These problems are mixed on purpose. None says which chapter it belongs to, most need two or three chapters at once, and several ask for a decision about the reference launcher that has to be defended with numbers. Data that a problem needs are stated in it. Solutions, hints, and answers are keyed R.1 to R.20 in the worked solutions.

1. The customer of Revision C raises the payload from \(5000\) to \(6500\ \mathrm{kg}\) for the same orbit. Keep the budget of \(9259.8\ \mathrm{m/s}\) , the exhaust velocities \(3027.1\) and \(3383.3\ \mathrm{m/s}\) , a first-stage hardware coefficient of \(0.06\) with a fairing fixed at \(570\ \mathrm{kg}\) counted as first-stage inert mass, and a second-stage coefficient of \(0.08246\) , which includes the disposal reserve. (a) Re-run the optimal staging, iterating on the first stage’s effective coefficient, and give the liftoff mass, both mass ratios, and the stack mass at second-stage ignition. (b) Explain why the liftoff mass grows by less than the payload’s \(30\) percent. (c) Revision C has three first-stage engines of \(692.2\ \mathrm{kN}\) each at sea level; decide how many engines of that size the new vehicle should carry and give the thrust-to-weight ratio at liftoff. (d) State which lines of the budget your choice in (c) would move, and in which direction.

   Solution

   Method. Theorem 2.4 needs a structural coefficient for each stage, and the fairing is a fixed mass, so the first stage’s effective coefficient depends on how large the stage turns out to be. Iterate: guess the effective coefficient; solve \(\Phi(\eta)=9259.8\ \mathrm{m/s}\) by bisection; size the stacks from the payload with \(m_{0,2}=m_L/\pi_2\) and \(m_{0,1}=m_{0,2}/\pi_1\) ; split the first stage into propellant and inert mass; set the hardware to \(0.06/0.94\) of the propellant; and recompute the coefficient as hardware plus \(570\ \mathrm{kg}\) over hardware plus \(570\ \mathrm{kg}\) plus propellant. Repeat until it stops moving.

   Check of the procedure. For \(5000\ \mathrm{kg}\) the loop returns an effective coefficient of \(0.06403\) and a liftoff mass of \(162\,891\ \mathrm{kg}\) , which is Revision C to within the rounding of its inputs.

   (a) For \(6500\ \mathrm{kg}\) the loop converges to an effective coefficient of \(0.06311\) , \(\eta=4.525\times10^{-4}\ \mathrm{s/m}\) , and

   \[ \mathcal R_1=4.278,\qquad \mathcal R_2=4.206,\qquad m_{0,2}=38\,403\ \mathrm{kg},\qquad m_{0,1}=210\,830\ \mathrm{kg}. \]

   The first stage holds \(161\,545\ \mathrm{kg}\) of propellant and \(10\,311\ \mathrm{kg}\) of hardware.

   (b) The mass grows by the factor \(1.294\) . A vehicle scaled in proportion would weigh \(1.3\) times as much, \(928\ \mathrm{kg}\) more than this one. Every mass scales with the payload except the fairing, which stays at \(570\ \mathrm{kg}\) , so the first stage’s effective coefficient falls from \(0.06403\) to \(0.06311\) , and a lower coefficient is a lighter vehicle. The optimum also shifts a little velocity back to the first stage: its mass ratio rises from \(4.251\) to \(4.278\) .

   (c) A thrust-to-weight ratio of \(1.30\) needs \((1.30)(210\,830\ \mathrm{kg})(9.80665\ \mathrm{m/s^2})=2687.8\ \mathrm{kN}\) , which is \(2687.8/692.2=3.88\) engines. Three engines give a ratio of \(1.004\) and the vehicle would hang on the pad. Four give \(2768.8\ \mathrm{kN}\) and \(\psi_0=1.339\) . Four engines, then, with a small surplus of thrust.

   (d) The gravity loss falls, because the steep part of the flight is shorter; scaling the \(222\ \mathrm{m/s}\) that Example 7.7 found for a change of \(0.2\) in \(\psi_{0,1}\) gives roughly \(40\ \mathrm{m/s}\) . The drag loss and the pitch-over impulse rise a little, and so do the maximum dynamic pressure and the acceleration at burnout, which are loads and not budget lines. The budget must then be taken again from the new vehicle’s own flight, which is one more pass of the loop of Section 12.8.

2. A reviewer is handed this velocity budget for the reference mission (\(5000\ \mathrm{kg}\) to a \(400\ \mathrm{km}\) circular orbit, due east from latitude \(28.5^\circ\) , through a \(200\ \mathrm{km}\) parking orbit), with every line in metres per second: speed of the target orbit, \(7668.6\) ; gravity loss, taken as \(g_0\) times the total burn time of \(461\ \mathrm s\) , \(4520.9\) ; energy of altitude, taken as \(\sqrt{2g_0z}\) for \(z=400\ \mathrm{km}\) , \(2800.9\) ; drag, \(49.0\) ; steering, \(3.5\) ; rotation credit, \(-465.1\) ; transfer, \(115.7\) ; disposal burn, \(102.3\) ; margin, \(191.7\) ; total, \(14\,987.5\) . (a) Find the six errors and say what principle each one violates. (b) Rebuild the budget line by line, using the model’s losses of \(1470.8\) , \(49.0\) , and \(3.5\ \mathrm{m/s}\) and the reference mission’s reserve of \(100\ \mathrm{m/s}\) , and state the total. (c) With stage exhaust velocities of \(3028.4\) and \(3383.3\ \mathrm{m/s}\) and structural coefficients of \(0.06\) and \(0.08246\) , compute the liftoff mass that each total implies for the \(5000\ \mathrm{kg}\) payload, and say what the flawed figure would have told a manager.

   Solution

   (a) The orbital speed is that of the wrong orbit. The ascent ends in the \(200\ \mathrm{km}\) parking orbit, at \(\sqrt{\mu_E/r}=7784.3\ \mathrm{m/s}\) ; the climb to \(400\ \mathrm{km}\) is the transfer line. Definition 9.2 takes the speed of the orbit entered at the end of the ascent.

   The gravity loss is the vertical-burn formula. The product \(g\,t_b\) is the integral \(\int g\sin\theta\,\mathrm dt\) only for a path that stays vertical. On the flown path the model gives \(1470.8\ \mathrm{m/s}\) , so the line is too large by \(3050.1\ \mathrm{m/s}\) .

   The altitude is charged twice. Since \(\sin\theta\,\mathrm dt=\mathrm dz/v\) , the gravity loss already is the velocity given up to climbing. Proposition 7.2 has no separate term for height.

   The rotation credit is the equator’s. The ground moves at \(\omega_ER_E\cos\phi\) , which is \(408.7\ \mathrm{m/s}\) at \(28.5^\circ\) and \(465.1\ \mathrm{m/s}\) only at the equator. The credit is too generous by \(56.4\ \mathrm{m/s}\) .

   Disposal is not a budget line. The burn is made by the empty stage after the payload has gone. It is charged as inert mass through \(\varepsilon_{2,\mathrm{eff}}=0.08246\) , and a budget line would price it as if the payload were carried through it.

   The allowance is called a margin, and there is no reserve. The \(191.7\ \mathrm{m/s}\) is an estimate of real losses that the model understates. It is not spare. A reserve of \(100\ \mathrm{m/s}\) against the unknown has to be added.

   (b) Line by line,

   \[ 7784.3+1470.8+49.0+3.5-408.7+115.7+191.7+100.0=9306.3\ \mathrm{m/s}. \]

   This is the budget of Example 9.7. The flawed total is \(5681\ \mathrm{m/s}\) higher. The two gravity lines alone overstate the budget by \(3050.1+2800.9=5851\ \mathrm{m/s}\) ; the other four errors partly cancel and take \(170\ \mathrm{m/s}\) back.

