Systems with Varying Mass: A Rocket — Question 4

Question 4

Gravity loss depends on how long the engine burns

A rocket already free of its launch support burns vertically upward from rest in uniform gravity $g=9.8\,\mathrm{m/s^2}$. Its mass falls from $1000$ to $800\,\mathrm{kg}$ in $20\,\mathrm s$ at a constant rate. Relative exhaust speed is $u=2000\,\mathrm{m/s}$; ignore air resistance.

1. Find initial acceleration and burnout velocity.

2. Compare with a $40\,\mathrm s$ burn using the same mass endpoints and exhaust speed. Check that both initially accelerate upward.

Question 4 - Solution

Gravity loss depends on how long the engine burns

(a) Derivation and calculation

For the 20 s burn, $q=(1000-800)/20=10$ kg/s: $$\boxed{a_i=\frac{uq}{m_i}-g=20-9.8=10.2\ \mathrm{m/s^2}}.$$ Integrating $\dot v=uq/(m_i-qt)-g$ gives $$\boxed{v_f=u\ln(m_i/m_f)-gt_b=2000\ln1.25-196=250.29\ \mathrm{m/s}}.$$

(b) Result and interpretation

For 40 s, $q=5$ kg/s and $$a_i=10-9.8=0.20\ \mathrm{m/s^2}>0,$$ $$\boxed{v_f=2000\ln1.25-392=54.29\ \mathrm{m/s}}.$$ Both initially accelerate upward. The slower burn loses an additional $g(40-20)=196$ m/s to gravity.

Physical check

The ideal propulsive term is unchanged because the mass endpoints are identical.