Worked Examples: Rockets & Spacecraft Performance

J. Gordon Leishman

87 Worked Examples: Rockets & Spacecraft Performance

Many of these worked examples have been fielded as homework problems or exam questions.

Worked Example #1 – Propellant Needed for a Single-Stage Rocket

A single-stage rocket must achieve an ideal velocity increment of $\Delta V$ = 6,000 m s $^{-1}$ with a payload mass $M_L$ of 12,000 kg. The vehicle’s structural mass coefficient is $\epsilon = 0.06$ , where $\epsilon = M_S/(M_S+M_P)$ . The propellant used in the engine has a specific impulse, $I_{\rm sp}$ , of 325 s. What must be the rocket’s initial mass, ${M_0}$ , and its propellant mass, $M_P$ , to meet these requirements?

The equivalent exhaust velocity is

(1) $\begin{equation*} V_{\rm eq} = I_{\rm sp} \, g_0 = 325 \times 9.807 = 3{,}187~\mbox{m s$^{-1}$} \end{equation*}$

The required initial mass-to-burnout mass ratio is

(2) $\begin{equation*} R = \exp \left( \frac{\Delta V}{V_{\rm eq}} \right) = \exp \left( \frac{6{,}000}{3{,}187} \right) = 6.57 \end{equation*}$

For a single-stage rocket, the payload ratio is

(3) $\begin{equation*} \lambda = \frac{M_L}{M_S+M_P} = \frac{1 - \epsilon R}{R - 1} \end{equation*}$

Therefore,

(4) $\begin{equation*} \lambda = \frac{1 - 0.06 \times 6.57}{6.57 - 1} = 0.109 \end{equation*}$

The initial mass of the rocket is then

(5) $\begin{equation*} M_0 = \left( \frac{1+\lambda}{\lambda} \right) M_L = \left( \frac{1+0.109}{0.109} \right) 12{,}000 = 122{,}000~\mbox{kg} \end{equation*}$

The burnout mass is

(6) $\begin{equation*} M_b = \frac{M_0}{R} = \frac{122{,}000}{6.57} = 18{,}600~\mbox{kg} \end{equation*}$

Finally, the propellant mass is

(7) $\begin{equation*} M_P = M_0 - M_b = 122{,}000 - 18{,}600 = 103{,}400~\mbox{kg} \end{equation*}$

Therefore, even with a payload mass of only 12,000 kg, the rocket’s initial mass must be about 122,000 kg, of which about 103,400 kg is propellant. This result illustrates why high velocity increments place severe demands on single-stage rocket designs.

Worked Example #2

A small rocket is launched vertically from the Earth’s surface with an initial mass of 1,000 kg. If the rocket engine has a thrust of 20 kN and a specific impulse of 250 s, calculate: (a) the rocket’s acceleration at liftoff, and (b) the flight velocity of the rocket after 30 seconds, neglecting aerodynamic drag.

(a) At liftoff, the thrust must overcome the weight of the rocket, so the initial acceleration is

$a_0 = \frac{T - M_0 g_0}{M_0} = \frac{20{,}000 - 1{,}000 \times 9.81}{1{,}000} = 10.19~\mbox{m s$^{-2}$}$

(b) The equivalent exhaust velocity is

$V_{\rm eq} = I_{\rm sp} g_0 = 250.0 \times 9.81 = 2{,}452.5~\mbox{m s$^{-1}$}$

and the thrust is

$T = \overbigdot{m} \, V_{\rm eq}$

so the propellant mass flow rate is

$\overbigdot{m} = \frac{T}{V_{\rm eq}} = \frac{20{,}000}{2{,}452.5} = 8.16~\mbox{kg s$^{-1}$}$

Neglecting aerodynamic drag, the vertical velocity after time $t$ is

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

Therefore, after 30 seconds,

$V(30) = 2{,}452.5 \, \ln \left( \frac{1{,}000}{1{,}000 - 8.16 \times 30.0} \right) - 9.81 \times 30.0 = 394~\mbox{m s$^{-1}$}$

Therefore, the rocket reaches a velocity of approximately 394 m s $^{-1}$ after 30 seconds, neglecting aerodynamic drag.

Worked Example #3

Consider a small single-stage rocket with the following design characteristics: propellant mass = 7,200 kg; structural mass = 800 kg; payload mass = 50 kg. The specific impulse for this rocket is 275 s. The rocket launches from Earth and ascends vertically. The fuel burns steadily, with a burnout time of 60 seconds. Aerodynamic drag can be neglected, but gravity should be accounted for. (a) What will be the value of the burnout velocity? (b) Find the thrust generated by the rocket. (c) What is the initial acceleration of the rocket? (d) What will the speed and acceleration of the rocket be 30 seconds into the flight?

(a) The equivalent velocity from the rocket engine will be

$V_{\rm eq} = I_{\rm sp} \, g_0 = 275 \times 9.81 = 2,698~\mbox{m s$^{-1}$}$

The burnout mass $M_b$ has no fuel left, so that this mass will be the sum of the structural mass and payload mass, i.e.,

$M_b = M_S + M_L = 800 + 50 = 850 \mbox{kg}$

The initial mass of the rocket ${M_0}$ has all the unburned fuel, so

$M_0 = M_S + M_L + M_P = 800 + 50 + 7,200 = 8,050~\mbox{kg}$

With the gravity loss term included, the burnout velocity of the rocket will be

$\Delta V = V_{\rm eq} \ln \left( \frac{M_0}{M_b} \right) - g_0 \, t_b = 2,698 \times \ln(8050/850) - 9.81 \times 60 = 5,477~\mbox{m s$^{-1}$}$

(b) The thrust generated by the rocket will be

$T = \overbigdot{m}_P \, V_{\rm eq} = \frac{M_P}{t_b} V_{\rm eq} = 7,200/60 \times 2,698 = 323.76~\mbox{kN}$

(c) With the consideration of gravity, the initial acceleration will be

$a = \frac{T - M_0 \, g_0}{M_0} = \frac{323{,}760 - 8{,}050 \times 9.81}{8{,}050} = 30.41~\mbox{m s$^{-2}$}$

which is about $3.10g_0$ as a net upward acceleration.