   (c) Theorem 2.4 with the stated stages gives \(162\,988\ \mathrm{kg}\) for \(9306.3\ \mathrm{m/s}\) , which is Revision B to rounding. The same stages could supply at most \(c_1\ln(1/\varepsilon_1)+c_2\ln(1/\varepsilon_{2,\mathrm{eff}})=16\,963\ \mathrm{m/s}\) with no payload at all, so \(14\,987.5\ \mathrm{m/s}\) is close to the edge of what they can do: the optimum has mass ratios of \(11.99\) and \(9.08\) and a liftoff mass of \(6.67\times10^{6}\ \mathrm{kg}\) , forty-one times Revision B. A manager shown that figure would conclude that a two-stage kerosene launcher for five tonnes must be more than twice the mass of a Saturn V, and cancel a project that in fact needs \(163\ \mathrm t\) . The exponential punishes an inflated budget as hard as it punishes a real one.

3. The reference mission is changed to a circular orbit at \(800\ \mathrm{km}\) , still inclined at \(28.5^\circ\) and still reached through the \(200\ \mathrm{km}\) parking orbit. Start from Revision B: a budget of \(9306.3\ \mathrm{m/s}\) , of which \(115.7\ \mathrm{m/s}\) is the transfer to \(400\ \mathrm{km}\) ; exhaust velocities of \(3028.4\) and \(3383.3\ \mathrm{m/s}\) ; structural coefficients of \(0.06\) and \(0.08\) ; and a liftoff mass of \(162\,985\ \mathrm{kg}\) . (a) Compute the two transfer impulses, the coast time between them, and the period of the new orbit. (b) Compute the burn that lowers the empty stage’s perigee from \(800\ \mathrm{km}\) to \(50\ \mathrm{km}\) , and the effective second-stage coefficient that carries the propellant for it. (c) Assemble the new budget and re-run the staging for the new liftoff mass. (d) Separate the increase into the part due to the budget, estimated with the multiplier \(4.491\times10^{-4}\ \mathrm{s/m}\) , and the part due to disposal, and explain why the disposal burn only doubled although the transfer nearly tripled.

   Solution

   (a) With \(r_1=6578.137\ \mathrm{km}\) , \(r_2=7178.137\ \mathrm{km}\) , and \(a_{\mathrm{tr}}=6878.137\ \mathrm{km}\) , Proposition 9.1 gives

   \[ \Delta v_{\mathrm I}=167.95\ \mathrm{m/s},\qquad \Delta v_{\mathrm{II}}=164.32\ \mathrm{m/s},\qquad \Delta v_{\mathrm I}+\Delta v_{\mathrm{II}}=332.27\ \mathrm{m/s}. \]

   The coast is half the period of the transfer ellipse, \(\pi\sqrt{a_{\mathrm{tr}}^3/\mu_E}=47.31\ \mathrm{min}\) , and the period of the \(800\ \mathrm{km}\) orbit is \(100.87\ \mathrm{min}\) .

   (b) Circular speed at \(800\ \mathrm{km}\) is \(7451.8\ \mathrm{m/s}\) . The ellipse from \(800\ \mathrm{km}\) down to \(50\ \mathrm{km}\) has semi-major axis \(6803.137\ \mathrm{km}\) and, by the vis-viva equation, an apogee speed of \(7243.5\ \mathrm{m/s}\) . So

   \[ \Delta v_{\mathrm{dis}}=7451.8-7243.5=208.3\ \mathrm{m/s},\qquad \varepsilon_{2,\mathrm{eff}}=(0.08)\exp\Bigl(\frac{208.3}{3383.3}\Bigr)=0.08508 . \]

   (c) Only the transfer line changes: \(9306.3-115.7+332.3=9522.9\ \mathrm{m/s}\) . Theorem 2.4 with \(c=(3028.4,\ 3383.3)\ \mathrm{m/s}\) and \(\varepsilon=(0.06,\ 0.08508)\) gives mass ratios of \(4.684\) and \(4.189\) and a liftoff mass of \(182\,352\ \mathrm{kg}\) , which is \(11.9\) percent more than Revision B.

   (d) The budget rose by \(216.6\ \mathrm{m/s}\) , and the multiplier predicts \(162\,985\exp\bigl[(4.491\times10^{-4})(216.6)\bigr]=179\,633\ \mathrm{kg}\) ; staging for the new budget with the old coefficient gives \(179\,766\ \mathrm{kg}\) , a little more, because the tangent underestimates a convex function. The larger disposal reserve accounts for the remaining \(2586\ \mathrm{kg}\) . For small changes of radius each tangential impulse is about a quarter of the circular speed times the fractional change of the radius it moves. The transfer raises the far side of the orbit by \(600\ \mathrm{km}\) in place of \(200\ \mathrm{km}\) , three times as far, and costs \(2.87\) times as much. The disposal lowers the perigee by \(750\ \mathrm{km}\) in place of \(350\ \mathrm{km}\) , a factor of \(2.14\) , and costs \(2.04\) times as much. The \(50\ \mathrm{km}\) target is fixed, so most of the drop was already being paid for.

4. An engine burns this course’s model chamber gas (\(T_c=3600\ \mathrm K\) , molar mass \(23\ \mathrm{g/mol}\) , \(\gamma=1.2\) ) at \(p_c=7.0\ \mathrm{MPa}\) through a throat \(0.30\ \mathrm m\) in diameter and a nozzle of area ratio \(30\) (illustrative values). (a) Find the characteristic velocity and the mass flow. (b) Invert the area-Mach relation for the exit Mach number, and find the exit pressure, temperature, and velocity. (c) Find the ideal thrust coefficient, thrust, and specific impulse in vacuum. (d) In the exponential atmosphere with \(p_{\mathrm{SL}}=101.325\ \mathrm{kPa}\) and \(H=8.5\ \mathrm{km}\) , find the altitude above which the nozzle flows full by the rule \(p_e\ge0.4\,p_a\) , the altitude at which it is matched, and the thrust and specific impulse at \(20\ \mathrm{km}\) . (e) Say what kind of stage this engine suits, and say which of the four results of (d) would change if the chamber ran hotter at the same pressure, and why the others would not.

   Solution

   (a) The gas constant is \(R=8.3145/0.023=361.5\ \mathrm{J/(kg\cdot K)}\) and \(\Gamma=0.6485\) for \(\gamma=1.2\) , so by Theorem 5.3 \(c^\ast=\sqrt{(361.5)(3600)}/0.6485=1759\ \mathrm{m/s}\) , whatever the chamber pressure. The throat area is \(\pi(0.15\ \mathrm m)^2=0.07069\ \mathrm{m^2}\) , and

   \[ \dot m=\frac{p_cA_t}{c^\ast}=\frac{(7.0\times10^{6}\ \mathrm{Pa})(0.07069\ \mathrm{m^2})}{1759\ \mathrm{m/s}}=281.3\ \mathrm{kg/s}. \]

   (b) Bisection on the supersonic branch of \(\mathcal A(M)=30\) gives \(M_e=4.039\) . Then \(1+0.1M_e^2=2.6314\) , and

   \[ \frac{p_e}{p_c}=(2.6314)^{-6}=0.003012,\qquad p_e=21.08\ \mathrm{kPa},\qquad T_e=\frac{3600}{2.6314}=1368\ \mathrm K,\qquad v_e=M_e\sqrt{\gamma RT_e}=3112\ \mathrm{m/s}. \]

   (c) Theorem 5.5 in vacuum gives \(C_F=1.859\) , so

   \[ F_{\mathrm{vac}}=C_F\,p_cA_t=(1.859)(7.0\times10^{6})(0.07069)\ \mathrm N=920.0\ \mathrm{kN},\qquad I_{sp}=\frac{F_{\mathrm{vac}}}{\dot m\,g_0}=333.5\ \mathrm s . \]

   As a check, \(v_e+p_eA_e/\dot m\) gives the same effective exhaust velocity.

