Question 14.8: A rocket of initial mass m0 (including shell and fuel) is fi......
A rocket of initial mass m_0 (including shell and fuel) is fired vertically at time t = 0. The fuel is consumed at a constant rate q = dm/dt and is expelled at a constant speed u relative to the rocket. Derive an expression for the magnitude of the velocity of the rocket at time t, neglecting the resistance of the air.
At time t, the mass of the rocket shell and remaining fuel is m=m_0-q t, and the velocity is v. During the time interval Δt, a mass of fuel Δm = q Δt is expelled with a speed u relative to the rocket. Denoting by \text v_e the absolute velocity of the expelled fuel, we apply the principle of impulse and momentum between time t and time t + Δt.
We write
\left(m_0-q t\right) v-g\left(m_0-q t\right) \Delta t=\left(m_0-q t-q~ \Delta t\right)(v+\Delta v)-q~ \Delta t(u-v)Dividing through by Δt and letting Δt approach zero, we obtain
-g\left(m_0-q t\right)=\left(m_0-q t\right) \frac{d}{d t}-q uSeparating variables and integrating from t = 0, v = 0 to t = t, v = v,
d v=\left(\frac{q u}{m_0-q t}-g\right) d t \quad \int_0^v d v=\int_0^t\left(\frac{q u}{m_0-q t}-g\right) d tv=\left[-u~ \ln \left(m_0-q t\right)-g t\right]_0^t \quad\quad v=u~ \ln \frac{m_0}{m_0-q t}-g t
Remark. The mass remaining at time t_f, after all the fuel has been expended, is equal to the mass of the rocket shell m_s=m_0-q t_f, and the maximum velocity attained by the rocket is v_m=u ~\ln \left(m_0 / m_s\right)-g t_f. Assuming that the fuel is expelled in a relatively short period of time, the term g t_f is small and we have v_m \approx u~ \ln \left(m_0 / m_s\right). In order to escape the gravitational field of the earth, a rocket must reach a velocity of 11.18 km/s. Assuming u = 2200 m/s and v_m = 11.18 km/s, we obtain m_0 / m_s=161. Thus, to project each kilogram of the rocket shell into space, it is necessary to consume more than 161 kg of fuel if a propellant yielding u = 2200 m/s is used.