(d) After 30 seconds from liftoff, the mass of the rocket will be

$M_{30} = M_0 - \overbigdot{m}_P \, t = 8,050 - 7,200 \times 30 /60 = 4,450~\mbox{kg}$

and so the acceleration at 30 seconds will be

$a_{30} = \frac{T - M_{30} \, g_0}{M_{30}} = \frac{ 323,760 - 4,450 \times 9.81}{4,450} = 62.94~\mbox{m s$^{-2}$}$

so about 6.4 ${g_0}$ .

$\Delta V_{30} = V_{\rm eq} \ln \left( \frac{M_0}{M_{30}} \right) - g_0 \, t = 2,698 \times \ln(8050/4450) - 9.81 \times 30 = 1,305~\mbox{m s$^{-1}$}$

Including aerodynamic drag on the rocket slightly reduces this value.

Worked Example #4

A single-stage rocket has a total mass of $1.14 \times 10^5$ kg and a burnout mass of $1.11 \times 10^4$ kg, including engines, structural shell, and payload. The rocket launches from Earth and ascends vertically, exhausting its propellant in 2 minutes and 20 seconds. The propellant burns at a steady rate, and the propulsion system’s specific impulse is 240 s. (a) If air resistance and gravity are neglected, what will the rocket’s velocity be at burnout conditions? (b) What thrust does the rocket engine develop at liftoff? (c) What is the initial acceleration of the rocket if gravity is not neglected? (d) What is the acceleration of the rocket at 60 seconds into the flight?

(a) The specific impulse $I_{\rm sp}$ is 240 s, the initial mass $M_0$ is $1.14 \times 10^5$ kg, and the burnout mass $M_b$ is $1.11 \times 10^4$ kg. Therefore, the propellant mass $M_P$ is

$M_P = M_0 - M_b = 1.14 \times 10^5 - 1.11 \times 10^4 = 1.029 \times 10^5~\mbox{kg}$

The rocket equation, neglecting gravity loss and aerodynamic drag, gives the ideal velocity increment, i.e.,

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

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

$V_{\rm eq} = I_{\rm sp} g_0 = 240.0 \times 9.81 = 2{,}354.4~\mbox{m s$^{-1}$}$

The ideal burnout velocity increment is therefore

$\Delta V = V_{\rm eq} \ln \left( \frac{M_0}{M_b} \right) = 2{,}354.4 \ln \left( \frac{114{,}000}{11{,}100} \right) = 5{,}484~\mbox{m s$^{-1}$} = 5.48~\mbox{km s$^{-1}$}$

if gravity loss and aerodynamic drag are not included, as stated.

(b) We need to find the propellant mass flow rate to the engine. Assuming the rate of propellant consumption is constant, then the mass of the rocket $M$ varies over time as

$M = M_0 - M_P \left( \frac{t}{t_b} \right) = M_0 - \left( \frac{M_P}{t_b} \right) t = M_0 - \overbigdot{m}_P \, t$

where $t_b$ is the burnout time and $\overbigdot{m}_P$ is the propellant mass flow rate. Because $t_b = 2$ minutes and 20 seconds = 140 s, then

$\overbigdot{m}_P = \frac{M_P}{t_b} = \frac{1.029 \times 10^5}{140.0} = 735~\mbox{kg s$^{-1}$}$

The thrust produced, $T$ , is

$T = \overbigdot{m}_P V_{\rm eq}$

Therefore,

$T = 735 \times 2{,}354.4 = 1.730 \times 10^6~\mbox{N} = 1.730~\mbox{MN}$

(c) The initial acceleration, including gravity, is

$a_0 = \frac{T - M_0 g_0}{M_0}$

Substituting the values gives

$a_0 = \frac{1.730 \times 10^6 - 1.14 \times 10^5 \times 9.81}{1.14 \times 10^5} = 5.37~\mbox{m s$^{-2}$}$

(d) The mass of the rocket at 60 seconds into the flight is

$M_{60} = M_0 - \overbigdot{m}_P t = 1.14 \times 10^5 - 735 \times 60.0 = 69{,}900~\mbox{kg}$

So, the acceleration at this time is

$a_{60} = \frac{T - M_{60} g_0}{M_{60}}$

Substituting the values gives

$a_{60} = \frac{1.730 \times 10^6 - 69{,}900 \times 9.81}{69{,}900} = 14.94~\mbox{m s$^{-2}$}$

Worked Example #5

Use the rocket equation to determine the burnout velocity and the maximum achievable height, assuming the rocket is launched vertically. Include the gravity loss during powered flight, assume constant gravitational acceleration $g_0$ throughout, and neglect aerodynamic drag. Solve for the burnout velocity and maximum altitude given a burnout time of 60 s. The specific impulse is 250 s, the initial mass is 12,700 kg, and the propellant mass is 8,610 kg.

We are given that the specific impulse $I_{\rm sp}$ is 250 seconds, the initial mass ${M_0}$ is 12,700 kg, and the propellant mass $M_P$ is 8,610 kg. The rocket equation gives us 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 exhaust velocity $V_{\rm eq}$ is given in terms of the specific impulse, i.e.,

$V_{\rm eq} = I_{\rm sp} g_0 = 250 \times 9.81 = 2,452.5~\mbox{m s$^{-1}$}$

The burnout mass $M_b$ is given by

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

So now we have the ideal velocity increment, which is

$\Delta V_{\rm ideal} = V_{\rm eq} \ln \left( \frac{M_0}{M_b} \right) = 2,452.5 \ln \left( \frac{12,700}{4,090} \right)$

so

$\Delta V_{\rm ideal} = 2,452.5 \times 1.133 = 2,778~\mbox{m s$^{-1}$} = 2.78~\mbox{km s$^{-1}$}$

if we do not include the gravity loss term. Including the gravity loss during the 60-second vertical burn gives

$V_b = \Delta V_{\rm ideal} - g_0 t_b = 2,778 - 9.81 \times 60 = 2,190~\mbox{m s$^{-1}$}$

The height reached at burnout is

$H_b = \int_0^{t_b} V dt$

which gives

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

Inserting the values gives

$H_b = 2,452.5 \times 60 \left[ 1 - \frac{4,090}{8,610} \ln \left( \frac{12,700}{4,090} \right) \right] - \frac{1}{2} \times 9.81 \times 60^2$

so

$H_b = 50,291~\mbox{m} = 50.3~\mbox{km}$

The additional altitude gained during the unpowered coast can then be determined by equating the rocket’s kinetic energy at burnout with the change in potential energy between that point and the maximum height, i.e.,

$H_c = \frac{V_b^2}{2g_0} = \frac{2,190^2}{2 \times 9.81} = 244,500~\mbox{m}$

Therefore, under the constant- $g_0$ assumption, the maximum altitude is

$H_{\rm max} = H_b + H_c = 50,291 + 244,500 = 294,791~\mbox{m} \simeq 295~\mbox{km}$

Because gravitational acceleration decreases with altitude, this is an approximate constant-gravity result.