   (d) At sea level \(p_e/p_a=0.208\) , well under \(0.4\) , so the attached-flow figure of \(705.1\ \mathrm{kN}\) means nothing: the flow separates. The nozzle flows full once \(p_a\le p_e/0.4=52.7\ \mathrm{kPa}\) , that is above \(H\ln(101.325/52.7)=5.56\ \mathrm{km}\) , and it is matched where \(p_a=21.08\ \mathrm{kPa}\) , at \(13.34\ \mathrm{km}\) . At \(20\ \mathrm{km}\) the ambient pressure is \(9.63\ \mathrm{kPa}\) , and Corollary 3.5 with the exit area of \(2.121\ \mathrm{m^2}\) gives \(F_{\mathrm{vac}}-p_aA_e=899.5\ \mathrm{kN}\) and \(326.1\ \mathrm s\) .

   (e) It is an engine for a stage that lights at altitude: an upper stage, or a core that ignites after boosters have carried it above five or six kilometres. Its bell is \(1.64\ \mathrm m\) across. Of the four results of (d), only the specific impulse would change. Every pressure in the nozzle is a fixed fraction of \(p_c\) that depends only on \(\gamma\) and the area ratio, and separation and matching are comparisons of pressures, so the two altitudes stay where they are. The thrust is \(C_F\,p_cA_t-p_aA_e\) , which contains no temperature either. The specific impulse is \(F/(\dot m\,g_0)\) with \(\dot m=p_cA_t/c^\ast\) , and \(c^\ast\) rises as \(\sqrt{T_c}\) : a chamber at \(4000\ \mathrm K\) would give the same \(899.5\ \mathrm{kN}\) at \(20\ \mathrm{km}\) from \(5\) percent less flow, at \(343.7\ \mathrm s\) .

5. Should Revision C have had three stages? Keep its budget of \(9259.8\ \mathrm{m/s}\) and its first stage, with \(c_1=3027.1\ \mathrm{m/s}\) and an effective coefficient of \(0.06403\) , held at that value although a smaller first stage would carry the same fairing. Give the second and third stages \(c=3383.3\ \mathrm{m/s}\) . The second, which is now dropped short of orbit, has a coefficient of \(0.08\) and needs no disposal reserve; the small third stage has a hardware coefficient of \(0.12\) (illustrative), raised by the factor \(\exp(102.3/3383.3)\) to carry its disposal burn. (a) Solve the three-stage optimum for the multiplier, the three mass ratios, the three velocity increments, and the liftoff mass for a \(5000\ \mathrm{kg}\) payload. (b) Compare with Revision C’s \(162\,889\ \mathrm{kg}\) . (c) Repeat with a third-stage hardware coefficient of \(0.14\) . Then estimate the hardware of the third stage found in (a) by the rules this course used for its second stage: tanks at \(15.1\ \mathrm{kg}\) per cubic metre of propellant, at a bulk density of \(1015\ \mathrm{kg/m^3}\) ; an engine whose thrust is \(0.9\) of the stack’s weight at ignition, with a thrust-to-weight ratio of \(74.4\) and a mounting of a quarter of its mass; \(400\ \mathrm{kg}\) of fixed items; \(12\ \mathrm{kg}\) of pressurization; residual propellant of \(1.5\) percent of the load; and \(20\) percent growth on the hardware (illustrative values). Use the estimate to argue which coefficient is the more believable. (d) State what the comparison decides, and what it leaves to considerations that a mass budget cannot see.

   Solution

   (a) The third stage’s effective coefficient is \((0.12)(1.0307)=0.1237\) . Theorem 2.4 for three stages, solved by bisection, gives \(\eta=3.965\times10^{-4}\ \mathrm{s/m}\) and mass ratios of \(2.607\) , \(3.183\) , and \(2.059\) . The stages supply \(2900.1\) , \(3916.9\) , and \(2442.8\ \mathrm{m/s}\) , and the liftoff mass is \(139\,204\ \mathrm{kg}\) . The stack at second-stage ignition is \(47\,534\ \mathrm{kg}\) and at third-stage ignition \(12\,101\ \mathrm{kg}\) .

   (b) Three stages are \(14.5\) percent lighter than Revision C. The problem held the first stage’s effective coefficient at \(0.06403\) ; with the same \(570\ \mathrm{kg}\) fairing on the smaller first stage it would be \(0.0659\) , and the loop of Exercise R.1 then gives \(139\,727\ \mathrm{kg}\) , which is \(14.2\) percent lighter, so the simplification flatters three stages by three tenths of a point. The saving is the one Section 2.3 described: less dead mass is carried to high speed. Here the middle stage, the best-built of the three, takes the largest share of the velocity.

   (c) With \(0.14\) the liftoff mass is \(143\,507\ \mathrm{kg}\) , still \(11.9\) percent lighter. To judge the coefficient, size the third stage of case (a): it is \(7101\ \mathrm{kg}\) , of which \(6223\ \mathrm{kg}\) is propellant. By the stated rules, which are those of Example 6.9, its \(6.13\ \mathrm{m^3}\) of tank weighs \(93\ \mathrm{kg}\) ; an engine for a thrust-to-weight ratio of \(0.9\) , that is \(106.8\ \mathrm{kN}\) , weighs \(146\ \mathrm{kg}\) and its mounting \(37\ \mathrm{kg}\) ; the fixed items are \(400\ \mathrm{kg}\) ; pressurization is about \(12\ \mathrm{kg}\) ; and residual propellant is \(93\ \mathrm{kg}\) . That is \(781\ \mathrm{kg}\) , a coefficient of \(0.112\) , and with \(20\) percent growth on the hardware about \(0.13\) . The fixed \(400\ \mathrm{kg}\) is more than half of the hardware, which is the small-stage penalty of Section 6.7. A value between the two cases is the believable one, and the saving is then \(12\) to \(14\) percent.

   (d) The comparison decides one thing: on mass alone, a third stage pays, even with a poor coefficient. It does not decide the architecture. The mass budget cannot see a third engine that must start in flight, a second separation, a second set of avionics to qualify, or the fact that the saving is in first-stage propellant, the cheapest mass on the vehicle. Section 2.3 quoted a manufacturer who chose two stages to keep separation events to a minimum. A defensible answer keeps two stages and says what was given up for it: about an eighth of the liftoff mass.

6. At staging, the reference ascent of Revision A leaves its spent first stage at an altitude of \(78.7\ \mathrm{km}\) , moving at \(3237\ \mathrm{m/s}\) on a path \(19.2^\circ\) above the local horizontal, \(132\ \mathrm{km}\) downrange of the pad, over an Earth that does not rotate. Ignore the atmosphere. (a) Compute the stage’s specific energy and specific angular momentum, and the semi-major axis and eccentricity of its path. (b) Find its highest altitude and its speed there. (c) Use the orbit equation to find the angle it travels around the Earth’s centre before reaching the surface, and the distance of the impact point from the pad. (d) A colleague applies the flat-Earth range formula \(v^2\sin2\theta/g\) with \(g=9.6\ \mathrm{m/s^2}\) and gets \(678\ \mathrm{km}\) from the staging point. Name the two things that formula leaves out, and say which term of the ascent equations carries the second. (e) Say in which direction the atmosphere changes the real figure.

   Solution

   (a) With \(r=6378.137+78.7=6456.837\ \mathrm{km}\) , Section 8.7’s procedure gives

   \[ \mathcal E=\frac{(3237)^2}{2}-\frac{3.986\times10^{14}}{6.456837\times10^{6}}=5.24-61.73=-56.49\ \mathrm{MJ/kg},\qquad \ell=r\,v\cos\theta=1.9738\times10^{10}\ \mathrm{m^2/s}, \]

   \[ a=-\frac{\mu_E}{2\mathcal E}=3527.8\ \mathrm{km},\qquad e=\sqrt{1+\frac{2\mathcal E\ell^2}{\mu_E^2}},\qquad e=0.8503 . \]

   The semi-major axis is barely half the Earth’s radius, and the perigee, \(a(1-e)=528\ \mathrm{km}\) from the centre, is deep inside the planet. The stage is on a long thin ellipse of which only the tip is above ground.

