocity near apogee (e.g., by a circularization burn).

Show solution/hide solution.

The rocket equation gives the change in the velocity of the vehicle ${\Delta V}$ , i.e.,

$\Delta V = V_{\rm eq} \ln \left( \frac{M_0}{M_b} \right)$

where ${M_0}$ is the initial mass of the vehicle and $M_b$ is the final or burnout mass. If gravity is included (but no aerodynamic drag), then

$\Delta V = V_{\rm eq} \ln \left( \frac{M_0}{M_b} \right) - g_0 \, t_b$

where $t_b$ is the burnout time, which is 60 seconds in this case.

The equivalent velocity $V_{\rm eq}$ is given in terms of the specific impulse, i.e.,

${ V_{\rm eq} } = I_{\rm sp} \, g_0 = 250 \times 9.81 = 2,452.5~\mbox{m/s}$

and the burnout mass $M_b$ is given by

$M_b = M_0 - M_P = 12,700 - 8,610 = 4,090~\mbox{kg}$

The burnout velocity is given by the rocket equation, i.e.,

$\Delta V = V_{\rm eq} \ln \left( \frac{M_0}{M_b} \right) - g_0 \, t_b$

Therefore, ${\Delta V}$ at the burnout is

$\Delta V = 2,778.82 - 9.81 \times 60 = 2,778.82 - 588.6 = 2,190.22~\mbox{m/s}$

which is 2.19 km/s. Assuming the rate of propellant consumption is constant, then the mass of the rocket $M$ varies over time as

$M = M_0 - M_P \left( \frac{t}{t_b} \right) = M_0 - \left( M_0 - M_b \right) \left( \frac{t}{t_b} \right)$

The velocity of the rocket is

$V = V_{\rm eq} \ln \left( \frac{M_0}{M} \right) - g_0 \, t$

The height achieved at the burnout time, $t_b$ , is

${ H_b } = V_{\rm eq} t_b \left[ \frac{ \displaystyle{\frac{M_0}{M_b}} - 1 - \ln \left( \displaystyle{\frac{M_0}{M_b}} \right) }{ \displaystyle{\frac{M_0}{M_b}} - 1 } \right] - \frac{1}{2} g_0 \, t_b^2$

Inserting the values gives

${ H_b } = 2,452.5 \times 60 \left[ \frac{ \displaystyle{\frac{12,700}{4,090}} - 1 - \ln \left( \displaystyle{\frac{12,700}{4,090}} \right) }{ \displaystyle{\frac{12,700}{4,090}} - 1 } \right] - \frac{1}{2} \times 9.81 \times 60^2$

Therefore, the height achieved is

$H_b = 2,452.5 \times 60 \left( \frac{3.105-1-1.133}{3.105-1} \right) - \frac{1}{2} \times 9.81 \times 60^2$

so that $H_b \approx$ 50.3 km. After burnout, the rocket coasts upward until its vertical velocity becomes zero. Neglecting drag and assuming constant gravitational acceleration, the additional coasting height is

$\Delta H_c = \frac{V_b^2}{2 g_0} = \frac{(2,190.22)^2}{2 \times 9.81} = 244.5~\mbox{km}$

Therefore, the maximum altitude is

$H_{\max} = H_b + \Delta H_c = 50.3 + 244.5 = 294.8~\mbox{km}$

Launching into Orbit

Rocket launches are hugely exciting events to watch! The synchronized launch process, with its 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 they enable, and the thrill of human endeavor all contribute to the excitement. Additionally, watching a rocket launch live or on YouTube can be an educational and inspiring experience, especially for students and younger people interested in science and space. Even those who live in Florida never grow tired of watching rocket launches!

Using the Rocket Equation

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 given as

(58) $\begin{equation*} \Delta V = V_{\rm eq} \ln \left( \frac{M_0}{M_b} \right) \end{equation*}$

where $V_{\rm eq}$ is the equivalent exit velocity from the particular rocket engine, ${M_0}$ is the initial mass of the rocket or spacecraft at the beginning of the burn, and $M_b$ is the final (burnout) mass after the burn at burnout time, $t_b$ , during which a propellant mass $M_P$ is consumed. For simplicity, it is assumed that the vehicle is a single-stage vehicle (i.e., no staging).

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

(59) $\begin{equation*} V_{\rm eq} = I_{\rm sp} \, g_0 \end{equation*}$

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 $^2$ . Therefore, the rocket equation can also be written as

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

where the value of $I_{\rm sp}$ depends on the type of rocket engine and its propellant. If the value of $I_{\rm sp}$ is known, then the rocket equation can be used to determine the propellant 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 Figure 7.

Illustrating of rocket reaching orbital height around Earth. — Orbital mechanics can be used to estimate the propellant mass needed to lift a payload into orbit.

For a launch, several factors can influence the ${\Delta V}$ required, i.e.,

The needed orbital altitude above the surface of the Earth, $h_s$ .
The orbital inclination relative to the Earth’s equatorial plane.
The launch latitude from the Earth (this affects the initial energy).
The effects of overcoming gravity (when the rocket goes vertically).
The aerodynamic drag on the rocket in the lower atmosphere.
Steering losses because of small misalignments of the thrust vector.

Kinetic Energy

For the satellite or spacecraft to reach the required orbital altitude, $h_s$ , the required orbital velocity is

(61) $\begin{equation*} V_{\rm orb} = \sqrt{ \frac{G \, M_E}{R_E + h_s} } \end{equation*}$

noting that this result does not depend on the mass of the satellite or spacecraft. This velocity represents the dominant component of the total $\Delta V$ required to reach orbit.

Potential Energy

The additional velocity increment associated with overcoming the gravitational potential to reach the same orbital altitude is

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

This result follows directly from conservation of mechanical energy. The ideal launch velocity required to reach the orbital altitude is

(63) $\begin{equation*} V_{\rm ideal} = \sqrt{ G M_E \left( \frac{2}{R_E} - \frac{1}{R_E+h_s} \right) } \end{equation*}$

The additional velocity increment associated with increasing the vehicle’s gravitational potential energy is therefore

(64) $\begin{equation*} \Delta V_U = V_{\rm ideal} - V_{\rm orb} \end{equation*}$

so that

(65) $\begin{equation*} V_{\rm ideal} = V_{\rm orb} + \Delta V_U \end{equation*}$

Therefore, the kinetic- and potential-energy requirements are first combined through conservation of energy, and the resulting ideal velocity is then written as the sum of two scalar contributions, $V_{\rm orb}$ and $\Delta V_U$ . These terms are not vector components of the spacecraft velocity. For low orbital altitudes with $h_s \ll R_E$ , then

(66) $\begin{equation*} \Delta V_U \approx \frac{h_s}{R_E} \sqrt{ \frac{G \, M_E}{R_E} } \end{equation*}$

which is generally small compared to the orbital velocity requirement.

Gravity Loss Effect

A gravitational effect must be added to the total needed ${\Delta V}$ , usually called gravity loss. For a rocket going up vertically, then

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

where $t_b$ is the burnout time. If the rocket follows a curved trajectory with local flight-path angle $\gamma(t)$ (measured with respect to the horizon), then the gravity-loss contribution to the required velocity increment may be written as

(68) $\begin{equation*} \Delta V_g = \int_{0}^{t_b} g_0 \sin\gamma(t)\, dt \end{equation*}$

where $\gamma=90^{\circ}$ corresponds to vertical flight and $\gamma=0^{\circ}$ to horizontal flight. This expression shows that gravity loss depends not only on the burnout time $t_b$ but also on the launch profile through $\gamma(t)$ . In general, reducing $\Delta V_g$ requires minimizing the duration over which the thrust must oppose the component of weight along the flight path, which can be achieved by shortening $t_b$ and/or by pitching over such that $\gamma$ decreases earlier, subject to aerodynamic and structural constraints.

Aerodynamic Loss Effect

There is aerodynamic drag on the rocket as it flies through the lower atmosphere, which contributes to the propellant required to reach a given ${\Delta V}$ . This drag force, $D$ , can be expressed as

(69) $\begin{equation*} D = \frac{1}{2} \varrho_{\infty} V_{\infty}^2 \, C_D \, A_{\rm ref} = q_{\infty} \, C_D \, A_{\rm ref} \end{equation*}$

where ${q_{\infty}}$ is the dynamic pressure, defined as $q_{\infty} = \frac{1}{2} \varrho_{\infty} V_{\infty}^2$ , where $\varrho_{\infty}$ is the local air density and ${V_{\infty}}$ is the true airspeed, both of which depend on the launch profile. The reference area, $A_{\rm ref}$ , in Eq. 69 is usually taken as the projected frontal area of the rocket. The drag coefficient, $C_D$ , will also depend on the overall shape of the rocket and its flight Mach number.

The rocket’s drag given by Eq. 69 is proportional to air density and the square of the airspeed. An ISA model, applicable to the troposphere, stratosphere, and higher altitudes, can be used to represent air density.
This result indicates that minimizing aerodynamic losses requires reducing the time spent in the dense lower atmosphere, thereby favoring rapid initial acceleration. However, this must be balanced against gravity losses, leading to an optimal ascent profile rather than a purely vertical or slow trajectory.

Nevertheless, the aerodynamic loss on the achievable ${\Delta V}$ is relatively small compared to gravity losses for most launch vehicles. More importantly, the dynamic pressure during launch requires particular emphasis, as it affects the aerodynamic-induced structural loads on the rocket. The value of ${q_{\infty}}$ increases rapidly after launch, reaching a maximum and then decreasing as the altitude increases and the density of the air diminishes. In many cases, the “maximum $q$ ” value on the rocket will be limited, requiring the rocket engines to be temporarily throttled down to prevent excessive aerodynamic loads on the vehicle.

Aerodynamic drag coefficients for launch vehicles are not generally available. However, some research has been published, including Ballistic Research Laboratories, Memorandum Report No. 545, and NASA TM X-53770. The latter report presents the drag coefficients of the NASA Saturn V launch vehicle, as measured in several wind tunnels over a wide range of Mach numbers, with a sample shown in Figure 8. The reference length used in the results is the diameter of the primary launch vehicle, where $D_{\rm ref}$ = 10.06 meters or 33 feet. The total drag (axial) force is split into forebody drag and base drag. Notice the rapid increase in drag coefficient during the transonic region near Mach = 1, followed by a progressive reduction in drag coefficient in the supersonic regime.

