Answer to Question #96952 in Classical Mechanics for Sayed

Question #96952
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.
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Expert's answer
2019-10-22T10:09:39-0400

Let the instant velocity of a rocket be vv. In the instantaneous rest frame of the rocket, particles of the gas cloud collide with it head-on with the same absolute velocity vv, and bounce-off with the same velocity. Hence, in the gas frame, they change velocity from zero to 2v2v. 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 Δt\Delta t, the mass of new particles involved in collisions is Δm=ρSvΔt\Delta m = \rho S v \Delta t, where SS is the cross-sectional area of the rocket, and ρ\rho is the gas mass density. The momentum loss by the rocket is then 2Δmv=2ρSv2Δt2 \Delta m v = 2 \rho S v^2 \Delta t. If the mass of the rocket is MM, this should be equal to MΔv- M \Delta v, where Δv\Delta v is the change in the rocket velocity. Equating these quantities and dividing by Δt\Delta t, we obtain a differential equation Mv˙=2ρSv2M \dot v = - 2 \rho S v^2. Its solution v(t)v (t) with the initial condition v(0)=v0v (0) = v_0 is

v(t)=(2ρSMt+1v0)1.v (t) = \left( \frac{2 \rho S}{M} t + \frac{1}{v_0} \right)^{-1} \, .

Requiring, by the statement of the problem, v(t0)=v0/2v (t_0) = v_0 / 2, we obtain the equation

2v0=2ρSMt0+1v0,\frac{2}{v_0} = \frac{2 \rho S}{M} t_0 + \frac{1}{v_0} \, ,

whence we find the answer:

t0=M2ρSv0.t_0 = \frac{M}{2 \rho S v_0} \, .

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