   (b) The apogee is at \(a(1+e)=6527.4\ \mathrm{km}\) , an altitude of \(149.2\ \mathrm{km}\) , where the speed is \(\ell/r_a=3023.9\ \mathrm{m/s}\) , all of it horizontal.

   (c) The semi-latus rectum is \(\ell^2/\mu_E=977.4\ \mathrm{km}\) , and Theorem 8.3 gives \(\cos\nu=(977.4/r-1)/e\) . At staging this is \(-0.99808\) , so the stage is \(3.55^\circ\) short of apogee. At \(r=R_E\) it is \(-0.99588\) , which is \(5.20^\circ\) past apogee. The stage travels \(8.76^\circ\) around the Earth’s centre, and along the surface that is

   \[ (6378.137\ \mathrm{km})(8.76^\circ)\frac{\pi}{180^\circ}=975\ \mathrm{km} \]

   from the point below staging, \(1107\ \mathrm{km}\) from the pad. It arrives at \(3464\ \mathrm{m/s}\) , descending at \(26.7^\circ\) .

   (d) The flat formula returns the body to the height it started from and ignores the \(78.7\ \mathrm{km}\) it still has to fall. It also ignores the curvature term: the stage’s horizontal speed is \(3057\ \mathrm{m/s}\) , so \(v^2\cos^2\theta/r=1.45\ \mathrm{m/s^2}\) , which is \(15\) percent of the local gravity of \(9.56\ \mathrm{m/s^2}\) . In the turning equation of Proposition 7.1 that is the term \(v^2/r\) inside \(g-v^2/r\) : the ground curves away beneath the stage, and its fall relative to the local horizon is slower by that much.

   (e) Shorter. Drag acts on the way down through the lower atmosphere and takes horizontal speed from the stage, so the real impact point is nearer the pad than the vacuum figure.

7. Decide whether Revision C should have had a hydrogen upper stage. Keep the budget of \(9259.8\ \mathrm{m/s}\) , the first stage (\(c_1=3027.1\ \mathrm{m/s}\) , hardware coefficient \(0.06\) , fairing \(570\ \mathrm{kg}\) ), and the disposal burn of \(102.3\ \mathrm{m/s}\) . Give the hydrogen stage a specific impulse of \(450\ \mathrm s\) , a hardware coefficient of \(0.12\) , and a propellant bulk density of \(343.3\ \mathrm{kg/m^3}\) (illustrative values, close to the hydrogen figures used earlier in the course). (a) Compute the stage’s effective coefficient, and re-run the staging for the liftoff mass and the two stage increments. (b) Compute the second stage’s propellant mass, its tank volume, and the length of a column \(3.0\ \mathrm m\) in diameter, and compare with the kerosene stage’s \(22\,754\ \mathrm{kg}\) in \(22.4\ \mathrm{m^3}\) . (c) Find the hardware coefficient at which the hydrogen stage would save nothing. (d) Make the decision and defend it, saying which input you trust least.

   Solution

   (a) The hydrogen stage has \(c_2=(450)(9.80665)=4413\ \mathrm{m/s}\) , and its effective coefficient is \((0.12)\exp(102.3/4413)=0.1228\) . Running the loop of Exercise R.1 with the fairing on the first stage gives a liftoff mass of \(111\,262\ \mathrm{kg}\) , which is \(31.7\) percent below Revision C. The mass ratios are \(3.135\) and \(3.723\) , and the stages supply \(3458.6\) and \(5801.2\ \mathrm{m/s}\) : the optimum hands the better engine nearly a kilometre per second more than the kerosene stage carried.

   (b) The stack at second-stage ignition is \(30\,087\ \mathrm{kg}\) , and the stage’s propellant is \(22\,006\ \mathrm{kg}\) , almost the same as the kerosene stage’s \(22\,754\ \mathrm{kg}\) . Its volume is not: \(22\,006/343.3\) is \(64.1\ \mathrm{m^3}\) , a column \(9.07\ \mathrm m\) long, against \(22.4\ \mathrm{m^3}\) and \(3.17\ \mathrm m\) . The tank is \(2.86\) times the size for the same mass of propellant.

   (c) Raising the hardware coefficient to \(0.14\) leaves a saving of \(23.9\) percent, and \(0.16\) leaves \(15.8\) percent. Bisection on the coefficient puts the break-even at \(0.197\) : the hydrogen stage’s coefficient would have to be two and a half times the kerosene stage’s \(0.08\) before it stopped paying.

   (d) On mass the case for hydrogen is strong and it survives any plausible coefficient. The input to distrust is the coefficient itself. It was assumed, the stage’s tank is nearly three times the volume and holds a liquid at \(20\ \mathrm K\) that needs insulation, and Section 6.8 found that even the kerosene stage’s \(0.08\) had no margin against the one real stage of its size. What the mass does not show is the second propellant, its ground equipment, and an engine that shares nothing with the first stage. A reasonable decision is the one the course made, kerosene in both stages, stated with its price: about a quarter to a third of the liftoff mass.

8. A proposal for Revision A flies the first stage at a planned angle of attack of \(3^\circ\) . It saves \(37.5\ \mathrm{m/s}\) of ideal velocity and loads the structure with \(q\,\alpha=2.02\ \mathrm{kPa\cdot rad}\) at maximum dynamic pressure. The structure is allowed \(4.0\ \mathrm{kPa\cdot rad}\) (illustrative). At maximum dynamic pressure on the unmodified ascent the vehicle flies at \(542\ \mathrm{m/s}\) on a path at \(50.8^\circ\) in air of density \(0.2235\ \mathrm{kg/m^3}\) , with \(M_\alpha=7.869\times10^{6}\ \mathrm{N\cdot m/rad}\) , \(M_\delta=3.473\times10^{7}\ \mathrm{N\cdot m/rad}\) , and a proportional gain of \(2.087\) . (a) For a head wind of \(30\ \mathrm{m/s}\) at that point, compute the angle of attack the wind produces, the steady angle of attack under attitude feedback, the gimbal angle, and the load, using the airspeed for the dynamic pressure. (b) The same vehicle met \(3.38\ \mathrm{kPa\cdot rad}\) in a \(50\ \mathrm{m/s}\) jet at \(10\ \mathrm{km}\) . Add the planned load to each wind load and compare the sums with the allowance. (c) Convert the saving to payload at \(2.37\ \mathrm{kg}\) per metre per second, and find the mass that a stronger first stage could add, at \(0.132\ \mathrm{kg}\) of payload per kilogram, before the proposal loses. (d) Decide.

   Solution

   (a) In still air the dynamic pressure is \(\tfrac12(0.2235)(542)^2=32.83\ \mathrm{kPa}\) . A head wind adds \(30\cos50.8^\circ=18.96\ \mathrm{m/s}\) to the air’s speed along the path, making it \(560.96\ \mathrm{m/s}\) , and \(30\sin50.8^\circ=23.25\ \mathrm{m/s}\) across it. The airspeed is the resultant, \(\sqrt{(560.96)^2+(23.25)^2}=561.4\ \mathrm{m/s}\) , so \(q=35.23\ \mathrm{kPa}\) . The component across the path tilts the relative wind by

   \[ \alpha_w=\arctan\frac{23.25}{560.96}=0.04142\ \mathrm{rad}=2.37^\circ . \]

   Under proportional feedback the steady angle of attack is larger by the factor \(M_\delta K_p/(M_\delta K_p-M_\alpha)=1.122\) (Section 11.4), so \(\alpha=2.66^\circ\) , held by a gimbal angle of \((M_\alpha/M_\delta)\alpha=0.60^\circ\) . The load is

   \[ q\,\alpha=(35.23\ \mathrm{kPa})(0.04646\ \mathrm{rad})=1.64\ \mathrm{kPa\cdot rad}. \]

   (b) The wind’s angle of attack adds to the planned one when the wind blows from the unfavorable side, so in the worst case the loads add: \(2.02+1.64=3.66\ \mathrm{kPa\cdot rad}\) in the \(30\ \mathrm{m/s}\) wind, inside the allowance, and \(2.02+3.38=5.40\ \mathrm{kPa\cdot rad}\) in the jet, which is \(1.35\) times the allowance. Without the planned angle the jet alone uses \(3.38\) of the \(4.0\) . The sums are estimates, since the two flights have slightly different dynamic pressures, but the second is not a near miss.

