Question

Suppose the gravitational force varies inversely as the $n^{th}$ power of distance. Then the time period of a planet in circular orbit of radius $R$ around the sun will be proportional to

• A. $R^{\left(\frac{n+1}{2}\right)}$
• B. $R^{\left(\frac{n-1}{2}\right)}$
• C. $R^{n}$
• D. $R^{\left(\frac{n-2}{2}\right)}$

Answer 1

Answer: (A)

$R^{\left(\frac{n+1}{2}\right)}$

Answer 2

Gravitational force, $F \propto \frac{1}{r^{n}}$
$F=\frac{k}{r^{n}}$ where $k$ is a constant.
For a planet, moving in a circular orbit of radius $R$,
$F=\frac{k}{R^{n}}$
But, $F=m \omega^{2}R$
$\Rightarrow \frac{k}{R^{n}}= m R \cdot\left(\frac{2 \pi}{T}\right)^{2}$
$\Rightarrow \frac{k}{R^{n+1}}=\frac{m(2 \pi)^{2}}{T^{2}}$
$\Rightarrow T^{2} \propto R^{n+1}$
$\therefore T \propto R^{\frac{n+1}{2}}$