NASA measurements of the aerodynamic drag coefficient on the Saturn V launch vehicle. Notice that the drag is decomposed into forebody and base drag components.

In the absence of detailed aerodynamic data, an approximate empirical representation for the drag coefficient of a slender launch vehicle may be used. The drag coefficient, $C_D$ , for a rocket is often represented as

(70) $\begin{equation*} C_D = \left\{ \begin{array}{ll} 0.2 & \mbox{for $0 < M_{\infty} \leq 0.85$} \\[8pt] 0.11 + \displaystyle{\frac{0.82}{M_{\infty}^2}} - \displaystyle{\frac{0.55}{M_{\infty}^4}} & \mbox{for $M_{\infty} > 0.85$} \end{array} \right. \end{equation*}$

The velocity increment required to overcome aerodynamic drag is given by

(71) $\begin{equation*} \Delta V_d = \int_0^{t_a} \frac{D}{M(t)} \, dt \end{equation*}$

where $M(t)$ is the instantaneous mass of the rocket and $t_a$ is the time over which aerodynamic forces are significant. For most launch vehicles, the cumulative velocity loss from aerodynamic drag is relatively small, typically on the order of 0.05 to 0.15 km/s. However, for smaller rockets or high-speed missiles operating in denser atmospheric regions, aerodynamic losses can become more significant and must be accounted for in performance estimates.

Launch Latitude

The required ${\Delta V}$ also depends on the launch latitude. The rotation of the Earth provides an initial velocity to the rocket, which is greatest at launch sites closer to the equator. The tangential velocity due to the Earth’s rotation is

(72) $\begin{equation*} V_{\rm rot} = \Omega_E \, R_E \cos \phi \end{equation*}$

where $\Omega_E$ is the angular velocity of the Earth, $R_E$ is the Earth’s radius, and $\phi$ is the launch latitude. The required ${\Delta V}$ is reduced if the rocket is launched in the direction of the Earth’s rotation (toward the east). For launches from Cape Canaveral, for example, the effective $\Delta V_{\rm lat}$ contribution is about 0.3 km/s, which represents a reduction in the propulsion $\Delta V$ requirement and is not insignificant in terms of propellant savings.

Total Delta-V

Therefore, the total needed $\Delta V_T$ may be written as

(73) $\begin{equation*} \Delta V_T = V_{\rm ideal} + \Delta V_g + \Delta V_d - \Delta V_{\rm lat} \end{equation*}$

where $V_{\rm ideal}$ is the ideal velocity requirement obtained from conservation of mechanical energy. The relation

(74) $\begin{equation*} V_{\rm ideal} = V_{\rm orb} + \Delta V_U \end{equation*}$

is an algebraic decomposition used to identify the additional velocity increment associated with the increase in gravitational potential energy; $V_{\rm orb}$ and $\Delta V_U$ are not separate vector components of the spacecraft velocity.

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

(75) $\begin{equation*} \Delta V_T = I_{\rm sp} \, g_0 \ln \left( \frac{M_0}{M_b} \right) = I_{\rm sp} \, g_0 \ln \left( \frac{M_0}{M_0 - M_P} \right) \end{equation*}$

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

(76) $\begin{equation*} M_P = M_0 \left[ 1 - \exp \left( -\frac{\Delta V_T}{I_{\rm sp} \, g_0} \right) \right] \end{equation*}$

If staging is used, the ${\Delta V}$ for each stage is calculated as the stage is depleted of propellant, and the empty stage is discarded.

Estimating the Escape Velocity from Earth

The Earth’s mass, $M_E$ , is 5.97 $\times 10^{24}$ kg. The radius of an 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, we know that the universal gravitational constant $G$ = 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 it can be shown that the minimum escape velocity is given by

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

which is the minimum speed required for a spacecraft to escape the Earth’s gravitational field in the ideal two-body problem.

Steering a Rocket

A rocket must be steered along a prescribed flight path to achieve the desired altitude and velocity. In modern launch vehicles, steering is accomplished primarily by gimbaling the main engine(s), which allows the thrust vector to be rotated slightly relative to the vehicle’s centerline, as shown in Figure 9. Tilting the thrust direction produces a torque that changes the rocket’s orientation and, hence, the direction of its flight path.

Rocket steering methods diagram — Several methods can be used to steer a rocket, the most common being the gimbaled thrust design, in which the nozzle is swiveled to redirect the thrust vector.

Historically, other techniques were also used. Early rockets and most air-to-air missiles employed movable aerodynamic fins to generate side forces and moments for steering. Some designs used small vernier engines for fine attitude control, or even thrust vanes inserted into the exhaust stream to deflect the jet. These systems have been largely replaced by thrust vector control (TVC) for its efficiency and mechanical simplicity.

Stability & Control

A rocket must also remain stably oriented so that the thrust direction follows the commanded flight path. At low altitudes, aerodynamic forces can help restore the vehicle’s alignment depending on where the center of pressure lies with respect to the center of gravity, i.e., for aerodynamic stability, then $x_{\rm c.p.} > x_{\rm c.g.}$ , which produces an aerodynamic restoring moment that tends to oppose disturbances, as shown in Figure 10. However, as the propellant burns and the rocket ascends, the center of gravity and the center of pressure shift, thereby affecting aerodynamic stability.

Directional control and stability of a rocket can be achieved aerodynamically and/or with thrust vectoring.

At subsonic velocities, the center of pressure is usually near the mid-body, about 50–60% of the length from the nose, but as the vehicle passes through transonic and into supersonic flight, the center of pressure shifts aft to about 80–90% of the body length, remaining near the fin plane until aerodynamic forces vanish in near-vacuum conditions. However, positive control can always be achieved primarily through thrust vectoring. Tilting the engine nozzle by a small angle $\delta$ generates a control moment

(77) $\begin{equation*} M_T = T\,\ell_{\text{tvc}}\,\delta \end{equation*}$

where $T$ is thrust and $\ell_{\text{tvc}}$ is the lever arm between the nozzle pivot point and the center of gravity. Even deflections of one or two degrees are sufficient because the thrust is so large.

At high altitude or in a vacuum, reaction control jets provide the necessary forces and torques for attitude control. Because aerodynamic damping is typically weak, every launch vehicle employs an automatic attitude control system. Gyros measure angular rates, and the controller commands tiny gimbal or thruster deflections to maintain pointing stability many times per second. The control system effectively supplies the stiffness and damping that the vehicle itself lacks.

Steering Losses

When a rocket changes direction, the thrust vector cannot remain perfectly aligned with the velocity vector. The component of thrust used to bend the trajectory does not contribute to increasing forward speed, leading to a small loss in total velocity capability known as the steering loss $\Delta V_{\rm steer}$ . The overall velocity increment provided by the propulsion system can be written as

(78) $\begin{equation*} \Delta V_T = V_{\rm ideal} + \Delta V_g + \Delta V_d + \Delta V_{\rm steer} - \Delta V_{\rm lat} \end{equation*}$

where $V_{\rm ideal}$ is the ideal velocity requirement obtained from conservation of mechanical energy, and the remaining terms account for gravity, drag, steering, and launch-latitude effects. Steering losses are usually much smaller than gravity or drag losses, but can become noticeable when large trajectory turns are required.

Staged Rocket Vehicles

Staging a launch vehicle aims to maximize the payload-to-mass ratio achievable in space. The goal is to launch the largest payload to the required burnout velocity with the least non-payload mass (defined as the rocket’s structural weight plus propellant). As shown in Figure 11, there are two types of staging:

Serial staging, where the stages are ignited, used, and jettisoned in serial sequence.
Parallel staging, where all stages are ignited and used, but the stages are jettisoned as they burn out, e.g., solid rocket boosters.

Staging is essential for maximizing a payload’s orbital velocity and height.

To develop an effective staged launch vehicle, several general considerations apply:

Lower stages usually require a high thrust-to-weight ratio and a robust structure because they operate near sea level and must lift the entire vehicle stack.
Upper stages often benefit from higher values of $I_{\rm sp}$ because they operate in near-vacuum conditions and accelerate a much smaller remaining mass.
Each successive stage is usually smaller than the previous stage because the lower stages have already discarded propellant and structural mass.
The optimum $\Delta V$ split between stages is not fixed by a simple rule; it depends on $I_{\rm sp}$ , structural mass fraction, thrust-to-weight ratio, gravity and drag losses, and the mission requirement.

Serial Staged Rocket

For a serial staged launch vehicle, then

(79) $\begin{equation*} R_i = \frac{M_{0_{i}}}{M_{b_{i}}} ; \quad \lambda_i = \frac{M_{L_{i}}}{M_{0_{i}} - M_{L_{i}}} ; \quad \epsilon_i = \frac{M_{S_{i}}}{M_{P_{i}} + M_{S_{i}}} \end{equation*}$

where the index $i$ refers to the stage number. Also

(80) $\begin{equation*} R_i = \frac{1 + \lambda_i}{\epsilon_i + \lambda_i} \end{equation*}$

For a staged vehicle overall, the ${\Delta V}$ values are added for each stage, i.e.,

(81) $\begin{equation*} \Delta V = \sum_{i=1}^{N} \Delta V_i = \sum_{i=1}^{N} V_{\rm eq, i} \ln R_i \end{equation*}$

It can be assumed that this velocity is redirected into a tangential component to achieve an orbital trajectory. The structural mass coefficients for a staged launch vehicle are usually similar to those of a single stage. However, payload ratios are generally higher for a staged vehicle. In some cases, determining the maximum allowable structural mass to meet specific payload requirements may be desirable, along with establishing certain structural design goals and constraints.

Sequence of staging for a three-stage launch vehicle.

With reference to Figure 12, this is a step-by-step breakdown of the general procedure for calculating the total burnout velocity or time for a multi-stage rocket with serial staging:

Divide the rocket as a system into its stages. A specific propulsion system or engine typically characterizes each stage and will have its own parameters, such as mass, specific impulse, thrust, and propellant weight or burn time.
For each stage, calculate the initial mass, which is the total mass of the rocket at the beginning of the stage burn, and the final mass, which is the rocket’s mass at the end of the stage burn after the propellant has been burned.
Calculate the burnout velocity for each stage using the rocket equation. The gravitational effects on the rocket must also be accounted for.
Add each stage’s burnout velocity to the previous stage’s initial velocity. Assuming that each stage occurs immediately after the previous one, the burnout velocity of one stage becomes the initial velocity for the next stage.
Repeat steps 2–4 for the final stage of the rocket system until the burnout time and/or burnout velocity have been calculated for the final stage carrying the payload mass.