   (c) The saving is worth \((37.5)(2.37)=88.9\ \mathrm{kg}\) of payload. A stronger first stage costs \(0.132\ \mathrm{kg}\) of payload per kilogram, so it may add up to \(88.9/0.132\) , which is \(673\ \mathrm{kg}\) , before the proposal loses.

   (d) Reject the proposal as it stands. The planned angle spends half of the structure’s allowance on every flight and leaves the vehicle unable to fly through a wind that the cited source calls typical. It could be accepted only with a structure rated for at least \(5.4\ \mathrm{kPa\cdot rad}\) , and only if raising the rating by a third costs less than \(673\ \mathrm{kg}\) , which is about \(8\) percent of the first stage’s hardware; or with a rule that the vehicle does not launch into a strong jet. This is the judgment of Section 10.9: the gravity turn keeps the whole allowance for disturbances that cannot be planned.

9. An upper stage is stranded in the \(200\ \mathrm{km}\) parking orbit with \(150\ \mathrm{m/s}\) of velocity change left, and spends all of it in one retrograde burn. (a) Find the perigee of the new orbit, and say what its negative altitude means. (b) Find the speed and flight-path angle at the entry interface, \(120\ \mathrm{km}\) . (c) Use Kepler’s equation to find the time from the burn to the interface, and find the angle travelled around the Earth in that time. (d) Evaluate the straight-path formula for the peak deceleration with \(H=8.5\ \mathrm{km}\) and say, from what is known about shallow entries, whether to believe it. (e) Compare the entry angle with the \(-1.22^\circ\) of the reference disposal, and say what a steeper angle buys.

   Solution

   (a) The burn is made at \(r=6578.137\ \mathrm{km}\) , where the circular speed is \(7784.3\ \mathrm{m/s}\) , and leaves the stage at \(7634.3\ \mathrm{m/s}\) , still horizontal. That point is now the apogee. The vis-viva equation gives \(1/a=2/r-v^2/\mu_E\) , so \(a=6336.3\ \mathrm{km}\) , and the perigee radius is \(2a-r=6094.4\ \mathrm{km}\) , an altitude of \(-283.7\ \mathrm{km}\) , with \(e=0.03817\) . A negative perigee altitude means the ellipse passes through the Earth: the stage cannot complete half a revolution.

   (b) At \(r=6498.137\ \mathrm{km}\) the vis-viva equation and \(\ell=r\,v\cos\theta\) , with \(\ell\) fixed at its apogee value, give

   \[ v=7731.4\ \mathrm{m/s},\qquad \cos\theta=\frac{(6578.137)(7634.3)}{(6498.137)(7731.4)},\qquad \theta=-1.63^\circ , \]

   negative because the stage is descending.

   (c) From \(r=a(1-e\cos E)\) the eccentric anomaly at the interface is \(E=2.3039\ \mathrm{rad}\) , measured from perigee, and \(e\sin E=0.0284\) . The mean motion is \(n=\sqrt{\mu_E/a^3}=1.2518\times10^{-3}\ \mathrm{rad/s}\) . By Theorem 8.6 the time from the interface to perigee would be \((E-e\sin E)/n\) , and by symmetry the time from apogee to the interface is half a period less that:

   \[ t=\frac{\pi-(E-e\sin E)}{n}=\frac{3.1416-2.2755}{1.2518\times10^{-3}\ \mathrm{rad/s}}=692\ \mathrm s , \]

   or \(11.5\ \mathrm{min}\) . The orbit equation puts the interface at a true anomaly of \(133.6^\circ\) , so the stage has travelled \(46.4^\circ\) around the Earth from the burn.

   (d) Proposition 12.1 gives

   \[ \lvert\dot v\rvert_{\max}=\frac{(7731.4)^2\sin1.63^\circ}{2(8500)}\exp(-1)=36.7\ \mathrm{m/s^2}=3.74\,g_0 . \]

   It should not be believed. Example 12.3 showed that below about \(5^\circ\) the straight path is the assumption that fails: the path steepens as the stage slows, and the peak deceleration of a ballistic entry from low orbit does not fall below seven to eight times \(g_0\) . Integrating the entry equations for this state, with the method of that example, gives \(7.9\,g_0\) .

   (e) The angle is a third steeper than the reference disposal’s \(-1.22^\circ\) . Example 12.6 found what that buys: a shorter, better-defined footprint, because bodies with different ballistic coefficients come down closer together. The price here was the whole \(150\ \mathrm{m/s}\) .

10. A spacecraft of \(1200\ \mathrm{kg}\) in a \(400\ \mathrm{km}\) circular orbit is to reach the geostationary radius, \(42\,164\ \mathrm{km}\) , in the same plane. Option A is a chemical engine of specific impulse \(320\ \mathrm s\) (illustrative) flying a two-impulse transfer. Option B is an ion thruster of \(237\ \mathrm{mN}\) and \(4170\ \mathrm s\) , the figures of a real thruster cited earlier, thrusting continuously along the velocity. (a) Compute the two impulses of option A, its propellant, and the mass delivered. (b) Show that a thrust small enough to keep the orbit nearly circular changes the specific orbital energy at the rate \(v\) times the thrust acceleration, and deduce that the ideal velocity of a slow spiral between two circular orbits equals the difference of their circular speeds. (c) Compute option B’s ideal velocity, propellant, delivered mass, and thrusting time. (d) Explain why B needs more ideal velocity than A, and say which option suits a payload that must be on station within a month.

    Solution

    (a) With \(r_1=6778.137\ \mathrm{km}\) and \(r_2=42\,164\ \mathrm{km}\) , Proposition 9.1 gives

    \[ \Delta v_{\mathrm I}=2397.5\ \mathrm{m/s},\qquad \Delta v_{\mathrm{II}}=1456.5\ \mathrm{m/s},\qquad \Delta v_{\mathrm I}+\Delta v_{\mathrm{II}}=3854.0\ \mathrm{m/s}. \]

    The coast takes \(5.29\ \mathrm h\) . With \(c=(320)(9.80665)=3138.1\ \mathrm{m/s}\) the propellant is \((1200)\bigl[1-\exp(-3854.0/3138.1)\bigr]=849\ \mathrm{kg}\) , and \(351\ \mathrm{kg}\) arrives.

    (b) Thrust along the velocity does work on the spacecraft at the rate thrust times speed, and gravity’s work is already inside the orbital energy, so \(\mathrm d\mathcal E/\mathrm dt=v\,(F/m)\) . If the orbit stays nearly circular, \(\mathcal E=-\mu/(2r)=-\tfrac12v^2\) with \(v\) the circular speed, so \(\mathrm d\mathcal E=-v\,\mathrm dv\) . Equating, \(\mathrm dv=-(F/m)\,\mathrm dt\) : the circular speed falls at exactly the rate of the thrust acceleration. Integrating, the ideal velocity \(\int(F/m)\,\mathrm dt\) equals the decrease of circular speed. The spacecraft thrusts forward and slows down, as in the drag paradox of Section 8.12 run in reverse.

    (c) The circular speeds are \(7668.6\) and \(3074.7\ \mathrm{m/s}\) , so

    \[ \Delta v=7668.6-3074.7=4593.9\ \mathrm{m/s}. \]

    With \(c=(4170)(9.80665)=40.89\ \mathrm{km/s}\) the propellant is \((1200)\bigl[1-\exp(-4593.9/40\,894)\bigr]=127.5\ \mathrm{kg}\) , and \(1072.5\ \mathrm{kg}\) arrives. The thruster passes \(F/c=5.796\times10^{-6}\ \mathrm{kg/s}\) , so the spiral takes \(127.5/5.796\times10^{-6}\ \mathrm s\) , which is \(255\) days. The thrust is \(2\times10^{-5}\) of the spacecraft’s weight.