Worked Example #6 – Effect of staging on rocket initial mass

A rocket must provide an ideal velocity increment of $\Delta V$ = 6,000 m s $^{-1}$ to a payload of 12,000 kg. Instead of using a single-stage vehicle, suppose the rocket uses two identical stages, each with a structural mass coefficient $\epsilon = 0.06$ and a specific impulse $I_{\rm sp}$ = 325 s. Assume that the total velocity increment is split equally between the two stages. Estimate the initial mass of the rocket and the total propellant mass required. Compare the result with the single-stage case.

The equivalent exhaust velocity is

$V_{\rm eq} = I_{\rm sp} g_0 = 325 \times 9.807 = 3{,}187~\mbox{m s$^{-1}$}$

Because the total velocity increment is split equally between the two stages, each stage must provide

$\Delta V_s = \frac{6{,}000}{2} = 3{,}000~\mbox{m s$^{-1}$}$

The required mass ratio for each stage is therefore

$R_s = \exp \left( \frac{\Delta V_s}{V_{\rm eq}} \right) = \exp \left( \frac{3{,}000}{3{,}187} \right) = 2.56$

Let $M_L$ be the payload carried by a given stage, meaning the mass above that stage. Let $M_S + M_P$ be the total stage mass, consisting of the structural mass plus the propellant mass. The structural mass coefficient is

$\epsilon = \frac{M_S}{M_S + M_P}$

For one stage, the mass ratio is

$R_s = \frac{M_L + M_S + M_P}{M_L + M_S}$

Because $M_S = \epsilon (M_S + M_P)$ , this becomes

$R_s = \frac{M_L + (M_S + M_P)}{M_L + \epsilon(M_S + M_P)}$

Solving for the ratio of stage mass to payload mass gives

$\frac{M_S + M_P}{M_L} = \frac{R_s - 1}{1 - \epsilon R_s}$

Substituting the values gives

$\frac{M_S + M_P}{M_L} = \frac{2.56 - 1}{1 - 0.06 \times 2.56} = 1.85$

For the upper stage, the payload is the useful payload, so

$M_{L,2} = 12{,}000~\mbox{kg}$

Therefore, the upper-stage mass is

$M_{S,2} + M_{P,2} = 1.85 \times 12{,}000 = 22{,}200~\mbox{kg}$

The initial mass of the upper stage plus payload is therefore

$M_{0,2} = 12{,}000 + 22{,}200 = 34{,}200~\mbox{kg}$

For the lower stage, the payload is the fully fueled upper stage plus the useful payload, so

$M_{L,1} = 34{,}200~\mbox{kg}$

The lower-stage mass is then

$M_{S,1} + M_{P,1} = 1.85 \times 34{,}200 = 63{,}200~\mbox{kg}$

Therefore, the initial mass of the complete two-stage rocket is

$M_0 = 34{,}200 + 63{,}200 = 97{,}400~\mbox{kg}$

The total stage mass is

$(M_S + M_P)_{\rm total} = 22{,}200 + 63{,}200 = 85{,}400~\mbox{kg}$

Because the structural coefficient is $\epsilon = 0.06$ , the propellant fraction of each stage is $1-\epsilon = 0.94$ . Therefore, the total propellant mass is

$M_P = 0.94 \times 85{,}400 = 80{,}300~\mbox{kg}$

Therefore, the two-stage rocket has an initial mass of about 97,400 kg, of which about 80,300 kg is propellant. This is substantially less than the corresponding single-stage result for the same payload and ideal velocity increment, which required an initial mass of about 122,000 kg. The result illustrates why staging is so important in rocket design: after the lower stage has burned its propellant, its empty structure is discarded rather than carried for the rest of the flight.

Worked Example #7 – Gravity losses during vertical rocket ascent

A vertically launched rocket has an initial mass of 12,700 kg and a propellant mass of 8,610 kg. The rocket engine has a specific impulse of 250 s. Neglect aerodynamic drag. Compare the burnout velocity if the same propellant mass is burned in (a) 60 s and (b) 120 s, assuming a constant mass flow rate and constant thrust within each case. Explain why the burn time affects the burnout velocity even though the ideal rocket equation gives the same ideal velocity increment in both cases.

The burnout mass is

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

The equivalent exhaust velocity is

$V_{\rm eq} = I_{\rm sp} g_0 = 250 \times 9.81 = 2{,}452.5~\mbox{m s$^{-1}$}$

The ideal velocity increment from the rocket equation is

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

Substituting the values gives

$\Delta V_{\rm ideal} = 2{,}452.5 \ln \left( \frac{12{,}700}{4{,}090} \right) = 2{,}779~\mbox{m s$^{-1}$}$

This result is the ideal velocity increment before accounting for gravity losses. For a vertical launch with no aerodynamic drag, the burnout velocity is reduced by the gravitational acceleration acting during the powered ascent, so

$V_b = \Delta V_{\rm ideal} - g_0 t_b$

where $t_b$ is the burn time.

For the 60-second burn,

$V_b = 2{,}779 - 9.81 \times 60 = 2{,}190~\mbox{m s$^{-1}$}$

For the 120-second burn,

$V_b = 2{,}779 - 9.81 \times 120 = 1{,}602~\mbox{m s$^{-1}$}$

Therefore, burning the same propellant over a longer time gives the same ideal rocket-equation velocity increment but a lower actual burnout velocity during vertical ascent because gravity has more time to act. The gravity loss is

$\Delta V_g = g_0 t_b$

which is 589 m s $^{-1}$ for the 60-second burn and 1,177 m s $^{-1}$ for the 120-second burn.

