Question

Consider a rocket that is in deep space and at rest relative to an inertial reference frame. The rocket's engine is to be fired for a certain interval. What must be the rocket's mass ratio (ratio of initial to final mass) over that interval if the rocket's final speed relative to the inertial frame is to be equal to (a) the exhaust speed (speed of the exhaust products relative to the rocket) and (b)7.00 times the exhaust speed? (The exhaust products from the rocket escape at a constant velocity relative to the rocket)

Answer 1

Answer:

$\frac{m_i}{m_f}=2.7182$

$\frac{m_i}{m_f}=1096.633$

Explanation:

$m_i$ = Initial mass of rocket

$m_f$ = Final mass of rocket

u = Initial velocity

$v_r$ = Relative velocity

v = Velocity

From the rocket equation

$v=u+v_{r}\ln {\frac {m_{i}}{m_{f}}}\\\Rightarrow v=v_{r}\ln {\frac {m_{i}}{m_{f}}}\\\Rightarrow \frac{m_i}{m_f}=e^{\frac{v}{v_{r}}}\\\Rightarrow \frac{m_i}{m_f}=e^{\frac{v}{v}}=e^1\\\Rightarrow \frac{m_i}{m_f}=2.7182$

$\frac{m_i}{m_f}=2.7182$

when $v=7v_r$

$v=v_{r}\ln {\frac {m_{i}}{m_{f}}}\\\Rightarrow \frac{m_i}{m_f}=e^{\frac{v}{v_{r}}}\\\Rightarrow \frac{m_i}{m_f}=e^{\frac{7v_r}{v_r}}=e^7\\\Rightarrow \frac{m_i}{m_f}=1096.633$

$\frac{m_i}{m_f}=1096.633$