    (d) The spiral needs \(19\) percent more ideal velocity because most of it is spent high up, where the spacecraft is slow. Section 9.3’s rule is that a burn buys most energy where the speed is greatest, and the two impulses of option A are placed to exploit it, while continuous thrust cannot be. The exhaust velocity more than pays for the difference: option B delivers three times the mass. It takes eight and a half months. A payload due on station within a month has to take option A, or a larger launcher.

11. Revision A’s first stage has a sea-level thrust of \(2267\ \mathrm{kN}\) , a propellant flow of \(811.1\ \mathrm{kg/s}\) , a total throat area of \(0.1427\ \mathrm{m^2}\) , a total exit area of \(2.355\ \mathrm{m^2}\) , and a thrust-to-weight ratio of \(1.30\) at liftoff. Compare three engines with five engines of the same total thrust. (a) For each layout give the thrust, flow, throat diameter, and exit diameter of one engine, and say whether the exit pressure, the specific impulse, or the total engine mass at a fixed engine thrust-to-weight ratio changes. (b) Find the thrust-to-weight ratio and the net acceleration at liftoff with one engine out, for each layout. (c) With five engines, an outboard engine \(1.0\ \mathrm m\) from the axis fails. Taking the gimbal plane \(14.1\ \mathrm m\) from the centre of mass, find the gimbal angle at which the remaining four cancel the moment of the unbalanced thrust, and compare it with a travel of \(5^\circ\) . (d) Decide between the layouts, and name one cost of five engines that none of these numbers shows.

    Solution

    (a) Dividing the totals by the number of engines:

    Per engine	Three engines	Five engines
    Sea-level thrust	\(755.7\ \mathrm{kN}\)	\(453.4\ \mathrm{kN}\)
    Flow	\(270.4\ \mathrm{kg/s}\)	\(162.2\ \mathrm{kg/s}\)
    Throat diameter	\(0.246\ \mathrm m\)	\(0.191\ \mathrm m\)
    Exit diameter	\(1.000\ \mathrm m\)	\(0.774\ \mathrm m\)

    Nothing else in the ideal analysis changes. The area ratio is still \(16.5\) , so by Theorem 5.4 the exit Mach number and exit pressure are the same; by Theorem 5.5 so are the thrust coefficient and the specific impulse; and at a fixed engine thrust-to-weight ratio of \(100\) the engines together weigh \(2267\ \mathrm{kN}/(100\,g_0)=2312\ \mathrm{kg}\) either way.

    (b) With one engine out at liftoff the ratio is

    \[ \psi_0=(1.30)\tfrac23=0.867\quad\text{for three},\qquad \psi_0=(1.30)\tfrac45=1.04 . \]

    Three engines less one cannot lift the vehicle: the net acceleration, \((\psi_0-1)g_0\) , would be \(-1.31\ \mathrm{m/s^2}\) . Five less one leave the pad at \(0.39\ \mathrm{m/s^2}\) , against \(2.94\ \mathrm{m/s^2}\) with all five running, with a heavy gravity loss, but they leave it.

    (c) The four live engines are no longer symmetric about the axis: the thrust of the engine opposite the dead one is unbalanced, with a moment of one engine’s thrust times \(1.0\ \mathrm m\) . Tilting all four by \(\delta\) gives a moment of four times the thrust times \(14.1\ \mathrm m\) times \(\sin\delta\) (Section 11.3). Setting them equal,

    \[ \sin\delta=\frac{1.0}{(4)(14.1)},\qquad \delta=1.02^\circ . \]

    That is a fifth of the \(5^\circ\) travel. Added to the \(1.4^\circ\) that the jet-stream wind of Example 11.7 demanded it comes to \(2.42^\circ\) , still under half.

    (d) Five. The ideal performance and the engine mass are the same, the gimbals can hold the vehicle straight with an engine out, and the vehicle survives the failure that three engines cannot. The cost that the numbers do not show is count: five engines are five sets of pumps, valves, and gimbals to build and to start, and with more engines an engine failure is more likely on any given flight. Section 6.7’s fixed items also grow with the number of engines, which would nudge \(\varepsilon_1\) upward.

12. Re-size Revision C for methane and liquid oxygen in both stages. Tabulated ideal exhaust velocities with a vacuum nozzle are \(3615\ \mathrm{m/s}\) for methane and \(3510\ \mathrm{m/s}\) for kerosene, and the bulk densities are \(820\) and \(1015\ \mathrm{kg/m^3}\) at the mixture ratios used in this course. (a) Scale both of Revision C’s exhaust velocities, \(3027.1\) and \(3383.3\ \mathrm{m/s}\) , by the ratio of the tabulated values. (b) In the kerosene mass estimate the tanks are \(1.86\) percent of the propellant mass on stage 1 and \(1.49\) percent on stage 2. Scale those shares inversely with bulk density. The shares are per kilogram of propellant, so convert the hardware coefficients of \(0.06\) and \(0.08\) to hardware mass per unit of propellant mass, \(\varepsilon/(1-\varepsilon)\) , add the increase there, and convert back. Then apply the disposal factor for \(102.3\ \mathrm{m/s}\) to the second. (c) With the budget of \(9259.8\ \mathrm{m/s}\) and the \(570\ \mathrm{kg}\) fairing, re-run the staging for the liftoff mass, and say how much of the gain the heavier tanks took back. (d) Compute the propellant volume of each stage and the change in tank length at \(3.0\ \mathrm m\) diameter. (e) Say whether the result settles the choice of propellant.

    Solution

    (a) The ratio is \(3615/3510=1.0299\) , so the exhaust velocities become \(3117.7\) and \(3484.5\ \mathrm{m/s}\) , that is \(317.9\) and \(355.3\ \mathrm s\) .

    (b) Methane’s bulk density is lower by the factor \(1015/820=1.238\) , and by Proposition 6.1 tank mass per kilogram of propellant rises in the same proportion: from \(1.86\) to \(2.30\) percent on stage 1 and from \(1.49\) to \(1.84\) percent on stage 2. A coefficient \(\varepsilon\) corresponds to hardware over propellant of \(\varepsilon/(1-\varepsilon)\) , which is \(0.0638\) and \(0.0870\) for the two stages; adding the increases gives \(0.0683\) and \(0.0905\) , and converting back, hardware coefficients of \(0.0639\) and \(0.0830\) . The disposal factor \(\exp(102.3/3484.5)\) raises the second to \(0.0855\) . The conversion matters because the tank shares are fractions of the propellant mass and a coefficient is a fraction of the whole stage; adding the increases to \(0.06\) and \(0.08\) directly would give \(0.0644\) and \(0.0835\) and overstate the penalty.

    (c) The loop of Exercise R.1 with these inputs gives \(149\,396\ \mathrm{kg}\) at liftoff, which is \(8.3\) percent below Revision C, with mass ratios of \(4.020\) and \(4.107\) . Had the coefficients stayed at \(0.06\) and \(0.08\) the saving would have been \(11.2\) percent, so the heavier tanks take back about a quarter of what the exhaust velocity gave.

    (d) The first stage burns \(112\,229\ \mathrm{kg}\) of propellant, which fills \(136.9\ \mathrm{m^3}\) , a column of \(19.36\ \mathrm m\) ; Revision C’s first stage needs \(122.7\ \mathrm{m^3}\) and \(17.36\ \mathrm m\) . The second stage holds \(21\,891\ \mathrm{kg}\) in \(26.7\ \mathrm{m^3}\) , a column of \(3.78\ \mathrm m\) against \(3.17\ \mathrm m\) . The methane vehicle carries \(9\) percent less propellant in \(13\) percent more tank and is about \(2.6\ \mathrm m\) longer.

