Question

Imagine a rocket with mass is moving toward a gas cloud with initial velocity . Naturally the rocket will be colliding with the gas particles and gradually losing its kinetic energy. Now for the sake of simplicity, we can model this gas cloud as tiny particles with very small mass and density /. The cross sectional area of the rocket is . Now find out how much time it will take to decrease the rocket’s velocity to half of its original.

Answer 1

Let the instant velocity of a rocket be $v$. In the instantaneous rest frame of the rocket, particles of the gas cloud collide with it head-on with the same absolute velocity $v$, and bounce-off with the same velocity. Hence, in the gas frame, they change velocity from zero to $2v$. The momentum is thus constantly transferred from the rocket to gas particles, as a result of which, the rocket loses the same amount of momentum. In a small time $\Delta t$, the mass of new particles involved in collisions is $\Delta m = \rho S v \Delta t$, where $S$ is the cross-sectional area of the rocket, and $\rho$ is the gas mass density. The momentum loss by the rocket is then $2 \Delta m v = 2 \rho S v^2 \Delta t$. If the mass of the rocket is $M$, this should be equal to $- M \Delta v$, where $\Delta v$ is the change in the rocket velocity. Equating these quantities and dividing by $\Delta t$, we obtain a differential equation $M \dot v = - 2 \rho S v^2$. Its solution $v (t)$ with the initial condition $v (0) = v_0$ is

$$v (t) = \left( \frac{2 \rho S}{M} t + \frac{1}{v_0} \right)^{-1} \, .$$

Requiring, by the statement of the problem, $v (t_0) = v_0 / 2$, we obtain the equation

$$\frac{2}{v_0} = \frac{2 \rho S}{M} t_0 + \frac{1}{v_0} \, ,$$

whence we find the answer:

$$t_0 = \frac{M}{2 \rho S v_0} \, .$$