The result illustrates an important design tradeoff. A short, high-thrust burn reduces gravity losses, but it requires a higher thrust level and usually a heavier propulsion structure. A longer, lower-thrust burn may reduce engine size and structural loads, but it increases gravity losses and lowers the useful velocity gained during ascent.

Worked Example #8

Consider a two-stage rocket with a payload mass = 60 kg. The first stage has two solid rocket boosters attached. First stage: propellant mass = 7,200 kg, structural mass = 800 kg, and mass flow rate = 80.0 kg s $^{-1}$ . For each booster, the propellant mass is 1,400 kg, the structural mass is 200 kg, and the burn time is 45 seconds. For the second stage: propellant mass = 5,400 kg, structural mass = 600 kg. The specific impulse for the first stage is 250 s, and for the boosters, 290 s. Find the increase in velocity at first-stage separation when using boosters compared to when not using them.

Here is a summary of the information provided:

Propellant mass for stage 1, $M_{P_{1}}$ = 7,200 kg
Propellant mass for stage 2, $M_{P_{2}}$ = 5,400 kg
Propellant mass for each booster, $M_{P_{b}}$ = 1,400 kg
Structural mass for stage 1, $M_{S_{1}}$ = 800 kg
Structural mass for stage 2, $M_{S_{2}}$ = 600 kg
Structural mass for each booster, $M_{S_{b}}$ = 200 kg
Payload mass, $M_{L}$ = 60 kg
Specific impulse for stage 1, $I_{{\rm sp}_{1}}$ = 250.0 s
Specific impulse for boosters, $I_{{\rm sp}_{b}}$ = 290.0 s
Burnout time for each booster, $t_{b_{b}}$ = 45 s
Mass flow rate for stage 1, $\overbigdot{m}_1$ = 80.0 kg s $^{-1}$

The equivalent exhaust velocity for the first stage is

$V_{{\rm eq}_{1}} = I_{{\rm sp}_{1}} g_0 = 250.0 \times 9.81 = 2{,}452.5~\mbox{m s$^{-1}$}$

The initial mass of the rocket without the boosters, $M_{0,\rm noB}$ , is

$M_{0,\rm noB} = M_{P_{1}} + M_{P_{2}} + M_{S_{1}} + M_{S_{2}} + M_{L}$

and inserting the values gives

$M_{0,\rm noB} = 7{,}200.0 + 5{,}400.0 + 800.0 + 600.0 + 60.0 = 14{,}060.0~\mbox{kg}$

The burnout mass without boosters, $M_{b,\rm noB}$ , just before first-stage separation is

$M_{b,\rm noB} = M_{P_{2}} + M_{S_{1}} + M_{S_{2}} + M_{L}$

which gives

$M_{b,\rm noB} = 5{,}400.0 + 800.0 + 600.0 + 60.0 = 6{,}860.0~\mbox{kg}$

The first-stage burn time without boosters is

$t_{b_1} = \frac{M_{P_1}}{\overbigdot{m}_1} = \frac{7{,}200}{80.0} = 90.0~\mbox{s}$

Therefore, the velocity increment for the first stage without the boosters is

$\Delta V_{1,\rm noB} = V_{{\rm eq}_{1}} \ln \left( \frac{M_{0,\rm noB}}{M_{b,\rm noB}} \right) - g_0 \, t_{b_1}$

and inserting the values gives

$\Delta V_{1,\rm noB} = 2{,}452.5 \ln \left( \frac{14{,}060.0}{6{,}860.0} \right) - 9.81 \times 90.0 = 877.08~\mbox{m s$^{-1}$}$

Therefore, for the first stage without boosters, the burnout velocity is

$V_{b_1,\rm noB} = 877.08~\mbox{m s$^{-1}$}$

The equivalent exhaust velocity for each booster is

$V_{{\rm eq}_{b}} = I_{{\rm sp}_{b}} \, g_0 = 290.0 \times 9.81 = 2{,}844.9~\mbox{m s$^{-1}$}$

The mass flow rate for each booster is

$\overbigdot{m}_b = \frac{M_{P_b}}{t_{b_b}} = \frac{1{,}400}{45.0} = 31.1~\mbox{kg s$^{-1}$}$

The net thrust produced by the two boosters is

$T_b = 2 \overbigdot{m}_b V_{{\rm eq}_{b}}$

so

$T_b = 2 \times 31.1 \times 2{,}844.9 = 177{,}016~\mbox{N} = 177.0~\mbox{kN}$

remembering that there are two boosters. Notice that the boosters nearly double the thrust at launch.

For the first 45 seconds, the first stage and the boosters burn together. The initial mass with the two boosters attached is

$M_{0,\rm par} = 7{,}200.0 + 5{,}400.0 + 800.0 + 600.0 + 60.0 + 2(1{,}400.0 + 200.0) = 17{,}260.0~\mbox{kg}$

The total propellant mass burned during the first 45 seconds is

$M_P = 80.0 \times 45.0 + 2(1{,}400.0) = 6{,}400.0~\mbox{kg}$

The effective exhaust velocity during this parallel burn is

$V_{{\rm eq}_{\rm eff}} = \frac{\overbigdot{m}_1 V_{{\rm eq}_{1}} + 2\overbigdot{m}_b V_{{\rm eq}_{b}}} {\overbigdot{m}_1 + 2\overbigdot{m}_b}$

or

$V_{{\rm eq}_{\rm eff}} = \frac{80.0 \times 2{,}452.5 + 2(31.1)(2{,}844.9)} {80.0 + 2(31.1)} = 2{,}624.2~\mbox{m s$^{-1}$}$

Therefore, the velocity increment during the parallel booster burn is

$\Delta V_{\rm par} = V_{{\rm eq}_{\rm eff}} \ln \left( \frac{M_{0,\rm par}}{M_{f,\rm par}} \right) - g_0 t_{b_b}$

which gives

$M_{f,\rm par} = 17{,}260.0 - 6{,}400.0 = 10{,}860.0~\mbox{kg}$

Therefore,

$\Delta V_{\rm par} = 2{,}624.2 \ln \left( \frac{17{,}260.0}{10{,}860.0} \right) - 9.81 \times 45.0 = 774.3~\mbox{m s$^{-1}$}$