Check Your Understanding #2 – Two-stage rocket calculation

Consider a two-stage rocket with the following design characteristics. Payload mass = 60 kg. First stage: propellant mass = 7,200 kg, structural mass = 800 kg, and the mass flow rate is $\overbigdot{m}$ = 80.0 kg s $^{-1}$ . Second stage: propellant mass = 5,400 kg, structural mass = 600 kg, and the burn time is 100 s. The specific impulse, $I_{\rm sp}$ , for the first and second stages is 275 s. Assume the launch is vertical, ignoring curvature and drag, and solve for the required velocity, assuming it will be redirected to tangential velocity near apogee (e.g., by a circularization burn). Calculate the following:

For the first stage:

The equivalent exhaust velocity.
The thrust produced.
The total burn time.
The burnout velocity.

For the second stage:

The equivalent exhaust velocity.
The mass flow rate.
The thrust produced.
The final burnout velocity.

Show solution/hide solution.

For the first stage:

1. The equivalent exhaust velocity, $V_{\rm eq, 1}$ , is

$V_{\rm eq, 1} = I_{\rm sp_{1}} \, g_0 = 275 \times 9.81 = 2,697.75~\mbox{m/s}$

2. The thrust, $T_1$ , produced is

${ T_1 } = \overbigdot{m}_1 \, V_{\rm eq, 1} = 80.0\times 2,697.75 = 215,820~\mbox{N}$

3. The total burn time, $t_{b_{1}}$ , is

$t_{b_{1}} = \frac{M_{P_{1}}}{\overbigdot{m}_1}= \frac{7,200}{80.0} = 90.0~\mbox{s}$

4. The initial mass, $M_{0_{1}}$ , is

$M_{0_{1}} = M_{P_{1}} +M_{P_{2}} + M_{S_{1}} + M_{S_{2}} + M_{L}$

and inserting the values gives

$M_{0_{1}} = 7,200 +5,400 + 800 +600 +60 = 14,060~\mbox{kg}$

The burnout mass, $M_{b_{1}}$ , is

$M_{b_{1}} = M_{0_{1}} - M_{P_{1}} = M_{P_{2}} + M_{S_{1}} + M_{S_{2}} + M_{L}$

and with the given values leads to the burnout mass of the first stage as

$M_{b_{1}} = 5,400 + 800 + 600 + 60 = 6,860~\mbox{kg}$

The ${\Delta V}$ increment for the first stage is

$\Delta V_1 = V_{b_{1}} - 0 = V_{\rm eq, 1} \ln \left( \frac{M_{0_{1}}}{M_{b_{1}}} \right) - g_0 \, t_{b_{1}}$

and inserting the values gives

$V_{b_{1}} = 2,697.75 \ln \left( \frac{14,060}{6,860} \right) - 9.81\times 90 = 1,053.0~\mbox{m/s}$

Therefore, for the first stage, the burnout velocity is

$V_{b_{1}} = 1,053.0~\mbox{m/s}$

For the second stage:

1. Because the $I_{\rm sp}$ remains the same for stage 2, the equivalent exhaust velocity will also be the same, i.e.,

$V_{\rm eq, 2} = I_{\rm sp_{2}} \, g_0 = 275 \times 9.81 = 2,697.75~\mbox{m/s}$

2. The mass flow rate, $\overbigdot{m}_2$ , is

$\overbigdot{m}_2 = \frac{M_{P_{2}}}{t_{b_{2}}} = \frac{5,400}{100} = 54.0~\mbox{kg/s}$

3. The thrust $T_2$ produced is

$T_2 = \overbigdot{m}_2 \, V_{\rm eq, 2} = 54\times 2,697.75 = 145,678~\mbox{N}$

4. For the second stage, the initial mass is

$M_{0_{2}} = M_{P_{2}} + M_{S_{2}} + M_{L} = 5,400 + 600 + 60 = 6,060~\mbox{kg}$

The burnout mass for the second stage is

$M_{b_{2}} = M_{0_{2}} - M_{P_{2}} = M_{S_{2}} + M_{L} = 600 + 60 = 660~\mbox{kg}$

The ${\Delta V}$ for the second stage is

$\Delta V_2 = V_{\rm eq, 2} \ln \left( \frac{M_{0_{2}}}{M_{b_{2}}} \right) - g_0 \, t_{b_{2}}$

and inserting the values gives

$\Delta V_2 = 2,697.75 \ln \left( \frac{6,060}{660} \right) - 9.81 \times 100 = 5,000.5~\mbox{m/s}$

The final value of the burnout velocity, $V_f$ , will be

$V_f = \Delta V_1 + \Delta V_2 = 1,053.0 + 5,000.5 = 6,053.5~\mbox{m/s} = 6.053~\mbox{km/s}$

Launch Height – Two-Stage Serial Rocket

The burnout height, $H_b$ , of a two-stage serial launcher is determined similarly to that of a single-stage rocket. In this case, the launch heights obtained by each respective stage are added, i.e.,

(82) $\begin{equation*} H_b = \int_0^{t_{b_{1}}} V(t) \, dt + \int_{t_{b_{1}}}^{t_{b_{1}}+t_{b_{2}}} V(t) \, dt = H_{b_{1}} + H_{b_{2}} \end{equation*}$

For the first stage, the height achieved at $t_{b_{1}}$ will be

(83) $\begin{equation*} \hspace*{-5mm} H_{b_{1}} = \int_0^{t_{b_{1}}} V \, dt = V_{\rm eq, 1} \, t_{b_{1}} \left( \frac{R_1 - 1 - \ln R_1}{R_1 - 1} \right) - \frac{1}{2} g_0 \, t_{b_{1}}^2 \end{equation*}$

and for the second stage, the additional height gained during the second-stage burn is

(84) $\begin{equation*} \hspace*{-5mm} H_{b_{2}} = \int_0^{t_{b_{2}}} V_2(t) \, dt = V_{b_{1}} \, t_{b_{2}} + V_{\rm eq, 2} \, t_{b_{2}} \left( \frac{R_2 - 1 - \ln R_2}{R_2 - 1} \right) - \frac{1}{2} g_0 \, t_{b_{2}}^2 \end{equation*}$

Parallel Staged Launcher

In a parallel-staged launch vehicle, there are usually dissimilar rockets and rocket engines burning simultaneously, as shown in Figure 13. An example would be the NASA Space Shuttle, which used LH2/LOX for the “core” main engines on the Orbiter, with solid propellant rocket boosters being used to significantly augment the initial launch ${\Delta V}$ velocity. NASA’s SLS uses the same type of core and booster. Other launch vehicles may be configured with different numbers of solid rocket boosters, depending on the payload mass and desired orbital altitude. An exception is the SpaceX Falcon Heavy, which uses two additional liquid propellant boosters identical to the first (core) stage.

With a parallel-staged launch vehicle, dissimilar rockets (core stage plus boosters) usually burn simultaneously, and the boosters are separated from the core stage after a short time into the launch.

For a vertical launch, neglecting aerodynamic drag and assuming constant gravitational acceleration, the rocket equation for a core rocket system with one or more boosters can be written as

(85) $\begin{equation*} \Delta V = \overline{V}_{\rm eq} \ln \left( \frac{M_0}{M_b}\right) - g_0 \, t_b \end{equation*}$

where ${M_0}$ is the initial mass of the core and boosters, and $M_b$ is the mass of the launch vehicle at booster burnout. The mean equivalent exhaust velocity, $\overline{V}_{\rm eq}$ , is obtained by weighting the equivalent exhaust velocity of each propulsion system by its propellant mass flow rate.

For $N_b$ identical boosters,

(86) $\begin{equation*} \overline{V}_{\rm eq} = \frac{ \overbigdot{m}_c \, V_{\rm eq,c} + N_b \, \overbigdot{m}_b \, V_{\rm eq,b} }{ \overbigdot{m}_c + N_b \, \overbigdot{m}_b } \end{equation*}$

where the subscripts ${c}$ and ${b}$ refer to the core and each individual booster, respectively. Therefore, $\overbigdot{m}_c$ is the propellant mass flow rate of the core, $\overbigdot{m}_b$ is the propellant mass flow rate of one booster, and $N_b$ is the number of boosters.

For the core stage and two identical boosters shown in Figure 14, $N_b=2$ , so

(87) $\begin{equation*} \overline{V}_{\rm eq} = \frac{ \overbigdot{m}_c \, V_{\rm eq,c} + 2 \, \overbigdot{m}_b \, V_{\rm eq,b} }{ \overbigdot{m}_c + 2 \, \overbigdot{m}_b } \end{equation*}$

The principle behind the calculation of the equivalent mean velocity for a parallel-staged booster.

In this case, the initial mass at the launch point is

(88) $\begin{equation*} M_0 = M_{P_{c}} + M_{P_{b}} + M_{S_{c}} + M_{S_{b}} + M_{L} \end{equation*}$

where $M_{P_{c}}$ and $M_{S_{c}}$ are the propellant and structural masses of the core, respectively, $M_{P_{b}}$ and $M_{S_{b}}$ are the combined propellant and structural masses of all boosters, respectively, and $M_L$ is the payload mass.

Determining the final burnout mass, $M_b$ , of a launch vehicle with parallel boosters requires further consideration, as the core launcher will still have propellant left at booster burnout. Therefore, the final mass at the time of booster burnout, say $t_{b_{0}}$ , can be written as

(89) $\begin{equation*} M_b= \chi M_{P_{c}} + M_{S_{c}} + M_{S_{b}} + M_{L} \end{equation*}$

where $\chi$ is the fraction of propellant mass remaining in the core at booster burnout.

