Question

Suppose the gravitational force varies inversely as the $n^{th}$ power of distance, then the time period of a planet in a circular orbit of radius $r$ around the Sun will be proportional to

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

Answer 1

Answer: (A)

Given $F = \frac{k}{r^{n}}$
Now $F = \frac{m v^{2}}{r}$ (centripetal force) $= m \omega ^{2} r$
$\Rightarrow \frac{k}{r^{n}} = m \left(\right. \frac{2 \pi }{T} \left.\right)^{2} r$
$T^{2} \propto r^{n + 1}$
$T \propto r^{\frac{n + 1}{2}}$