The boosters are then discarded, leaving the first stage to burn for an additional 45 seconds, so $t_{b_{1,\rm rem}} = 45$ s. After booster separation, the new initial mass is

$M_{0,\rm rem} = 0.5 \times 7{,}200.0 + 5{,}400.0 + 800.0 + 600.0 + 60.0 = 10{,}460.0~\mbox{kg}$

and the burnout mass just before stage 1 separation is

$M_{b,1} = M_{P_{2}} + M_{S_{1}} + M_{S_{2}} + M_{L}$

Therefore,

$M_{b,1} = 5{,}400.0 + 800.0 + 600.0 + 60.0 = 6{,}860.0~\mbox{kg}$

and so the velocity increment for this part of the launch is

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

Inserting the values gives

$\Delta V_{\rm rem} = 2{,}452.5 \ln \left( \frac{10{,}460.0}{6{,}860.0} \right) - 9.81 \times 45.0 = 593.14~\mbox{m s$^{-1}$}$

Therefore, at stage 1 burnout, the velocity of the rocket with boosters is

$V_{b_1,\rm B} = \Delta V_{\rm par} + \Delta V_{\rm rem} = 774.3 + 593.14 = 1{,}367.5~\mbox{m s$^{-1}$}$

The increase in velocity at first-stage separation obtained by using the boosters is therefore

$\Delta V_{\rm extra} = V_{b_1,\rm B} - V_{b_1,\rm noB} = 1{,}367.5 - 877.08 = 490.4~\mbox{m s$^{-1}$}$

Therefore, under the assumptions used here, the two strap-on boosters increase the velocity at first-stage separation by approximately 490 m s $^{-1}$ . The gain comes from both the extra propellant carried in the boosters and their higher effective exhaust velocity, while the booster structures are discarded after 45 seconds rather than being carried to first-stage burnout.

Worked Example #9 – Idealized Space Shuttle launch velocity estimate

Consider the Space Shuttle as an example of a parallel-burn launch vehicle. The Space Shuttle used LH₂/LOX in the Space Shuttle Main Engines on the Orbiter, supplied from the external tank, together with two solid rocket boosters, or SRBs. It is desired to estimate the ideal velocity increment of the Shuttle stack using a simplified staged-rocket calculation. Neglect gravity losses, aerodynamic drag, steering losses, and the variation of specific impulse during ascent. The information available includes the following:

Orbiter and external tank:
- Orbiter structural mass = 110,000 kg
- Payload mass = 24,000 kg
- External tank structural mass = 30,000 kg
- External tank propellant mass = 720,000 kg
- Specific impulse of the main engines = 454 s
- Main-engine burn time = 480 s

SRB, each:
- Structural mass = 86,000 kg
- Propellant mass = 500,000 kg
- Specific impulse = 269 s
- Burnout time = 124 s

For the Orbiter main engines, the equivalent exhaust velocity, $V_{{\rm eq},c}$ , is

$V_{{\rm eq},c} = I_{{\rm sp},c} \, g_0 = 454 \times 9.81 = 4{,}453.7~\mbox{m s$^{-1}$}$

For the SRBs, the equivalent exhaust velocity, $V_{{\rm eq},b}$ , is

$V_{{\rm eq},b} = I_{{\rm sp},b} \, g_0 = 269 \times 9.81 = 2{,}638.9~\mbox{m s$^{-1}$}$

The fraction of external-tank propellant remaining at SRB burnout is

$\chi = \frac{480 - 124}{480} = 0.741667$

Therefore, the fraction of external-tank propellant burned during the parallel-burn phase is $1-\chi = 0.258333$ .

The mean equivalent exhaust velocity, $\overline{V}_{\rm eq}$ , during the parallel-burn portion is estimated by weighting the exhaust velocities by the propellant masses burned during this interval, i.e.,

$\overline{V}_{\rm eq} = \frac{ 2 M_{P_b} \, V_{{\rm eq},b} + (1 - \chi) \, M_{P_c} \, V_{{\rm eq},c} }{ 2 M_{P_b} + (1 - \chi) M_{P_c} }$

Substituting the values gives

$\overline{V}_{\rm eq} = \frac{ 2 \times 500{,}000 \times 2{,}638.9 + (1 - 0.742) \times 720{,}000 \times 4{,}453.7} {2 \times 500{,}000 + (1 - 0.742)\times 720{,}000} = 2{,}923.5~\mbox{m s$^{-1}$}$

The initial mass at launch is

$M_{0,0} = M_{P_c} + 2M_{P_b} + M_{S_{\rm orb}} + M_{S_{\rm tank}} + 2M_{S_b} + M_L$

and putting in the values gives

$M_{0,0} = 720{,}000 + 2 \times 500{,}000 + 110{,}000 + 30{,}000 + 2 \times 86{,}000 + 24{,}000 = 2{,}056{,}000~\mbox{kg}$

The mass of the vehicle at SRB burnout, just before SRB separation, is

$M_{f,0} = \chi M_{P_c} + M_{S_{\rm orb}} + M_{S_{\rm tank}} + 2M_{S_b} + M_L$

and putting in the values gives

$M_{f,0} = 0.741667 \times 720{,}000 + 110{,}000 + 30{,}000 + 2 \times 86{,}000 + 24{,}000 = 870{,}000~\mbox{kg}$

Therefore, the ideal velocity increment at SRB burnout is

$\Delta V_0 = \overline{V}_{\rm eq} \ln \left( \frac{M_{0,0}}{M_{f,0}} \right)$

which gives

$\Delta V_0 = 2{,}923.5 \ln \left( \frac{2{,}056{,}000}{870{,}000} \right) = 2{,}514.3~\mbox{m s$^{-1}$}$

After SRB separation, the initial mass for the second part of the burn is

$M_{0,1} = \chi M_{P_c} + M_{S_{\rm orb}} + M_{S_{\rm tank}} + M_L$

and putting in the values gives

$M_{0,1} = 0.741667 \times 720{,}000 + 110{,}000 + 30{,}000 + 24{,}000 = 698{,}000~\mbox{kg}$

