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

Answer: 0

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