Discussion
A rocket, with a total lift-off mass of 10000 kg, moves vertically upward from rest under a constant gravitational acceleration of 9.81 m/s2. The propellant mass of 8400 kg burns at a constant rate of 1200 kg/s. If the specific impulse of the rocket engine is 240 s, neglecting drag, the burnout velocity in m/s is
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.
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