The final mass at depletion of the external-tank propellant is

$M_{f,1} = M_{S_{\rm orb}} + M_{S_{\rm tank}} + M_L$

or

$M_{f,1} = 110{,}000 + 30{,}000 + 24{,}000 = 164{,}000~\mbox{kg}$

The ideal velocity increment for this part of the burn is

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

which gives

$\Delta V_1 = 4{,}453.7 \ln \left( \frac{698{,}000}{164{,}000} \right) = 6{,}450.6~\mbox{m s$^{-1}$}$

Therefore, the total ideal velocity increment estimated from this simplified calculation is

$\Delta V_{\rm ideal} = \Delta V_0 + \Delta V_1 = 2{,}514.3 + 6{,}450.6 = 8{,}964.9~\mbox{m s$^{-1}$} \approx 9.0~\mbox{km s$^{-1}$}$

This value should not be interpreted as the actual final inertial velocity of the Orbiter. It is an ideal rocket-equation estimate based on the specified masses and specific impulses. A real launch must also overcome gravity losses, aerodynamic drag, steering losses, and other trajectory losses, so the actual ascent analysis requires a full trajectory calculation. Nevertheless, the result is of the same order as the total velocity increment required to reach low Earth orbit.

Worked Example #10 – Change in kinetic energy of a spacecraft

A spacecraft with an initial mass of 10,000 kg travels with a velocity of 5,000 m/s. If a rocket engine with a specific impulse of 250 s is turned on and produces a constant thrust of 50 kN for 100 s, then calculate (a) the final velocity of the spacecraft and (b) the increase in kinetic energy of the remaining spacecraft mass. Neglect external forces. Do not attempt to account for the kinetic energy carried away by the expelled propellant.

(a) The propellant mass flow rate is

$\overbigdot{m} = \frac{T}{I_{\rm sp} \, g_0} = \frac{50{,}000}{250.0 \times 9.81} = 20.39~\mbox{kg s$^{-1}$}$

Therefore, after a burn time of 100 s, the final mass is

$M_f = M_i - \overbigdot{m} \, t_b = 10{,}000 - 20.39 \times 100.0 = 7{,}961~\mbox{kg}$

The equivalent exhaust velocity is

$V_{\rm eq} = I_{\rm sp} g_0 = 250.0 \times 9.81 = 2{,}452.5~\mbox{m s$^{-1}$}$

Using the ideal rocket equation, the velocity increment is

$\Delta V = V_{\rm eq} \ln \left( \frac{M_i}{M_f} \right) = 2{,}452.5 \ln \left( \frac{10{,}000}{7{,}961} \right) = 559.2~\mbox{m s$^{-1}$}$

Therefore, the final velocity of the spacecraft is

$V_f = V_i + \Delta V = 5{,}000 + 559.2 = 5{,}559~\mbox{m s$^{-1}$}$

(b) Because the question asks for the kinetic-energy increase of the mass that remains after the burn, the same final mass $M_f$ is used in both terms. The increase in kinetic energy of the remaining spacecraft mass is

$\Delta KE_{\rm rem} = \frac{1}{2} M_f V_f^2 - \frac{1}{2} M_f V_i^2 = \frac{1}{2} M_f \left( V_f^2 - V_i^2 \right)$

Substituting the values gives

$\Delta KE_{\rm rem} = \frac{1}{2} \times 7{,}961 \times \left(5{,}559^2 - 5{,}000^2\right) = 2.35 \times 10^4~\mbox{MJ}$

The remaining spacecraft mass has gained kinetic energy, as expected, because the rocket engine has increased its velocity. The expelled propellant also carries away kinetic energy. Hence, this value is not the total change in kinetic energy of all the material originally on board, nor is it the total energy released by the propulsion system.

Worked Example #11 – Effect of structural mass on rocket performance

A single-stage rocket carries a payload of 200 kg. It uses 3,000 kg of propellant and has a structural mass of 400 kg. The engine has a specific impulse of 300 s. For this ideal rocket-equation comparison, neglect gravity loss and aerodynamic drag.

Compute the rocket’s ideal increment in burnout velocity.
A design proposal reduces the structural mass to 300 kg while keeping all other parameters constant. What is the new ideal burnout velocity increment?
Based on your answers, briefly discuss whether it is more effective to reduce dry mass by 100 kg or increase propellant mass by 100 kg while keeping the same structure. You may assume all else remains constant.

The ideal burnout velocity increment can be found from the rocket equation, i.e.,
$\Delta V = I_{\rm sp} g_0 \ln\left( \frac{M_0}{M_b} \right)$

where the initial mass is

$M_0 = \mbox{payload mass} + \mbox{structural mass} + \mbox{propellant mass}$

and the burnout mass is

$M_b = \mbox{payload mass} + \mbox{structural mass}$

Therefore,

$M_0 = 200 + 400 + 3{,}000 = 3{,}600~\mbox{kg}$

and

$M_b = 200 + 400 = 600~\mbox{kg}$

Using the rocket equation gives

$\Delta V = 300 \times 9.81 \times \ln\left( \frac{3{,}600}{600} \right) = 5{,}273~\mbox{m s$^{-1}$}$
With the structural mass reduced to 300 kg, the new masses are
$M_0 = 200 + 300 + 3{,}000 = 3{,}500~\mbox{kg}$

and

$M_b = 200 + 300 = 500~\mbox{kg}$

so that

$\Delta V = 300 \times 9.81 \times \ln\left( \frac{3{,}500}{500} \right) = 5{,}727~\mbox{m s$^{-1}$}$

Reducing the structural mass by 100 kg increases the ideal burnout velocity increment from approximately 5,273 m s $^{-1}$ to 5,727 m s $^{-1}$ , a gain of about 454 m s $^{-1}$ .
If instead the propellant mass were increased by 100 kg, keeping the structural mass at 400 kg, the new masses would be
$M_0 = 200 + 400 + 3{,}100 = 3{,}700~\mbox{kg}$

and

$M_b = 200 + 400 = 600~\mbox{kg}$

so that

$\Delta V = 300 \times 9.81 \times \ln\left( \frac{3{,}700}{600} \right) = 5{,}354~\mbox{m s$^{-1}$}$

Increasing the propellant mass by 100 kg increases the ideal velocity increment from 5,273 m s $^{-1}$ to 5,354 m s $^{-1}$ , a gain of approximately 81 m s $^{-1}$ . Therefore, at this design point, reducing structural (dry) mass is much more effective than increasing propellant mass by the same amount. The reason is that dry mass remains with the vehicle after burnout, whereas propellant mass is expended during the burn.

