A satellite in force free space sweeps stationary interplanetary dust at a rate \[(dM/dt)=+\alpha v\]. The acceleration of satellite of mass M is:
A.) \[-3\alpha {{v}^{2}}/M\]
B.) \[-2\alpha {{v}^{2}}/M\]
C.) \[-\alpha {{v}^{2}}/M\]
D.) \[-\alpha {{v}^{2}}\]

Hint: Use Newton’s second law to get the acceleration of the satellite. The acceleration of dust will be the same as the acceleration of the satellite, since a satellite is the object that sweeps dust. The satellite is moving in space. Hence there are no other forces that will affect the dust or satellite during its motion. So, we can equate this in the relevant position of law of force and acceleration (Newton’s second law).

Complete step by step answer:

Mass of the satellite is M.
The rate of change of mass of interplanetary dust sweeps out by satellite is given.
\[\dfrac{dM}{dt}=+\alpha v\]

According to Newton's second law, the rate change of momentum is directly proportional to the applied force in the same direction of force applied.

\[F=ma\]

We can write this equation like this also, since here the mass is also changing. As we know, force is the rate of change of momentum.

\[F=\dfrac{d}{dt}(mv)\]
Here the mass of the dust is equal to M.
\[F=\dfrac{d}{dt}(Mv)\]
\[F=M\dfrac{dv}{dt}+v\dfrac{dM}{dt}\]
We can assign the given rate of change of mass value to the equation.
\[F=M\dfrac{dv}{dt}+v(\alpha v)\]
Satellites will be in space. Hence, there will not be any other forces to restrict the motion.
So, the force will be zero.
\[M\dfrac{dv}{dt}+v(\alpha v)=0\]
\[M\dfrac{dv}{dt}=-v(\alpha v)\]
\[\dfrac{dv}{dt}=\dfrac{-\alpha {{v}^{2}}}{M}\]
Acceleration is the rate of change of velocity. It is known as instantaneous acceleration. Thus, the acceleration of satellite of mass M will be,
\[a=\dfrac{-\alpha {{v}^{2}}}{M}\]

Hence the option (C) is correct.

Note: Stationary interplanetary dust has lots of meaning. There are no external forces. If there are any other forces, the acceleration will also change. So, the candidates are advised to make force as zero.