If the propellant mass flow rates remain constant over the common core-and-booster burn time, then the propellant mass consumed is $M_P = \overbigdot{m} \, t$ . Therefore, the mean equivalent exhaust velocity, $\overline{V}_{\rm eq}$ , can also be written as

(90) $\begin{equation*} \overline{V}_{\rm eq} = \frac{ (1-\chi) \, M_{P_{c}} \, V_{\rm eq,c} + M_{P_{b}} \, V_{\rm eq,b} }{ (1-\chi) \, M_{P_{c}} + M_{P_{b}} } \end{equation*}$

Therefore, the launch of parallel rocket stages can be presented using the sum of pseudo-serial stages, where for stage “0”, with the boosters and the core together, then

(91) $\begin{equation*} \Delta V_0 = \overline{V}_{\rm eq} \ln \left( \frac{M_{P_{c}} + M_{P_{b}} + M_{S_{c}} + M_{S_{b}} + M_{L} }{\chi M_{P_{c}} + M_{S_{c}} + M_{S_{b}} + M_{L}}\right) - g_0 \, t_{b_{0}} \end{equation*}$

For stage “1” after booster separation, the initial mass will be

(92) $\begin{equation*} M_{0_{1}} = \chi M_{P_{c}} + M_{S_{c}} + M_{L} \end{equation*}$

where $\chi$ is the fraction of propellant mass remaining in the core at booster burnout. The final mass after the stage 1 core burns out will be

(93) $\begin{equation*} M_{b_{1}} = M_{S_{c}} + M_{L} \end{equation*}$

Therefore, for stage 1, then

(94) $\begin{equation*} \Delta V_1 = V_{\rm eq,c} \ln \left( \frac{\chi \, M_{P_{c}} + M_{S_{c}} + M_L} {M_{b_{1}}} \right) - g_0 \, t_{b_{1}} \end{equation*}$

The ${\Delta V}$ values of the remaining stages, $2......N$ , are then calculated as a serial launcher, as before, i.e., the final launch velocity $V_f$ will be

(95) $\begin{equation*} V_f = \Delta V_0 + \Delta V_1 + \Delta V_2 + ......+ \Delta V_N \end{equation*}$

Again, it can be assumed that this velocity is redirected into tangential motion, allowing the payload to follow an orbital path.

Burnout Conditions for a Non-Vertical Launch

To place a payload into a stable circular orbit, a launch vehicle must impart sufficient energy to overcome gravity and atmospheric drag and achieve both the correct altitude and the correct velocity vector. In realistic scenarios, the trajectory is not vertical but gradually transitions from vertical to horizontal flight using what is known as a gravity turn, allowing the vehicle to align its velocity vector with the local horizon at burnout.

For a circular orbit at burnout, the vehicle must travel horizontally at velocity $V_{\rm orb}$ such that the gravitational acceleration provides the necessary centripetal acceleration, i.e., using the universal law of gravitation gives

(96) $\begin{equation*} \frac{G M_E}{(R_E + H_b)^2} = \frac{V_{\rm orb}^2}{R_E + H_b} \end{equation*}$

where $G$ is the gravitational constant, $M_E$ is the mass of Earth, $R_E$ is the Earth’s radius, and $H_b$ is the altitude at burnout. Solving for the required altitude gives

(97) $\begin{equation*} H_b = \frac{G M_E}{V_{\rm orb}^2} - R_E \end{equation*}$

Notice that this relationship sets a geometric constraint in that orbital velocity and altitude are not independent.

The total velocity increment required from the launch vehicle must equal or exceed the orbital velocity plus the applicable velocity losses, i.e.,

(98) $\begin{equation*} \Delta V_T = V_{\rm ideal} + \Delta V_g + \Delta V_d + \Delta V_{\rm steer} \end{equation*}$

In high-altitude ascent, aerodynamic drag is negligible, and steering losses can often be minimized. The dominant loss is from gravity, i.e.,

(99) $\begin{equation*} \Delta V_g \approx g_0 \int_0^{t_b} \sin \gamma(t) \, dt \end{equation*}$

where $\gamma(t)$ is the flight-path angle measured from the local horizontal and $t_b$ is the burn time. For an approximate analysis, then

(100) $\begin{equation*} \Delta V_g \approx g_0 \, \langle \sin \gamma \rangle \, t_b \end{equation*}$

where $\langle \sin \gamma \rangle$ is the average value of $\sin \gamma$ during the powered-flight time $t_b$ . Substituting into Eq. 98 gives the approximate required velocity increment as

(101) $\begin{equation*} \Delta V_T \approx V_{\rm ideal} + g_0 \, \langle \sin \gamma \rangle \, t_b \end{equation*}$

The achievable velocity increment is governed by the rocket equation, i.e.,

(102) $\begin{equation*} \Delta V_T = \sum_i V_{{\rm eq},i} \ln \left( \frac{M_{0,i}}{M_{b,i}} \right) \end{equation*}$

where $V_{{\rm eq},i} = I_{{\rm sp},i} \, g_0$ is the effective exhaust velocity of stage ${i}$ , $M_{0,i}$ is the initial mass, and $M_{b,i}$ is the burnout mass. For a two-stage vehicle, then

(103) $\begin{equation*} \Delta V_T = V_{{\rm eq},1} \ln \left( \frac{M_{0,1}}{M_{b,1}} \right) + V_{{\rm eq},2} \ln \left( \frac{M_{0,2}}{M_{b,2}} \right) \end{equation*}$

Equating this to Eq. 101 sets the design constraint for orbital insertion.

The altitude gained during thrust is computed by integrating the vertical velocity component, i.e.,

(104) $\begin{equation*} H_b = \int_0^{t_b} V(t) \sin \gamma(t) \, dt \end{equation*}$

A simple estimate using average values gives

(105) $\begin{equation*} H_b \approx \left\langle V \sin \gamma \right\rangle t_b \approx \langle V \rangle \, \langle \sin \gamma \rangle \, t_b \end{equation*}$

where $\left\langle V \sin \gamma \right\rangle$ is the average vertical velocity during the powered flight. The second approximation assumes that the average of the product can be represented by the product of the averages. This estimate shows how the burnout altitude depends on both the flight speed and the flight-path angle during the burn.

To design an ascent trajectory that leads to orbital insertion, one must determine the orbital altitude using Eq. 97, estimate gravity losses from Eq. 101, compute propulsion performance via Eq. 102, and confirm that the altitude and horizontal velocity at burnout are consistent with orbital conditions. This framework supports preliminary trajectory design, staging analysis, and mission feasibility studies.

Check Your Understanding #3 – Non-vertical launch

A two-stage launch vehicle is designed to place a 300 kg satellite into a circular low Earth orbit (LEO) at an altitude of 200 km. The first stage uses a propulsion system with a specific impulse of $I_{\rm sp,1}$ = 285 s, and the second stage has $I_{\rm sp,2}$ = 326 s. The vehicle consists of a first stage with 8,000 kg of propellant and 800 kg of structure, and a second stage with 2,500 kg of propellant and 250 kg of structure. The payload mass is 300 kg. Assume standard gravity ${g_0}$ = 9.81 m/s², negligible drag, and an average flight-path profile such that $\langle \sin \gamma \rangle$ = 0.85, where $\gamma$ is measured from the local horizontal. The total powered flight time is estimated to be $t_b$ = 85 s. Determine whether the rocket can deliver the required velocity increment and estimate the burnout altitude.

Show solution/hide solution.

The orbital velocity required for a circular orbit at $H_b = 200~\mbox{km}$ is

$V_{\rm orb} = \sqrt{ \frac{G M_E}{R_E + H_b} }$

Using $G M_E = 3.986 \times 10^{14}$ m $^3$ /s $^2$ , $R_E = 6.371 \times 10^6$ m, and $H_b = 2.00 \times 10^5$ m gives

$V_{\rm orb} = \sqrt{ \frac{3.986 \times 10^{14}} {6.571 \times 10^6} } = 7,788~\mbox{m/s}$

The ideal velocity requirement, including the increase in gravitational potential energy, is

$V_{\rm ideal} = \sqrt{ G M_E \left( \frac{2}{R_E} - \frac{1}{R_E+H_b} \right) }$

Therefore,

$V_{\rm ideal} = \sqrt{ 3.986 \times 10^{14} \left( \frac{2}{6.371 \times 10^6} - \frac{1}{6.571 \times 10^6} \right) } = 8,029~\mbox{m/s}$

The additional velocity increment associated with increasing the gravitational potential energy is

$\Delta V_U = V_{\rm ideal} - V_{\rm orb} = 8,029 - 7,788 = 241~\mbox{m/s}$

The approximate gravity loss is

$\Delta V_g = g_0 \langle \sin \gamma \rangle \, t_b$

so that

$\Delta V_g = 9.81 \times 0.85 \times 85 = 708.8~\mbox{m/s}$

The total required velocity increment is then

$\Delta V_T = V_{\rm ideal} + \Delta V_g = 8,029 + 708.8 = 8,738~\mbox{m/s}$

The effective exhaust velocities of the two stages are

$V_{{\rm eq},1} = I_{\rm sp,1} \, g_0 = 285 \times 9.81 = 2,795.9~\mbox{m/s}$

and

$V_{{\rm eq},2} = I_{\rm sp,2} \, g_0 = 326 \times 9.81 = 3,198.1~\mbox{m/s}$

The initial and burnout masses for the first stage are

$M_{0,1} = 8,000 + 800 + 2,500 + 250 + 300 = 11{,}850~\mbox{kg}$

and

$M_{b,1} = 800 + 2,500 + 250 + 300 = 3{,}850~\mbox{kg}$

The ideal velocity increment from the first stage is

$\Delta V_1 = V_{{\rm eq},1} \ln \left( \frac{M_{0,1}}{M_{b,1}} \right)$

which gives

$\Delta V_1 = 2,795.9 \ln \left( \frac{11{,}850}{3{,}850} \right) = 3,143~\mbox{m/s}$

For the second stage,

$M_{0,2} = 2,500 + 250 + 300 = 3{,}050~\mbox{kg}$

and

$M_{b,2} = 250 + 300 = 550~\mbox{kg}$

The ideal velocity increment from the second stage is

$\Delta V_2 = V_{{\rm eq},2} \ln \left( \frac{M_{0,2}}{M_{b,2}} \right)$

which gives

$\Delta V_2 = 3,198.1 \ln \left( \frac{3{,}050}{550} \right) = 5,478~\mbox{m/s}$

The total velocity increment delivered by the two stages is

$\Delta V_{\rm rocket} = \Delta V_1 + \Delta V_2 = 3,143 + 5,478 = 8,621~\mbox{m/s}$

The available margin is

$\Delta V_{\rm margin} = \Delta V_{\rm rocket} - \Delta V_T = 8,621 - 8,738 = -117~\mbox{m/s}$

Therefore, the rocket does not meet the simplified velocity requirement. It falls short by approximately 117 m/s, even before accounting for aerodynamic drag, steering losses, residual propellant, or other operational allowances.