Worked Example #12 – Falcon 9-class booster landing burn

A reusable first-stage booster similar in concept to the Falcon 9 first stage is returning for a vertical landing. During the final landing burn, assume the booster has an initial mass of $M_0$ = 25,000 kg, a downward vertical speed of $V_d$ = 180 m s $^{-1}$ , and uses one sea-level rocket engine producing a constant thrust of $T$ = 845 kN. The engine has a specific impulse of $I_{\rm sp}$ = 282 s. Neglect aerodynamic drag during the short terminal burn and assume the thrust acts vertically upward.

Determine the propellant mass flow rate during the landing burn.
Determine the burn time required to reduce the vertical speed to zero.
Determine the propellant mass used during the landing burn.
Estimate the altitude above the landing point at which the burn must begin.
Explain why the booster cannot hover efficiently and why a precisely timed landing burn is required.

The equivalent exhaust velocity of the engine is

$V_{\rm eq} = I_{\rm sp} g_0 = 282 \times 9.81 = 2{,}766~\mbox{m s$^{-1}$}$

The propellant mass flow rate is found from

$T = \overbigdot{m} V_{\rm eq}$

so that

$\overbigdot{m} = \frac{T}{V_{\rm eq}} = \frac{845{,}000}{2{,}766} = 305.5~\mbox{kg s$^{-1}$}$

During the vertical landing burn, the rocket’s upward velocity increment must overcome the initial downward velocity and the velocity loss caused by gravity during the burn. Therefore,

$V_d = V_{\rm eq} \ln \left( \frac{M_0}{M_0 - \overbigdot{m} t_b} \right) - g_0 t_b$

Substituting the known values gives

$180 = 2{,}766 \ln \left( \frac{25{,}000}{25{,}000 - 305.5 t_b} \right) - 9.81 t_b$

Solving this equation gives

$t_b = 7.05~\mbox{s}$

The propellant mass used during the landing burn is

$M_P = \overbigdot{m} t_b = 305.5 \times 7.05 = 2{,}153~\mbox{kg}$

Therefore, the booster mass at touchdown is

$M_f = M_0 - M_P = 25{,}000 - 2{,}153 = 22{,}847~\mbox{kg}$

The altitude at which the burn must begin can be estimated by integrating the downward velocity during the burn. Taking downward velocity as positive, the downward velocity during the burn is

$V_{\rm down}(t) = V_d - \left[ V_{\rm eq} \ln \left( \frac{M_0}{M_0 - \overbigdot{m}t} \right) - g_0 t \right]$

so the required starting height is

$h_b = \int_0^{t_b} V_{\rm down}(t) \, dt$

Using the values above gives

$h_b \approx 648~\mbox{m}$

Therefore, under these assumptions, the landing burn must begin at an altitude of about 650 m above the landing point.

Finally, the initial thrust-to-weight ratio at the beginning of the burn is

$\frac{T}{W_0} = \frac{845{,}000}{25{,}000 \times 9.81} = 3.45$

and at touchdown, it is

$\frac{T}{W_f} = \frac{845{,}000}{22{,}847 \times 9.81} = 3.77$

Because the thrust-to-weight ratio is much greater than 1, the booster cannot hover efficiently on a single engine. Instead, it must perform a short, carefully timed landing burn. If the burn starts too early, the vehicle will slow down too high above the landing point and will either waste propellant or begin to accelerate upward. If the burn starts too late, there will not be enough time or altitude to arrest the descent before touchdown. This is why the terminal landing burn of a reusable booster is a demanding problem in guidance, navigation, control, and propulsion.

Worked Example #13 – Entry burn and dynamic pressure of a reusable booster

A reusable first-stage booster enters the upper atmosphere at high speed before returning to land. To reduce aerodynamic and thermal loads, the booster performs a short retropropulsive entry burn. Assume that before the entry burn, the booster has a speed of $V_1$ = 1,800 m s $^{-1}$ at an altitude where the air density is $\varrho$ = 0.020 kg m $^{-3}$ . The entry burn reduces the speed to $V_2$ = 1,350 m s $^{-1}$ before the vehicle reaches the denser part of the atmosphere. Assume that the booster has a reference area of $S$ = 10.5 m ${^2}$ and an effective drag coefficient of $C_D = 1.2$ during this part of the descent.

Calculate the dynamic pressure before and after the entry burn.
Calculate the corresponding aerodynamic drag force before and after the entry burn.
Determine the percentage reduction in dynamic pressure and drag force caused by the entry burn.
Estimate the propellant mass required for the entry burn if the booster mass before the burn is $M_0$ = 38,000 kg and the engines have an effective specific impulse of $I_{\rm sp}$ = 300 s.
Explain the design trade between using propellant for the entry burn and reducing aerodynamic and thermal loads on the vehicle.