    (e) It does not. The saving is real in this model, but it rests on scaling a tabulated ideal ratio of three percent onto delivered specific impulses that Section 6.8 could pin down only to within two percent, and on a tank rule that carries an illustrative factor of two. The model also leaves out what separates the two fuels in practice: methane is a cryogenic liquid and needs insulation, where kerosene is stored at room temperature; a longer vehicle bends more easily, which Chapter 11 flagged as the unexamined load. The arithmetic says methane is worth a serious look, and that the look should begin with the two inputs just named.

13. For a two-stage vehicle at its staging optimum, compare two ways of charging for a disposal burn \(\Delta v_{\mathrm{dis}}\) made by the empty upper stage: as a line of the velocity budget, and as an increase of the upper stage’s structural coefficient from \(\varepsilon_2\) to \(\varepsilon_2\exp(\Delta v_{\mathrm{dis}}/c_2)\) . (a) Using the design sensitivities of the staging optimum, write the fractional growth of liftoff mass in each case to first order in \(\Delta v_{\mathrm{dis}}\) . (b) Show that the ratio of the two is \(\eta c_2\) divided by \(\varepsilon_2\,\partial J/\partial\varepsilon_2\) , and evaluate both growths for \(\eta=4.461\times10^{-4}\ \mathrm{s/m}\) , \(c_2=3383.3\ \mathrm{m/s}\) , \(\varepsilon_2=0.08\) , \(\mathcal R_2=4.217\) , and \(\Delta v_{\mathrm{dis}}=102.3\ \mathrm{m/s}\) . (c) Say in one sentence which of the two charges describes what the hardware does.

    Hint

    (a) Proposition 2.6 gives the growth of \(J=\ln(m_{0,1}/m_L)\) as \(\eta\,\Delta v_{\mathrm{dis}}\) for a budget line and as \((\partial J/\partial\varepsilon_2)\,\varepsilon_2\,\Delta v_{\mathrm{dis}}/c_2\) for the coefficient, using \(\exp(x)-1\approx x\) . (b) Divide one by the other. With the stated values \(\partial J/\partial\varepsilon_2=\mathcal R_2/(1-\varepsilon_2\mathcal R_2)-1/(1-\varepsilon_2)\) is \(5.277\) , the coefficient route costs \(1.3\) percent of liftoff mass, the budget line \(4.6\) percent, and the ratio is \(3.6\) . (c) The budget line prices the burn as if the payload were carried through it; the coefficient prices the propellant as inert mass on the upper stage, which is what it is until the payload has been released.

14. Show that for a body with no lift, on a path where the component of gravity along the path is negligible, the dynamic pressure equals the ballistic coefficient times the deceleration, \(q=\beta\lvert\dot v\rvert\) . The reference upper stage decelerates at a peak of \(7.36\,g_0\) with \(\beta\) between \(100\) and \(280\ \mathrm{kg/m^2}\) . Compute its dynamic pressure at that peak, compare it with the \(33.9\ \mathrm{kPa}\) that Revision C meets on the way up and with the \(0.4\ \mathrm{MPa}\) its tank was built to hold, and say what the comparison implies about whether pressure or heat destroys the stage.

    Answer

    The drag deceleration is \(D/m=qC_DS/m=q/\beta\) , and with gravity along the path neglected that is the whole of \(\lvert\dot v\rvert\) . At \(7.36\,g_0\) , which is \(72.2\ \mathrm{m/s^2}\) , the dynamic pressure is \(7.2\) to \(20.2\ \mathrm{kPa}\) for the two ballistic coefficients. That is less than the \(33.9\ \mathrm{kPa}\) of the ascent and about a twentieth of the pressure the tank holds from inside. The stage is not crushed. It is destroyed by heat, which by Example 12.6 melts the thin wall minutes before the deceleration peaks.

15. Audit the energy of Revision C. (a) Compute the specific orbital energy of the \(400\ \mathrm{km}\) target orbit, and that of a kilogram at rest on the pad at latitude \(28.5^\circ\) , where the ground moves at \(408.7\ \mathrm{m/s}\) , and take the difference. (b) Multiply by the \(5000\ \mathrm{kg}\) payload. (c) The vehicle burns \(124\,569+22\,754+61\ \mathrm{kg}\) of propellant, releasing \(9.71\ \mathrm{MJ/kg}\) ; find the fraction of the chemical energy that ends as orbital energy of the payload. (d) Estimate the fraction of the chemical energy that becomes kinetic energy of the jet measured in the vehicle’s frame, using \(\tfrac12c^2\) per kilogram with exhaust velocities of \(3027.1\) and \(3383.3\ \mathrm{m/s}\) . Say where the rest of the chemical energy goes, and then what the jet’s energy is divided among, given that the payload’s share in (b) is part of it.

    Answer

    (a) \(\mathcal E=-29.40\ \mathrm{MJ/kg}\) in the target orbit and \(\tfrac12(408.7)^2-\mu_E/R_E=-62.41\ \mathrm{MJ/kg}\) on the pad: a difference of \(33.0\ \mathrm{MJ/kg}\) . (b) \(165\ \mathrm{GJ}\) . (c) The propellant releases \((147\,384\ \mathrm{kg})(9.71\ \mathrm{MJ/kg})=1431\ \mathrm{GJ}\) , so \(11.5\) percent ends as orbital energy of the payload. (d) In the vehicle’s frame the jet carries \(701\ \mathrm{GJ}\) , the disposal propellant included, which is \(49\) percent. The other \(51\) percent never becomes directed motion: it leaves as heat in exhaust that is still hot (Example 4.5), or is never released, because the engines run fuel-rich and the products dissociate (Section 4.6). The jet’s \(701\ \mathrm{GJ}\) is the mechanical energy, and the payload’s \(165\ \mathrm{GJ}\) comes out of it. So does the energy given to the hardware: the second stage’s \(1984\ \mathrm{kg}\) carried to orbit, and the first stage’s \(7951\ \mathrm{kg}\) with its \(570\ \mathrm{kg}\) fairing carried to staging at \(73.3\ \mathrm{km}\) and \(3587\ \mathrm{m/s}\) (Section 12.9), hold about \(126\ \mathrm{GJ}\) between them. The remaining \(410\ \mathrm{GJ}\) or so stays with the exhaust as seen from outside the vehicle, apart from the little that drag dissipates.

16. The flat, airless ascent that arrived level at \(20\ \mathrm{km}\) with \(1650\ \mathrm{m/s}\) , under a uniform gravity of \(1.62\ \mathrm{m/s^2}\) , used \(1813.9\ \mathrm{m/s}\) of ideal velocity. The Moon’s gravitational parameter is \(4.90\times10^{12}\ \mathrm{m^3/s^2}\) and its mean radius is \(1737.4\ \mathrm{km}\) (NASA Moon Fact Sheet; see sources). (a) Compute the circular speed and the local gravity at \(20\ \mathrm{km}\) above the Moon. (b) Compute the curvature term \(v^2/r\) at the cutoff speed and the fraction of local gravity it balances. (c) Explain, using the turning equation of the ascent over a round body, what the flat model charged the vehicle for that a round Moon would not, and whether the flat model’s loss of \(163.9\ \mathrm{m/s}\) is therefore too large or too small.

    Answer

    (a) At \(r=1757.4\ \mathrm{km}\) the circular speed is \(1669.8\ \mathrm{m/s}\) and gravity is \(1.587\ \mathrm{m/s^2}\) . (b) At \(1650\ \mathrm{m/s}\) the curvature term is \(1.549\ \mathrm{m/s^2}\) , which balances \(97.6\) percent of gravity. (c) Proposition 7.1 turns the path at the rate set by \(g-v^2/r\) , which shrinks to almost nothing as the vehicle nears circular speed. The flat model keeps the full \(g\) to the end, so its thrust must go on holding the vehicle up through the second half of the burn when a round Moon would be doing that work. Its loss of \(163.9\ \mathrm{m/s}\) is too large. Its cutoff state is also \(20\ \mathrm{m/s}\) short of an orbit.