To estimate the burnout altitude, assume

$\overline{V} \approx \frac{1}{2}\Delta V_{\rm rocket} = 4,311~\mbox{m/s}$

The approximate burnout altitude is

$H_b \approx \overline{V} \langle \sin \gamma \rangle t_b - \frac{1}{2} g_0 t_b^2$

Substituting the values gives

$H_b \approx 4,311 \times 0.85 \times 85 - \frac{1}{2} \times 9.81 \times 85^2$

so that

$H_b \approx 2.76 \times 10^5~\mbox{m} = 276~\mbox{km}$

The assumed average flight-path factor $\langle \sin \gamma \rangle = 0.85$ produces an estimated burnout altitude above the specified 200 km orbit. However, the vehicle does not provide sufficient velocity increment to achieve the required circular orbit. The assumed flight-path profile is therefore not consistent with direct insertion into a 200 km circular orbit. A different trajectory, greater propulsion performance, or a reduced payload mass would be required.

Maximum Dynamic Pressure (“Max-q ”)

During the ascent of a launch vehicle, the aerodynamic loads on the structure are determined by the dynamic pressure, i.e.,

(106) $\begin{equation*} q = \frac{1}{2} \varrho_{\rm atm}(h) \, V^2 \end{equation*}$

where $\varrho_{\rm atm}(h)$ is the atmospheric density at altitude $h$ , and $V$ is the true flight velocity through the air. Immediately after liftoff, the vehicle moves relatively slowly, so the dynamic pressure is low despite the high air density. At higher altitudes, the vehicle may be traveling at several kilometers per second, but the density is so low that $q$ is again small. Between these two extremes, there is an intermediate flight condition where the opposing trends of decreasing $\varrho_{\rm atm}$ and increasing $V$ produce a maximum in $q$ . This condition is referred to as the point of maximum dynamic pressure, or “max- $q$ ,” and it is a critical point in the launch envelope.

The velocity, altitude, and time at which the max- $q$ occurs can be determined by differentiating Eq. 106 with respect to time, which gives

(107) $\begin{equation*} \overbigdot{q} = \varrho_{\rm atm} \, V \overbigdot{V} + \frac{1}{2} V^2 \, \overbigdot{\varrho}_{\rm atm} \end{equation*}$

Assuming the standard exponential atmosphere, then

(108) $\begin{equation*} \varrho_{\rm atm}(h) = \varrho_{\rm atm_{0}} \, e^{-h/H} \quad \text{and} \quad \overbigdot{\varrho}_{\rm atm} = -\frac{\varrho_{\rm atm}}{H} \, \overbigdot{h} \end{equation*}$

where $H$ is the density scale height, which is usually taken as $H \approx$ 7.2 km in the lower atmosphere. Substituting Eq. 108 into Eq. 107 gives

(109) $\begin{equation*} \overbigdot{q} = \varrho_{\rm atm} \, V \, \overbigdot{V} - \frac{\varrho_{\rm atm} \, V^2}{2 H} \, \overbigdot{h} \end{equation*}$

For a flight-path angle $\gamma$ , the vertical velocity is

(110) $\begin{equation*} \overbigdot{h} = V \, \sin\!\gamma \end{equation*}$

At the instant of maximum dynamic pressure, $\overbigdot{q} = 0$ , so combining Eq. 109 and Eq. 110 gives

(111) $\begin{equation*} \overbigdot{V} = \frac{V^2 \sin\gamma}{2H} \end{equation*}$

This result has a clear physical meaning: at max- $q$ , the acceleration along the trajectory, $\overbigdot{V}$ , is exactly balanced by the rate at which the decreasing air density reduces the aerodynamic loading along the flight path. If $\overbigdot{V}$ is greater than the right-hand side of Eq. 111, then $q$ is still increasing; if it is smaller, then $q$ has already begun to decrease. For a nearly vertical ascent, where ${\sin\gamma \approx 1}$ , the condition reduces to $\overbigdot{V} = V^2/(2H)$ . A shallower ascent, as in a gravity turn, delays the occurrence of max- $q$ to a higher altitude and later time in the flight trajectory.

The engineering importance of max- $q$ is that it defines the point at which the vehicle and payload experience peak aerodynamic forces and moments. The structural design requirements for the launch vehicle are often driven by the loads that occur at this point. In addition to structural concerns, the product $q\alpha$ , i.e., dynamic pressure times angle of attack, must remain below specified limits to prevent excessive lateral aerodynamic loading and structural bending. Guidance and control laws are therefore designed to schedule $\alpha$ appropriately. In most launch vehicles, the engines are throttled down in the seconds before max- $q$ to limit the dynamic pressure to a predetermined safe value, then throttled back up once the vehicle has climbed into the less-dense air.

A representative medium-lift launch vehicle is shown in Figure 15. Max- $q$ typically occurs about 60–80 s after liftoff, at an altitude of 11 to 15 km, when the velocity is approximately 0.4 to 0.55 km s $^{-1}$ and the dynamic pressure reaches values of 25 to 40 kPa (about 3.6 to 5.8 psi). These values depend on the ascent profile and the vehicle’s thrust-to-weight ratio, but they provide a helpful reference.

Representative launch profile for medium-lift launch vehicle showing the occurrence of “max- $q$ .”

Variable Thrust (Throttling) Conditions

The thrust of a liquid-propellant rocket engine can be varied during flight by changing the propellant mass flow rate. Neglecting gravity and aerodynamic drag, the equation of motion is

(112) $\begin{equation*} M \frac{dV}{dt} = T(t) \end{equation*}$

where the thrust is

(113) $\begin{equation*} T(t) = \overbigdot{m}(t) \, V_{\rm eq} \end{equation*}$

and $\overbigdot{m}(t)=-dM/dt$ is the positive propellant mass flow rate. Introducing a throttle setting $\tau(t)$ , defined as the ratio of the instantaneous thrust to the maximum thrust, gives

(114) $\begin{equation*} T(t) = \tau(t) \, T_{\max} \end{equation*}$

where $0 < \tau(t) \leq 1$ . If the equivalent exhaust velocity is assumed constant, then

(115) $\begin{equation*} \overbigdot{m}(t) = \frac{T(t)}{V_{\rm eq}} = \tau(t)\frac{T_{\max}}{V_{\rm eq}} = \tau(t)\,\overbigdot{m}_{\max} \end{equation*}$

Therefore, the rocket mass varies according to

(116) $\begin{equation*} \frac{dM}{dt} = -\tau(t)\,\overbigdot{m}_{\max} \end{equation*}$

so that

(117) $\begin{equation*} M(t) = M_0 - \overbigdot{m}_{\max} \int_0^t \tau(\xi)\,d\xi \end{equation*}$

The equation of motion may also be written as

(118) $\begin{equation*} dV = -V_{\rm eq}\frac{dM}{M} \end{equation*}$

Integrating from the initial mass $M_0$ to the final mass $M_f$ gives

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

Therefore, if $V_{\rm eq}$ and the initial and final masses remain unchanged, the ideal velocity increment does not depend on the throttle schedule. Throttling changes the rate at which the propellant is consumed rather than the ideal $\Delta V$ obtained from a given mass ratio.

For a constant throttle setting $\tau$ , the burnout time is

(120) $\begin{equation*} t_b = \frac{M_0-M_f}{\tau\,\overbigdot{m}_{\max}} \end{equation*}$

or

(121) $\begin{equation*} t_b = \frac{t_{b,\max}}{\tau} \end{equation*}$

where $t_{b,\max}$ is the burnout time at maximum thrust. Therefore, reducing the throttle setting increases the burnout time.

For a vertical launch, the gravity loss is approximately

(122) $\begin{equation*} \Delta V_g = g_0 t_b = \frac{g_0 t_{b,\max}}{\tau} \end{equation*}$

Consequently, throttling does not directly change the ideal rocket-equation $\Delta V$ when $V_{\rm eq}$ is constant, but it can reduce the actual burnout velocity by increasing the burn time and the associated gravity loss. During launch, this performance penalty may be accepted temporarily to limit dynamic pressure and structural loading near max- $q$ .

Missiles & Ballistics

Ballistic trajectories refer to the paths followed by missiles or artillery projectiles after their propulsion systems have stopped producing thrust. Predicting these trajectories is important for determining flight time, range, altitude, and target location. The trajectory depends on the initial velocity and launch angle, as well as atmospheric drag, changes in gravitational acceleration, the curvature and rotation of the Earth, and any guidance or control forces acting during flight.

Ballistic missiles are accelerated during a powered launch phase and then enter an unpowered ballistic phase governed primarily by gravity. Short-range trajectories can sometimes be approximated as parabolic using a flat-Earth model with constant gravitational acceleration. Long-range ballistic trajectories must be treated as portions of elliptical or other conic trajectories about the Earth, with atmospheric drag becoming important during ascent and re-entry.

Consider the equations of motion of a missile, as shown in Figure 16. With the flight-path angle $\theta$ measured from the local horizontal and the thrust inclined by an angle $\alpha$ relative to the velocity vector, the equation of motion parallel to the flight path is

(123) $\begin{equation*} M \frac{dV_{\infty}}{dt} = T \cos \alpha - D - M g_0 \sin \theta \end{equation*}$

where $M$ is the instantaneous mass of the missile and ${V_{\infty}}$ is its flight speed.

Consideration of the forces acting on a missile in flight can be used to derive its equations of motion.

The acceleration perpendicular to the flight path is

(124) $\begin{equation*} a_n = \frac{V_{\infty}^2}{R} = V_{\infty}^2 \frac{d\theta}{ds} = V_{\infty} \frac{d\theta}{dt} \end{equation*}$

because

(125) $\begin{equation*} \frac{d\theta}{ds} = \frac{1}{R} \qquad \mbox{and} \qquad \frac{ds}{dt} = V_{\infty} \end{equation*}$

where $s$ is distance measured along the flight path and $R$ is the instantaneous radius of curvature. The equation of motion perpendicular to the flight path is then

(126) $\begin{equation*} M V_{\infty} \frac{d\theta}{dt} = T \sin \alpha + L - M g_0 \cos \theta \end{equation*}$

In principle, Eqs. 123 and 126 can be integrated to determine the missile trajectory. However, realistic trajectory prediction is much more complicated because the mass, thrust, atmospheric density, aerodynamic coefficients, gravitational acceleration, and guidance commands may all vary during flight. Earth’s curvature and rotation must also be accounted for in long-range trajectories.

