Question 14.SP.9: aA rocket of initial mass m0 (including shell and fuel) is ......

STRATEGY: Because you have a system that is losing mass, apply the principle of impulse and momentum. This gives you an equation you can integrate to obtain the velocity.

MODELING: Choose the rocket shell and its fuel as your system. At time t, the mass of the rocket shell and remaining fuel is m = m_{0} − qt, 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. The impulse–momentum diagram for this system is shown in Fig. 1, where v_{e} is the absolute velocity of the expelled fuel.

ANALYSIS: Apply the principle of impulse and momentum between time t and time t + Δt to find

(m_{0}-q t)\nu-g(m_{0}-q t)\,\Delta t=(m_{0}-q t-q\Delta t)(\nu+\Delta\nu)-q\Delta t(u-\nu)

Divide through by Δt and let Δt approach zero for

-g(m_{0}-q t)=(m_{0}-q t){\frac{d\nu}{d t}}-q u

Separating variables and integrating from t = 0, v = 0 to t = t, v = v, you have

d\nu=\left({\frac{q u}{m_{0}-q t}}-g\right)d t

\int_{0}^{v}d\nu=\int_{0}^{t}\left({\frac{q u}{m_{0}-q t}}-g\right)d t

\nu=[-u\ln(m_{0}-q t)-g t]_{0}^{t}\qquad\qquad \nu=u\ln{\frac{m_{0}}{m_{0}-q t}}-g t  ◂

REFLECT and THINK: The mass remaining at time t_{f}, after all of the fuel has been expended, is equal to the mass of the rocket shell m_{s} = m_{0} − qt_{f}, and the maximum velocity attained by the rocket is v_{m} = u \;\text{ln} (m_{0}/m_{s}) − gt_{f}. Assuming that the fuel is expelled in a relatively short period of time, the term gt_{f} is small, and we have v_{m} ≈ u\text{ ln} (m_{0}/m_{s}). 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 we use a propellant yielding u = 2200 m/s.