Question

A test particle is moving in a circular orbit in the gravitational field produced by a mass-density $\rho(r)=\frac{k}{r^{2}}$, where $k$ is a constant. The relation between the radius $R$ of particle's orbit and its period $T$ is found to be $T^{N} R^{M}=$ constant. What is the value of $N + M$ ?

Answer 1

$\rho=\frac{ k }{ r ^{2}}$
$\Rightarrow M =\int \rho dv$
$=\oint \frac{ k }{ r ^{2}}\left(4 \pi r ^{2}\right) dr$
$=4 \pi kr$
Gravitational force on particle, $\frac{ GMm }{ r ^{2}}=\frac{ mv ^{2}}{ r }$
$\Rightarrow v^{2}=\frac{G M}{r}$
$\Rightarrow \frac{ v ^{2}}{(2 \pi r )^{2}}=\frac{ GM }{(2 \pi r )^{2} r }$
$\Rightarrow \frac{1}{ T ^{2}}=($ Constant $) \frac{1}{ r ^{2}} ...[\because M =4 \pi kr ]$
$\Rightarrow \frac{ T }{ R }=$ constant ...[Taking $r=R$]
$N =1, M =-1$
$\therefore N + M =0$