Question

An artificial satellite of the Moon revolves in a circular orbit whose radius exceeds the radius of the Moon $\eta$ times. In the process of motion the satellite experiences a slight resistance due to cosmic dust. Assuming the resistance force to depend on the velocity of the satellite as $F=\alpha v^2$, where $\alpha$ is a constant, find how long the satellite will stay in orbit until it falls onto the moon's surface.

Answer 1

For a satellite in a circular orbit about any massive body, the following relation holds between kinetic, potential & total energy,

$T=-E$, $U=2E$ (1)

Thus since total mechanical energy must decrease due to resistance of the cosmic dust, the kinetic energy will increase and the satellite will 'fall', We see then, by work energy theorem

$dT=-dE=-d{A}_{fr}$

So, $mvdv=\alpha v^{2}vdt$ or $\frac{\alpha dt}{m}=\frac{dv}{v^2}$

Now from Newton's law at an arbitrary radius $r$ from the moon's centre,$\frac{v^2}{r}=\frac{\gamma M}{r^2}$ or $v=\sqrt{\frac{\gamma M}{r}}$

($M$ is the mass of the moon.) Then,

$v_i=\sqrt{\frac{\gamma M}{\eta R}}$, $v_f=\sqrt{\frac{\gamma M}{R}}$

where $R=$ moon's radius. So

${\int }_{v_i}^{v_f}\frac{dv}{v^2}=\frac{\alpha }{m}{\int }_0^{\tau }dt=\frac{\alpha \tau }{m}$

or, $\tau =\frac{m}{\alpha }(\frac{1}{v_i}-\frac{1}{v_f})=\frac{m}{\alpha \sqrt{\frac{\gamma M}{R}}}(\sqrt{\eta }-1)=\frac{m}{\alpha \sqrt{gR}}(\sqrt{\eta }-1)$

where $g$ is moon's gravity. The averaging implied by equation (1) (for non-circular orbits) makes the result approximate.