For the following elementary analysis, assume that the rocket engine has stopped producing thrust, so $T = 0$ , as shown in Figure 17. Also assume that the aerodynamic lift is zero. The missile is then treated as an unpowered projectile acted upon only by gravity and aerodynamic drag.

A missile’s ballistic profile. The actual trajectory depends on the initial launch conditions, range, maximum altitude, and re-entry conditions.

Let the $x$ -axis be horizontal, and the $y$ -axis be vertically upward. The equations of motion become

(127) $\begin{eqnarray*} \mbox{\small $x$-direction:\quad} M \overbigddot{x} &=& -D \cos \theta \\[6pt] \mbox{\small $y$-direction:\quad} M \overbigddot{y} &=& -D \sin \theta - M g_0 \end{eqnarray*}$

The aerodynamic drag is

(128) $\begin{equation*} D = \frac{1}{2} \varrho_{\infty} V_{\infty}^2 C_D A \end{equation*}$

where $C_D$ is based on the reference area $A$ , which is usually the maximum cross-sectional area of the missile. A simplified analytical solution can be obtained by representing the drag over the trajectory in the approximate form

(129) $\begin{equation*} D = \mu V_{\infty} \overbigdot{x} \end{equation*}$

where $\mu$ is treated as a constant mean value defined by

(130) $\begin{equation*} \mu = \left[ \frac{ \displaystyle{ \frac{1}{2} \varrho_{\infty} V_{\infty} C_D A } }{ \overbigdot{x} } \right]_{\mbox{\small Mean value}} \end{equation*}$

This approximation replaces the actual variations in atmospheric density, speed, and drag coefficient by a representative average over the ballistic portion of the trajectory. Because

(131) $\begin{equation*} \overbigdot{x} = V_{\infty}\cos\theta \end{equation*}$

the horizontal equation of motion becomes

(132) $\begin{equation*} M \overbigddot{x} = -\mu V_{\infty} \overbigdot{x} \cos\theta = -\mu \overbigdot{x}^{\,2} \end{equation*}$

or

(133) $\begin{equation*} M \overbigddot{x} + \mu \overbigdot{x}^{\,2} = 0 \end{equation*}$

Defining the horizontal velocity as $u = \overbigdot{x}$ gives

(134) $\begin{equation*} M \frac{du}{dt} = -\mu u^2 \end{equation*}$

Separating the variables gives

(135) $\begin{equation*} \int_{u_0}^{u} \frac{du}{u^2} = -\frac{\mu}{M} \int_0^t dt \end{equation*}$

which gives

(136) $\begin{equation*} -\frac{1}{u} + \frac{1}{u_0} = -\frac{\mu}{M} t \end{equation*}$

and hence

(137) $\begin{equation*} u = \frac{dx}{dt} = \frac{u_0}{ 1 + \displaystyle{ \frac{\mu u_0}{M} t } } \end{equation*}$

Defining

(138) $\begin{equation*} \lambda = \frac{\mu u_0}{M} \end{equation*}$

gives

(139) $\begin{equation*} u = \frac{u_0}{1+\lambda t} \end{equation*}$

Integrating with $x = 0$ at $t = 0$ gives

(140) $\begin{equation*} x = \frac{u_0}{\lambda} \ln\left(1+\lambda t\right) \end{equation*}$

Now consider the equation of motion in the $y$ -direction. Because

(141) $\begin{equation*} \overbigdot{y} = V_{\infty}\sin\theta \end{equation*}$

the vertical drag component is

(142) $\begin{equation*} D\sin\theta = \mu V_{\infty} \overbigdot{x} \sin\theta = \mu \overbigdot{x} \overbigdot{y} \end{equation*}$

The vertical equation of motion becomes

(143) $\begin{equation*} M \overbigddot{y} = -\mu \overbigdot{x} \overbigdot{y} - M g_0 \end{equation*}$

Using

(144) $\begin{equation*} \overbigdot{x} = \frac{u_0}{1+\lambda t} \qquad \mbox{and} \qquad \lambda = \frac{\mu u_0}{M} \end{equation*}$

gives

(145) $\begin{equation*} \overbigddot{y} + \frac{\lambda}{1+\lambda t} \overbigdot{y} = -g_0 \end{equation*}$

Multiplying by the integrating factor $1+\lambda t$ gives

(146) $\begin{equation*} \frac{d}{dt} \left[ \left(1+\lambda t\right) \overbigdot{y} \right] = -g_0 \left(1+\lambda t\right) \end{equation*}$

Integrating and applying the initial condition $\overbigdot{y}(0) = v_0$ gives

(147) $\begin{equation*} \left(1+\lambda t\right) \overbigdot{y} = v_0 - g_0 t - \frac{1}{2} g_0 \lambda t^2 \end{equation*}$

so that

(148) $\begin{equation*} \overbigdot{y} = \frac{ v_0 - g_0 t - \displaystyle{ \frac{1}{2} g_0 \lambda t^2 } }{ 1+\lambda t } \end{equation*}$

An equivalent form is

(149) $\begin{equation*} \overbigdot{y} = -\frac{g_0 t}{2} - \frac{g_0}{2\lambda} + \frac{ \displaystyle{ v_0 + \frac{g_0}{2\lambda} } }{ 1+\lambda t } \end{equation*}$

Integrating with $y = 0$ at $t = 0$ gives

(150) $\begin{equation*} y = -\frac{g_0 t^2}{4} - \frac{g_0 t}{2\lambda} + \left( v_0 + \frac{g_0}{2\lambda} \right) \frac{ \ln\left(1+\lambda t\right) }{ \lambda } \end{equation*}$

Eqs. 140 and 150 provide an approximate analytical description of the ballistic trajectory under the assumed mean-drag model.

When aerodynamic drag is negligible, $\mu \rightarrow 0$ and hence $\lambda \rightarrow 0$ . The horizontal motion reduces to

(151) $\begin{equation*} \overbigdot{x} = u_0 \qquad \mbox{and} \qquad x = u_0 t = V_0 \cos\theta_0 \, t \end{equation*}$

The vertical motion becomes

(152) $\begin{equation*} \overbigdot{y} = v_0 - g_0 t \end{equation*}$

and

(153) $\begin{equation*} y = v_0 t - \frac{1}{2} g_0 t^2 = V_0 \sin\theta_0 \, t - \frac{1}{2} g_0 t^2 \end{equation*}$

where

(154) $\begin{equation*} u_0 = V_0 \cos\theta_0 \qquad \mbox{and} \qquad v_0 = V_0 \sin\theta_0 \end{equation*}$

These are the classical flat-Earth equations for the parabolic trajectory of an unpowered projectile. For level ground, with the launch point and impact point at the same height, the total flight time is

(155) $\begin{equation*} T_f = \frac{ 2 V_0 \sin\theta_0 }{ g_0 } \end{equation*}$

The corresponding range is

(156) $\begin{equation*} R_f = \frac{ V_0^2 \sin 2\theta_0 }{ g_0 } \end{equation*}$

The maximum range in this idealized constant-gravity, drag-free model occurs when $\theta_0 = 45^{\circ}$ because $\sin 2\theta_0 = 1$ .

These equations are sometimes referred to as the Laws of Ballistics or the Artilleryman’s Range Equations. They are appropriate only when the flight distance and altitude are sufficiently small that the Earth can be treated as flat and $g$ can be treated as constant. For long-range ballistic missiles, the curvature and rotation of the Earth, the inverse-square variation of gravity, and atmospheric drag during ascent and re-entry must be included. The resulting unpowered trajectory is a portion of an orbit about the Earth rather than a simple parabola.

Falcon 9 Booster Re-entry

Reusable launch vehicles are revolutionizing spaceflight and reducing the cost of launching payloads into orbit. The James Bond film You Only Live Twice, which was released in 1967, introduced the concept of a reusable rocket in popular culture, depicting a spacecraft launched into orbit by SPECTRE, completing its mission of capturing another spacecraft, and returning to Earth for reuse. Though fictional at the time, the concept became a reality with the Space Shuttle and is now being further advanced by companies such as SpaceX and Blue Origin.

Falcon 9’s first-stage booster recovery is a remarkable engineering achievement, combining an advanced knowledge of aerothermodynamics, flight dynamics, and guidance and control into a single, integrated system. The process begins with a drag-governed ballistic entry, in which aerodynamic forces dissipate most of the booster’s potential and kinetic energy. This is followed by two retro-burns of the rocket motors and a precisely timed landing maneuver. Guidance algorithms continuously update the flight trajectory based on real-time values of velocity, altitude, and atmospheric conditions. The goals in achieving a successful booster recovery are:

Accurately predict and control the descent trajectory so the recovered booster arrives at the landing site on the correct flight path, velocity, and attitude.
Determine the optimum amount of propellant required for each burn (boost-back, re-entry, landing) to ensure a safe, precise touchdown within the available propellant mass budget.
Manage the flight path of the booster to minimize kinetic heating and the associated thermal loads.

While predicting the details of the optimal re-entry and recovery trajectory requires numerical solutions of the equations of motion with a realistic atmospheric model (e.g., the exponential model or the piecewise atmospheric model for $\varrho_{\mathrm{atm}}(h)$ from the troposphere to the thermosphere), the sequence of steps can be described analytically to illustrate the key principles.

Stage Separation

After stage separation at an altitude of about 70 km and a velocity of about 8,000 km/h (about 2,220 m/s, which is in the Mach 6–8 hypersonic regime at this altitude), the booster continues on an upward ballistic trajectory to an apogee of about 120 km at a velocity of 7,000 km/h (1,944 m/s). Then it begins its return to Earth in an engines-first orientation, as shown in Figure 18. Cold-gas thrusters, which are small nitrogen attitude-control jets, control the booster’s orientation and prevent tumbling during the high-altitude, low-dynamic-pressure portion of re-entry.

The recovery of a first-stage booster involves a series of controlled maneuvers, including ballistic re-entry, a re-entry burn, and a landing burn. The Falcon 9 uses nine Merlin rocket motors.

