A satellite moving with velocity $v$ in a force free space collects stationary interplanetary dust at a rate…
A satellite moving with velocity $v$ in a force free space collects stationary interplanetary dust at a rate of $\frac{d M}{d t}=\alpha v$ where $M$ is the mass (of satellite + dust) at that instant. The instantaneous acceleration of the satellite is
$-\frac{\alpha v^2}{2 M}$
$-\frac{\alpha v^2}{M}$
$-\alpha v^2$
$-\frac{2 \alpha v^2}{M}$
Solution
$a_{\text {inst }}=\frac{\alpha \mathrm{V}^2}{\mathrm{M}}$
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The force acting on the satellite is due to the momentum of the dust particles being collected.
The rate of change of momentum of the satellite system is equal to the force acting on it.
The rate of change of momentum dP/dt = F, where F is the force.
The momentum P of the system (satellite + dust) is given by the product of the total mass and the velocity,
i.e., P = Mv.
The rate of change of momentum is dP/dt = d(Mv)/dt. Using the product rule,
we get dP/dt = v*(dM/dt) + M*(dv/dt).
Given that dM/dt = αv, the equation becomes dP/dt = v*(αv) + M*a, where a = dv/dt is the acceleration.
Given that there is no external force acting on the satellite, dP/dt = 0. So, 0 = v*(αv) + M*a.
Rearranging the terms for acceleration, we get a = - (αv^2)/M.
$
So, the instantaneous acceleration of the satellite is a = - (αv^2)/M.
The negative sign indicates that the acceleration is in the opposite direction to the velocity,
implying that the satellite is slowing down due to the collection of interplanetary dust