The dynamic pressure is

$q = \frac{1}{2} \varrho V^2$

Before the entry burn, the dynamic pressure is

$q_1 = \frac{1}{2} \times 0.020 \times 1{,}800^2 = 32{,}400~\mbox{Pa} = 32.4~\mbox{kPa}$

After the entry burn, the dynamic pressure is

$q_2 = \frac{1}{2} \times 0.020 \times 1{,}350^2 = 18{,}225~\mbox{Pa} = 18.2~\mbox{kPa}$

The aerodynamic drag force is

$D = q S C_D$

Before the entry burn, the drag force is

$D_1 = 32{,}400 \times 10.5 \times 1.2 = 408{,}240~\mbox{N} = 408~\mbox{kN}$

After the entry burn, the drag force is

$D_2 = 18{,}225 \times 10.5 \times 1.2 = 229{,}635~\mbox{N} = 230~\mbox{kN}$

The percentage reduction in dynamic pressure is

$\frac{q_1 - q_2}{q_1} \times 100 = \frac{32.4 - 18.2}{32.4} \times 100 = 43.8\%$

Because the reference area and drag coefficient are assumed unchanged, the drag force is reduced by essentially the same percentage, i.e.,

$\frac{D_1 - D_2}{D_1} \times 100 = \frac{408 - 230}{408} \times 100 = 43.8\%$

The velocity reduction produced by the entry burn is

$\Delta V = V_1 - V_2 = 1{,}800 - 1{,}350 = 450~\mbox{m s$^{-1}$}$

The equivalent exhaust velocity is

$V_{\rm eq} = I_{\rm sp} g_0 = 300 \times 9.81 = 2{,}943~\mbox{m s$^{-1}$}$

Using the rocket equation,

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

so that

$\frac{M_0}{M_f} = \exp \left( \frac{\Delta V}{V_{\rm eq}} \right) = \exp \left( \frac{450}{2{,}943} \right) = 1.165$

Therefore,

$M_f = \frac{M_0}{1.165} = \frac{38{,}000}{1.165} = 32{,}618~\mbox{kg}$

and the propellant mass used during the entry burn is

$M_P = M_0 - M_f = 38{,}000 - 32{,}618 = 5{,}382~\mbox{kg}$

This example shows why a reusable booster may spend propellant before reaching the densest part of the atmosphere. The entry burn reduces velocity, and because dynamic pressure varies as $V^2$ , even a modest reduction in speed can substantially lower aerodynamic loads. The same reduction in speed also reduces aerodynamic heating, which is especially important for protecting engines, tanks, control surfaces, and thermal protection systems. The trade-off in design is that every kilogram of propellant reserved for entry and landing is a kilogram that cannot be used to increase payload or ascent performance. Reusable launch vehicle design is therefore a balance between ascent performance, recovery propellant margin, structural loads, thermal protection, and landing reliability.

Worked Example #14 – Booster catch versus landing legs

A very large reusable booster is to be recovered vertically after launch. Two recovery concepts are being compared. In Concept A, the booster lands on deployable landing legs. In Concept B, the booster is caught by a launch tower just before touchdown, so the booster does not need to carry landing legs during ascent. Assume the dry booster mass, excluding landing legs, is $M_b$ = 185,000 kg. Landing legs would add $M_{\rm legs}$ = 8,000 kg to the booster. The terminal landing maneuver requires a velocity increment of $\Delta V$ = 140 m s $^{-1}$ , including gravity losses and control margin. The engines used for the landing maneuver have an effective specific impulse of $I_{\rm sp}$ = 330 s.

Calculate the propellant mass required for the terminal landing maneuver if landing legs are carried.
Calculate the propellant mass required for the same terminal maneuver if the tower catches the booster and the booster does not carry landing legs.
Determine the propellant savings during the terminal burn resulting from the removal of the landing legs.
Estimate the kinetic energy associated with a residual vertical speed of 2 m s $^{-1}$ at the catch point.
Discuss the design trade between carrying landing legs and using a tower-catch system.

For a given required velocity increment, the rocket equation gives

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

where

$V_{\rm eq} = I_{\rm sp} g_0$

In this case,

$V_{\rm eq} = 330 \times 9.81 = 3{,}237~\mbox{m s$^{-1}$}$

Therefore, the required mass ratio for the terminal maneuver is

$\frac{M_0}{M_f} = \exp \left( \frac{\Delta V}{V_{\rm eq}} \right) = \exp \left( \frac{140}{3{,}237} \right) = 1.0442$

For Concept A, with landing legs, the final mass after the terminal burn is

$M_{f,A} = M_b + M_{\rm legs} = 185{,}000 + 8{,}000 = 193{,}000~\mbox{kg}$

The initial mass before the terminal burn is therefore

$M_{0,A} = 1.0442 M_{f,A} = 1.0442 \times 193{,}000 = 201{,}531~\mbox{kg}$

so the propellant mass required for the terminal burn is

$M_{P,A} = M_{0,A} - M_{f,A} = 201{,}531 - 193{,}000 = 8{,}531~\mbox{kg}$

For Concept B, with tower catch and no landing legs, the final mass after the terminal burn is

$M_{f,B} = M_b = 185{,}000~\mbox{kg}$

The initial mass before the terminal burn is therefore

$M_{0,B} = 1.0442 M_{f,B} = 1.0442 \times 185{,}000 = 193{,}177~\mbox{kg}$

so the propellant mass required for the terminal burn is

$M_{P,B} = M_{0,B} - M_{f,B} = 193{,}177 - 185{,}000 = 8{,}177~\mbox{kg}$

The direct propellant saving during the terminal burn is therefore

$\Delta M_P = M_{P,A} - M_{P,B} = 8{,}531 - 8{,}177 = 354~\mbox{kg}$

However, the larger benefit is not only this terminal-burn propellant saving. If landing legs are removed, then the booster does not have to accelerate the landing-leg mass during ascent, boostback, entry, and landing. The apparent saving of 8,000 kg in dry mass, therefore, has a compounding effect throughout the mission.

The kinetic energy associated with a low residual vertical speed at the catch point is

$KE = \frac{1}{2} M_b V^2$

For a residual vertical speed of 2 m s $^{-1}$ , this gives

$KE = \frac{1}{2} \times 185{,}000 \times 2^2 = 370{,}000~\mbox{J} = 0.37~\mbox{MJ}$

This energy is small compared with the kinetic energy during atmospheric entry, but it must be absorbed or reacted by the catch system, vehicle structure, and control system without exceeding local structural loads.

The landing-leg concept is mechanically self-contained, allowing the booster to land on a prepared pad or ship. However, it requires the vehicle to carry landing-leg mass throughout ascent and recovery. The tower-catch concept removes this dry mass from the booster but transfers complexity to the ground system and requires highly accurate guidance, navigation, and control, engine throttling, and structural load management near the catch point. This trade illustrates a central design question in reusable launch vehicles: whether recovery hardware should be carried on the vehicle or moved to the ground infrastructure.

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

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