17. Air meets the returning upper stage at \(7894\ \mathrm{m/s}\) . (a) Treating air as a calorically perfect gas with \(c_p=1004.5\ \mathrm{J/(kg\cdot K)}\) , compute the stagnation temperature that the steady-flow energy equation predicts. (b) The result is absurd. Using what happens to the chamber gas of a kerosene engine above \(3000\ \mathrm K\) , say which assumption has failed and where the energy goes instead. (c) Compare the kinetic energy per kilogram of the oncoming air with the \(9.71\ \mathrm{MJ/kg}\) that kerosene and oxygen release, and explain why the stagnation-heating correlation of this course is written with enthalpy and not with temperature.

    Hint

    (a) \(T+v^2/(2c_p)\) with \(v=7894\ \mathrm{m/s}\) gives about \(31\,000\ \mathrm K\) above the ambient temperature. (b) The gas is not calorically perfect at such energies. Section 4.6 found that above about \(3000\ \mathrm K\) added energy goes into pulling molecules apart; the same happens to air behind the shock, so the temperature is thousands of kelvin, not tens of thousands. (c) The air arrives with \(31.2\ \mathrm{MJ/kg}\) , which is \(3.2\) times what kerosene and oxygen release. Enthalpy is conserved along the stream whatever the gas does with it, by Theorem 4.2, whereas temperature depends on how the energy is shared among motions and broken bonds.

18. Three change requests reach Revision C after its tanks are frozen: an upper-stage engine with \(8\ \mathrm s\) more specific impulse that weighs \(60\ \mathrm{kg}\) more; \(150\ \mathrm{kg}\) of stiffening on the first stage against wind loads; and a guidance update said to cut the ideal velocity of the ascent by \(20\ \mathrm{m/s}\) . Using Revision C’s exchange ratios (\(1\ \mathrm{kg}\) of payload per kilogram of second-stage hardware, \(0.14\ \mathrm{kg}\) per kilogram of first-stage hardware, \(33\ \mathrm{kg}\) per second of specific impulse, and \(2.3\ \mathrm{kg}\) per metre per second), find the payload effect of each request and of all three together, state what is held fixed in such an estimate, and say which request’s number you would examine most closely before accepting it.

    Answer

    The engine: \((8)(33)-60\) , that is \(+204\ \mathrm{kg}\) . The stiffening: \(-(150)(0.14)\) , that is \(-21\ \mathrm{kg}\) . The guidance update: \((20)(2.3)\) , that is \(+46\ \mathrm{kg}\) . Together \(+229\ \mathrm{kg}\) . The propellant loads, the budget, and everything not named are held fixed, and the ratios are linear, so the sum is good for changes of this size. Examine the guidance claim first. It is a statement about the model’s own flight, and Section 12.8’s allowance rule returns to the allowance whatever a better steering law saves in steering and drag loss; only a saving in gravity loss survives. The engine’s figure is the largest, and it should come with a test record and a weighed engine.

19. The first-stage nozzle of the reference launcher has an exit pressure of \(65.0\ \mathrm{kPa}\) at full chamber pressure. Use the exponential atmosphere, in which the ambient pressure is \(101.325\ \mathrm{kPa}\) at the ground and falls by the factor \(\exp(-z/H)\) with \(H=8.5\ \mathrm{km}\) . (a) Find the altitude at which the nozzle is matched, and say on which side of it the nozzle is overexpanded. (b) The reference ascent passes \(3.07\ \mathrm{km}\) at \(40\ \mathrm s\) and \(7.76\ \mathrm{km}\) at \(60\ \mathrm s\) ; estimate the time at which the nozzle is matched. (c) With the chamber throttled to \(60\) percent of its pressure, find the exit pressure and the altitudes above which it exceeds \(0.4\) and \(0.5\) of the ambient pressure. (d) Two throttle schedules have been priced for this vehicle, one to \(70\) percent from \(65\) to \(90\ \mathrm s\) and one that holds \(5\,g_0\) from \(156\ \mathrm s\) . Say whether flow separation constrains either, and what does limit them.

    Answer

    (a) \(z=H\ln(101.325/65.0)\) , which is \(3.77\ \mathrm{km}\) ; below it the nozzle is overexpanded, above it underexpanded. (b) Interpolating linearly between the two stated points, \(43.0\ \mathrm s\) after liftoff; the vehicle is accelerating, so the true time is a little later. (c) All nozzle pressures scale with chamber pressure, so \(p_e\) is \(39.0\ \mathrm{kPa}\) , which is \(0.385\) of sea-level pressure; it exceeds \(0.4\) and \(0.5\) of ambient above \(0.33\ \mathrm{km}\) and \(2.22\ \mathrm{km}\) . (d) Neither. From \(65\ \mathrm s\) the vehicle is above the \(7.76\ \mathrm{km}\) it passed at \(60\ \mathrm s\) , where the ambient pressure is already down to \(40.7\ \mathrm{kPa}\) , so even at \(70\) percent the exit pressure of \(45.5\ \mathrm{kPa}\) exceeds the ambient pressure, and at \(156\ \mathrm s\) the vehicle is in near vacuum. Separation limits throttling only in the first kilometres. What limits the two schedules is their cost in ideal velocity: \(84.9\ \mathrm{m/s}\) for the first and \(3.1\ \mathrm{m/s}\) for the second.

20. Each of the following statements is wrong. Correct each in one or two sentences and name the quantity or relation that settles it. (a) “A stage with a specific impulse of \(345\ \mathrm s\) burns for \(345\ \mathrm s\) .” (b) “A satellite at \(400\ \mathrm{km}\) stays up because there is no gravity there.” (c) “The hottest flame gives the highest exhaust velocity.” (d) “Of two bodies that enter the atmosphere at the same speed and angle, the heavier one suffers the lower peak deceleration.” (e) “The \(174\ \mathrm{m/s}\) of allowance in Revision C’s budget is spare capability.” (f) “An upper stage whose thrust is less than its weight cannot do its job.” (g) “Raising both gains of the attitude loop always makes it safer.”

    Hint

    Each is a mistake that a chapter warned against. (a) Specific impulse is impulse per unit weight of propellant; burn time is propellant mass over flow, and Revision C’s upper stage, at \(345\ \mathrm s\) , burns its \(22\,754\ \mathrm{kg}\) at \(77.74\ \mathrm{kg/s}\) in \(293\ \mathrm s\) . (b) Gravity at that height is \(89\) percent of its surface value; the satellite stays up because of its horizontal speed, through the term \(v^2/r\) . (c) Exhaust velocity follows \(T_c/\mathcal M\) , which is why hydrogen wins with the cooler flame: \(3013\ \mathrm K\) and \(4462\ \mathrm{m/s}\) in the tabulated values, against kerosene’s \(3676\ \mathrm K\) and \(3510\ \mathrm{m/s}\) . (d) Proposition 12.1’s peak does not contain \(\beta\) ; the heavier body peaks lower down at the same value. (e) An allowance is an estimate of losses the model understates; only the reserve of \(100\ \mathrm{m/s}\) is spare. (f) Gravity loss is \(\int g\sin\theta\,\mathrm dt\) , and the upper stage ignites on a nearly level path at high speed: on Revision A’s reference ascent, at \(19.2^\circ\) , gravity takes \(3.1\ \mathrm{m/s^2}\) along the path from a thrust acceleration of \(8.8\ \mathrm{m/s^2}\) . (g) Once the actuator lags, Proposition 11.4 requires \(K_d>t_{\mathrm{act}}K_p\) , a condition on the ratio of the gains: raising the proportional gain faster than the derivative gain breaks it, and raising both in step does not. What raising both in step does is raise the loop’s frequency, \(\sqrt{(M_\delta K_p-M_\alpha)/I_y}\) , toward the modes the model leaves out: four times the gains of Example 11.4 put it at \(5.2\ \mathrm{rad/s}\) , past the slosh frequency of \(5.07\ \mathrm{rad/s}\) .