The grid fins are also deployed (except during a boost-back maneuver), which help with aerodynamic braking and control. However, the density is so low at altitudes above 100 km that drag is not significant until the vehicle descends below approximately 80 km. The booster exhibits passive engine-first directional stability because its center of mass lies near the engine section, ahead of the aerodynamic center of pressure associated with the body and deployed grid fins. A disturbance then produces an aerodynamic restoring moment that tends to realign the booster with the relative flow. It also places the nine Merlin engines into the oncoming flow, creating a high-drag configuration that enables rapid deceleration and significantly slows the booster before it reaches the denser lower atmosphere. The engines and propulsion hardware, along with the surrounding base heat shield, are designed to withstand the kinetic heating during reentry.

The overall descent of the booster follows a nominally ballistic trajectory, governed by gravity and aerodynamic forces, and interrupted only by brief periods of thrust from the boost-back burn (if used) and the re-entry burn. In the initial ballistic phase, the deceleration of the booster can be represented by the sum of gravity and aerodynamic terms, i.e., according to the principles of ballistics, the equation of motion for the booster can be approximated by

(157) $\begin{equation*} M(t) \,\frac{dV}{dt} = M(t) \, g_0 \left(\frac{w}{V}\right) - \frac{1}{2}\,\varrho_{\rm atm}(h)\,V^2 \left( \underbrace{ A_{\mathrm{ref}} \, C_{D_b} }_{\text{Booster}} + \underbrace{ S_{\rm gf} \, C_{D_{\rm gf}} }_{\text{Grid-fins}} \right) \end{equation*}$

where $V$ is its velocity, $M(t)$ is the instantaneous mass of the booster, and $\varrho_{\rm atm}(h)$ is the atmospheric density profile for a given height $h$ . In addition, $A_{\mathrm{ref}}$ is the base area of the booster and $S_{\rm gf}$ is the area of the grid-fins. $C_{D_b}$ is the drag coefficient of the booster in the tail-first flight orientation, which is typically in the range $1.2$ to $1.5$ for cylindrical-like bluff bodies. The drag coefficient, $C_{D_{\rm gf}}$ , of the grid-fins is more complex to model and is considered next. For sub-orbital heights, as in this case, it is usually assumed that $g(h) = g_0$ .

The booster has both horizontal velocity $u$ and vertical (downward) velocity $w$ , so the total relative flow velocity is $V = \sqrt{u^2 + w^2}$ . Splitting into horizontal ( $x$ ) and height ( $h$ ) components gives

(158) $\begin{eqnarray*} M(t)\,\overbigdot{u} &= &-\frac{1}{2}\,\varrho_{\mathrm{atm}}(h)\,V^2 \left( A_{\mathrm{ref}}\,C_{D_b} + S_{\rm gf}\,C_{D_{\rm gf}} \right) \left( \frac{u}{V} \right) \\[6pt] M(t)\,\overbigdot{w} &= &\;M(t)\,g_0-\frac{1}{2}\,\varrho_{\mathrm{atm}}(h)\,V^2 \left( A_{\mathrm{ref}}\,C_{D_b} + S_{\rm gf}\,C_{D_{\rm gf}} \right) \left( \frac{w}{V} \right) \end{eqnarray*}$

with the kinematic relations $\overbigdot{x} = u$ and $\overbigdot{h} = -w$ , noting that the $u$ velocity, and hence the downrange distance, is affected by the amount of aerodynamic drag. These ODEs are then numerically solved to estimate the flight trajectory. When the booster is in flight, guidance algorithms continuously update the required trajectory to follow the estimated trajectory. In trajectory prediction, calculating the ballistic coefficient is a useful cross-check of how quickly the booster will decelerate, i.e., using

(159) $\begin{equation*} \beta = \frac{M(t)}{C_{D_b} \, A_{\mathrm{ref}}} \end{equation*}$

with lower values producing more rapid aerodynamic deceleration. Immediately after separation, the stage enters the hypersonic regime, where aerodynamic heating rates are high, but the dynamic pressure and aerodynamic drag remain relatively low.

Boost-Back Burn

The boost-back burn maneuver, if performed, alters the landing point from a downrange drone ship to a pad near the launch site by reversing the booster’s horizontal velocity after launch, thereby imparting a $\Delta V$ component opposite to that of the typical ballistic trajectory. After stage separation, the booster is commanded to perform a backflip maneuver; the exact separation velocity (on the order of 8,000 km/h) depends on the mission and recovery profile. A burn follows, which significantly reduces its downrange velocity and distance. The grid-fins are not deployed during a boost-back maneuver. During the burn, the vehicle continues to coast upward on a ballistic trajectory, reaching a peak altitude (apogee) of about 150 km at a velocity of roughly 1,000 km/h (0.28 km/s). Some of the booster’s initial kinetic energy is converted into gravitational potential energy, which is why its altitude increases even as its velocity decreases.

From an energy perspective, the booster initially possesses mechanical energy per unit mass given by

(160) $\begin{equation*} E_0 = \frac{1}{2} \left( u_0^2 + w_0^2 \right) + g_0 \, h_0 \end{equation*}$

where $u_0$ and $w_0$ are the horizontal and vertical velocity components, respectively, and $h_0$ is the nominal altitude at stage separation. The boost-back burn imparts a velocity change $\Delta V_{\mathrm{bb}}$ , which reduces the horizontal velocity and may increase the upward component of velocity if the thrust vector is pitched upward. Therefore,

(161) $\begin{equation*} \Delta V_{\mathrm{bb}} = \sqrt{\left(u_1 - u_0\right)^2 + \left(w_1 - w_0\right)^2} \end{equation*}$

Treating the burn as impulsive and neglecting aerodynamic drag, the change in specific kinetic energy across the burn is

(162) $\begin{equation*} \Delta K = \frac{1}{2}\left(u_1^2+w_1^2\right) - \frac{1}{2}\left(u_0^2+w_0^2\right) \end{equation*}$

so that

(163) $\begin{equation*} \Delta K = u_0(u_1 - u_0) + \frac{1}{2}(u_1 - u_0)^2 + w_0(w_1 - w_0) + \frac{1}{2}(w_1 - w_0)^2 \end{equation*}$

Recognizing that $(u_1 - u_0)^2 + (w_1 - w_0)^2 = \Delta V_{\mathrm{bb}}^{\,2}$ , then

(164) $\begin{equation*} \Delta K = u_0(u_1 - u_0) + w_0(w_1 - w_0) + \frac{1}{2}\,\Delta V_{\mathrm{bb}}^{\,2} \end{equation*}$

The post-burn mechanical energy per unit mass is

(165) $\begin{equation*} E_1 = \frac{1}{2} \left( u_1^2 + w_1^2 \right) + g_0 \, h_1 \end{equation*}$

During the subsequent ballistic coast (no thrust, drag neglected), energy is conserved, so $E$ remains constant as some kinetic energy is converted into gravitational potential energy. A reduction in $u$ from $u_0$ to $u_1$ , combined with an increase in the upward velocity component, can raise the apogee even as the total velocity decreases.

This boost-back maneuver requires a significant additional mass of propellant, reducing the propellant available for the launch and, therefore, lowering the maximum payload mass to orbit compared with a downrange drone-ship recovery. The required velocity change in the burn to follow the required trajectory can be estimated from the rocket equation using

(166) $\begin{equation*} \Delta V_{\mathrm{bb}} \approx g_0 \, I_{\mathrm{sp}} \, \ln \left(\frac{M_0}{M_0 - M_{P,\mathrm{bb}}}\right) \approx 5,000~\text{km/h} \approx 1,389~\text{m/s} \end{equation*}$

where $M_0$ is the booster’s mass immediately before the burn and $M_{P,\mathrm{bb}}$ is the mass of propellant needed for the burn, i.e.,

(167) $\begin{equation*} M_{P,\mathrm{bb}} = M_0 \left( 1 - \exp \left( -\frac{\Delta V_{\mathrm{bb}}}{g_0 \, I_{\mathrm{sp}}} \right) \right) \end{equation*}$

Grid-Fins

The four deployable grid fins (not a SpaceX invention), mounted near the top of the booster, as shown in the photograph in Figure 19, provide both aerodynamic drag (i.e., braking) and control during the supersonic, transonic, and subsonic phases of descent. When deployed, the booster flies with its engines at the front, so the fins are positioned at the top of the booster (now its tail), where they provide directional stability and drag. In this respect, they function like a steerable parachute, generating both stabilizing and controllable aerodynamic forces.

Grid-fins have relatively high drag coefficients when they are placed normal to the flow, typically in the range of about $C_D$ = 0.6–0.8 at subsonic speeds, rising to 0.8–1.5 in the transonic and supersonic regimes. At a local flow angle $\alpha_f$ , set by the vehicle attitude, and with a commanded fin deflection $\delta_f$ , the effective angle of attack is

(168) $\begin{equation*} \alpha_{\mathrm{eff}} = \alpha_f + \eta \, \delta_f \end{equation*}$

where $\eta$ is the control-effectiveness factor, as used for a flap or other aerodynamic control surface. A canonical drag coefficient model for a grid fin is

(169) $\begin{equation*} C_{D_{\rm gf}} = C_{D_0} + K \, \left|\sin \alpha_{\mathrm{eff}}\right|^3 \end{equation*}$

where $C_{D_0}$ is the minimum drag coefficient, typically about 0.2, and $K$ is an empirical constant that represents the behavior of a particular grid-fin design. Notice that the $\left|\sin \alpha_{\mathrm{eff}}\right|^3$ dependence arises from the combination of two effects: the projected frontal area of the fin grows approximately as $\left|\sin \alpha_{\mathrm{eff}}\right|$ , while the perpendicular dynamic pressure scales as $\sin^2 \alpha_{\mathrm{eff}}$ . Multiplying these effects gives a cubic dependence on the magnitude of the effective angle.

Therefore, the equation of motion becomes

(170) $\begin{equation*} M_1 \,\frac{dV}{dt} = M_1 \, g_0 - \frac{1}{2}\,\varrho_{\rm atm}(h)\,V^2 \left( \underbrace{ A_{\mathrm{ref}} \, C_{D_b} }_{\text{Booster}} + \underbrace{ S_{\rm gf} \Big(C_{D_0} + K \, \left|\sin \alpha_{\mathrm{eff}}\right|^3\Big)}_{\text{Grid-fins}} \right) \end{equation*}$

where $M_1 = M_0 - M_{P,\mathrm{bb}}$ with the boost-back burn or $M_1 = M_0$ without it.

