Rocket Equation and Staging Questions and Answers

Answer: D

The payload reaches about 385 km.

With all propellant burnt instantaneously there are no gravity losses during the burn, so the full rocket-equation velocity is reached at once. The mass ratio is \(m_{0}/m_{f} = 1/0.4 = 2.5\):

\[ \begin{aligned} \Delta V &= V_{e}\ln\frac{m_{0}}{m_{f}} = 3000\ln 2.5 \\ &= 3000 \times 0.9163 = 2749\ \text{m/s} \end{aligned} \]

The payload then coasts upward against constant gravity with no drag:

\[ \begin{aligned} h &= \frac{\Delta V^{2}}{2g} = \frac{2749^{2}}{2 \times 9.81} \\ &= 3.85 \times 10^{5}\ \text{m} \end{aligned} \]

which is 385 km. The smaller options would need a lower burnout velocity, for example from using a mass ratio below 2.5, which the 60% propellant fraction does not allow.

Answer: C

The burnout velocity is about 4245.9 m/s.

The burn time is \(8400/1200 = 7\) s, and the burnout mass is \(10000 - 8400 = 1600\) kg. With constant gravity and no drag, the rocket equation with a gravity loss term gives

\[ \begin{aligned} v_{b} &= I_{sp}\,g\ln\frac{m_{0}}{m_{b}} - g\,t_{b} \\ &= 240 \times 9.81 \times \ln 6.25 - 9.81 \times 7 \\ &= 2354.4 \times 1.8326 - 68.7 \\ &\approx 4314.6 - 68.7 = 4245.9\ \text{m/s} \end{aligned} \]

4314.6 m/s is the ideal velocity without the gravity loss, a deliberate trap. 4383.3 adds the gravity loss instead of subtracting it, and 3933.7 m/s implies a loss of about 380 m/s, far more than the 68.7 m/s of a 7-second burn.

Answer: C

The specific impulse is 200 s.

The rocket burns \(150 - 50 = 100\) kg of propellant in 10 s, a mass flow of 10 kg/s. Specific impulse is thrust per unit weight flow of propellant:

\[ \begin{aligned} I_{sp} &= \frac{F}{\dot{m}g} \\ &= \frac{19620}{10 \times 9.81} = 200\ \text{s} \end{aligned} \]

400 s would need a mass flow of only 5 kg/s, half the actual figure, and 100 s would need 20 kg/s. 300 s corresponds to about 6.7 kg/s, which is what dividing the 100 kg burnt by 15 s instead of 10 s would give. Only 10 kg/s matches the stated masses and burn time.

Answer: C

The final velocities are related by \(V_{A} = 0.5V_{B}\).

Without gravity or drag, each missile’s velocity gain follows the rocket equation \(V = I_{sp}g_{0}\ln(m_{0}/m_{f})\), and both have the same \(I_{sp}\) and liftoff mass. Only the propellant burnt differs:

\[ \begin{aligned} m_{p,A} &= 120 \times 30 = 3600\ \text{kg} \\ m_{p,B} &= 70 \times 70 = 4900\ \text{kg} \end{aligned} \]

\[ \begin{aligned} \frac{V_{A}}{V_{B}} &= \frac{\ln(5600/2000)}{\ln(5600/700)} \\ &= \frac{\ln 2.8}{\ln 8} = \frac{1.030}{2.079} \\ &= 0.495 \approx 0.5 \end{aligned} \]

Missile A burns faster, but the burn rate does not matter once gravity is neglected; only the mass ratio does. \(V_{A} = V_{B}\) would need equal propellant loads, and 0.7 is roughly the ratio of the propellant masses, 3600/4900 = 0.73, which forgets the logarithm. 4.1 has no basis.