The orientation of each grid fin can be independently controlled to provide precise steering, enabling the stage to be placed on the exact flight trajectory required for landing. The resulting aerodynamic forces and moments, combined with reaction-control thrusters in the near-vacuum upper portion of the trajectory, maintain the booster’s correct attitude for a well-controlled descent profile.

Re-Entry Burn

Between the boost-back burn (if performed) and the re-entry burn, the descent remains predominantly ballistic. The re-entry burn is performed at approximately 60 km and at a velocity of between 5,000 km/h and 7,000 km/h (1.39 km/s to 1.94 km/s, or at a Mach number between 4 and 6), as shown in the flight profile in Figure 20. This burn reduces the velocity by $\Delta V_{\mathrm{re}} \approx$ 3,000 km/h (0.83 km/s). Only three of the nine Falcon rocket motors are used for this burn. The required propellant mass for the burn can be estimated from the rocket equation using

(171) $\begin{equation*} \Delta V_{\mathrm{re}} = g_0 \, I_{\mathrm{sp}} \ln \left(\frac{M_1}{M_1 - M_{P,\mathrm{re}}}\right) \approx 3,000 \text{ km/h} \approx 833 \text{ m/s} \end{equation*}$

where $M_1$ is the booster’s mass immediately before the burn and $M_{P,\mathrm{re}}$ is the propellant mass required, i.e.,

(172) $\begin{equation*} M_{P,\mathrm{re}} = M_1 \left( 1 - \exp \left( -\frac{\Delta V_{\mathrm{re}}}{g_0 \, I_{\mathrm{sp}}} \right) \right) \end{equation*}$

The mass of the booster after the re-entry burn is now $M_2 = M_0 - M_{P,\mathrm{bb}} - M_{P,\mathrm{re}}$ when including the boost-back burn, or $M_2 = M_0 - M_{P,\mathrm{re}}$ without the boost-back burn.

Nominal altitude versus Mach number profile for the Falcon 9 booster re-entry and recovery to the landing zone.

As the booster stage descends into denser air, it progressively decelerates through the supersonic regime, where the peak dynamic pressure usually occurs at an altitude of around 15 km and a velocity of 3,000 km/h (Mach 2–3). Because the dynamic pressure is $\dfrac{1}{2} \, \varrho_{\rm atm}(h) \, V^{2}$ and the kinetic heating rate scales as $V^{3}$ , even a modest velocity reduction has a significant effect on the structural and thermal loads. Heating is most severe throughout the engine section, including the nozzles, support structures, the surrounding base heat shield, and the grid fins projecting into the flow. Early Falcon 9 flights used aluminum grid fins, but these were later replaced with cast-titanium fins to withstand higher re-entry temperatures.

Landing-Burn

While still supersonic, the bow and tail shock waves extend to the ground as sonic booms, which can be heard over considerable distances downrange of the flight path. As the vehicle passes through transonic conditions, rapid, nonlinear changes in aerodynamic coefficients occur, and shock waves propagate across the grid fins, necessitating careful control of the booster’s trajectory. Below Mach 0.8, the flow is fully subsonic, and the booster, which is now slowing down quickly in the thicker lower atmosphere, is stabilized for the start of the landing burn. The landing burn, which uses a single engine, commences at an altitude of approximately 2.5 km, with the booster traveling at approximately 900 km/h (0.25 km/s). In the final seconds of descent, the landing burn negates the remaining velocity of the booster, i.e., using the rocket equation gives

(173) $\begin{equation*} \Delta V_L = g_0 \, I_{\mathrm{sp}} \ln \left(\frac{M_2}{M_2 - M_{P,\mathrm{L}}}\right) \approx 900 \text{ km/h} \approx 250 \text{ m/s} \end{equation*}$

where $M_2$ is the booster’s mass immediately before the burn and $M_{P,\mathrm{L}}$ is the propellant mass required for the landing burn, i.e.,

(174) $\begin{equation*} M_{P,\mathrm{L}} = M_2 \left( 1 - \exp \left( -\frac{\Delta V_L}{g_0 \, I_{\mathrm{sp}}} \right) \right) \end{equation*}$

Therefore, the total propellant mass required for recovering the booster is

(175) $\begin{equation*} { M_{\rm total} = M_{P,\mathrm{bb}} + M_{P,\mathrm{re}} +M_{P,\mathrm{L}} } \end{equation*}$

including the boost-back maneuver, if one is used. The booster, just about to touch down, is shown in Figure 21.

Falcon 9 first stage booster in the landing-burn just about to touch down at Cape Canaveral, Florida. (SpaceX photo.)

The timing of the landing burn is critical. Starting too early wastes propellant and can result in hovering above the landing pad, whereas starting too late risks a hard landing. A “hover-slam” profile is therefore flown, in which the booster touches down with a low rate of descent and the engines shut down at the instant of touchdown. As of 2025, SpaceX has performed booster landings over 500 times.

Summary & Closure

Payloads launched into space encompass a wide range of missions, including communication satellites, weather satellites, scientific instruments, crewed spacecraft, and planetary exploration probes. Payload mass and size are critical considerations in the design and selection of a launch vehicle. Heavier payloads and missions requiring higher orbital altitudes require larger, more powerful rockets capable of delivering greater energy. Missions to low Earth orbit (LEO), such as those involving Earth observation or communication satellites, require less total energy than missions to geostationary orbit or interplanetary destinations. Launch vehicle configurations are carefully tailored to meet mission requirements, optimizing performance and minimizing costs.

Understanding launch vehicle performance requires an appreciation of the fundamental principles governing rocket flight. The rocket equation describes the ideal velocity increment achievable by a rocket, i.e.,

$\Delta V = V_{\rm eq} \ln \left( \frac{M_0}{M_b} \right)$

where ${M_0}$ and ${M_b}$ are the initial and burnout masses, respectively, and $V_{\rm eq}$ is the equivalent exhaust velocity. However, actual missions require a greater ${\Delta V}$ than the ideal value because of losses associated with gravity, aerodynamic drag, and steering. Gravity losses, which typically amount to 1.5 to 2.5 km/s, arise from the continuous downward pull of Earth’s gravitational field during ascent. Drag losses, associated with aerodynamic resistance in the atmosphere, typically contribute 0.1-0.3 km/s. Steering losses, incurred when the rocket’s thrust vector is not perfectly aligned with its velocity vector, typically range from 0.1 to 0.5 km/s.

Because of the inherent limitations of carrying all of the propellant in a single stage, launch vehicles often employ multiple stages. Staging allows the vehicle to discard empty structural mass and engines, improving the effective mass ratio and making orbit more attainable. The performance of each stage depends critically on its propellant mass fraction, i.e., the fraction of its initial mass that is usable propellant, and structural mass fraction, i.e., the fraction dedicated to tanks, engines, and support systems.

The efficiency of a rocket engine is characterized by its specific impulse, $I_{\rm sp}$ , which relates directly to the exhaust velocity through the expression $V_{\rm eq} = I_{\rm sp} \, g_0$ , where ${g_0}$ is the standard gravitational acceleration. Higher values of $I_{\rm sp}$ enable rockets to achieve a given ${\Delta V}$ with less propellant, increasing mission efficiency. Launch trajectories are carefully designed to minimize energy consumption. Rockets perform a gravity turn shortly after liftoff, arcing over gradually to build horizontal velocity while minimizing steering losses.

Commercial companies such as SpaceX and Blue Origin have contributed to the development of new rocket technologies, including reusable launch vehicles and crewed spacecraft operated in commercial service. SpaceX’s Crew Dragon spacecraft, for example, provides transportation to and from the International Space Station under contract with governmental agencies. Commercial satellite launches for communications, navigation, and Earth observation have also become an established sector of the space industry. Additionally, there is active development of human spaceflight for non-governmental purposes, including space tourism initiatives.

5-Question Self-Assessment Quickquiz

For Further Thought or Discussion

Consider a staged vehicle for which all $V_{\rm eq}$ values are the same. How can ${\Delta V}$ be maximized?
How does a rocket navigate and change its course during flight?
What is the role of aerodynamics in rocket design?
How do scientists and engineers calculate the optimal launch trajectory for a rocket?
What are the challenges of reusing rockets, and why is rocket reusability important?
Can you name some famous rockets and their notable achievements?
How has rocket technology evolved, and what are the prospects for rocketry?

Other Useful Online Resources

To learn more about rockets, launch vehicles, and other spacecraft, check out these helpful online resources:

An article on the history of rockets by NASA
Great video: The Evolution of Space Rockets.
The U.S.’s first person in space, Alan Shepard, launches on a Mercury Redstone rocket.
Saturn V: The rocket that took humans to the Moon – The Saturn V Story.
Experience the flight of Apollo 11.
STS-134 – The final launch of the Space Shuttle Endeavour.
Artemis I Launch to the Moon – Official NASA Broadcast – Nov. 16, 2022.
First test flight of the SpaceX Falcon Heavy.
SpaceX Falcon Heavy launches the NASA Psyche – see video here.
SpaceX Starship SN10 launch and landing – see video here.
Blastoff! Blue Origin launches 33 payloads.
Scott Manley’s excellent YouTube channel on everything to do with spaceflight!

License

Icon for the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License

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

Introduction

Types of Rockets

Launch Vehicles

Spacecraft

Missiles

Miscellaneous Types of Rockets

Rocket Equation

Derivation

Effects of Gravity

Burnout Height

Rocket Mass Breakdown

Structural Mass Breakdown

Impulse

Specific Impulse

Launching into Orbit

Using the Rocket Equation

Kinetic Energy

Potential Energy

Gravity Loss Effect

Aerodynamic Loss Effect

Launch Latitude

Total Delta-V

Steering a Rocket

Stability & Control

Steering Losses

Staged Rocket Vehicles

Serial Staged Rocket

Launch Height – Two-Stage Serial Rocket

Parallel Staged Launcher

Burnout Conditions for a Non-Vertical Launch

Maximum Dynamic Pressure (“Max-q ”)

Variable Thrust (Throttling) Conditions

Missiles & Ballistics

Falcon 9 Booster Re-entry

Stage Separation

Boost-Back Burn

Grid-Fins

Re-Entry Burn

Landing-Burn

Summary & Closure

License

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