Suppose the gravitational force varies inversely as the (n^ th )...

Suppose the gravitational force varies inversely as the (n^{\text{th }}) power of distance. Then the time period of a planet in circular orbit of radius

R

around the sun will be proportional to

Options

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)})

Correct answer

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

Step-by-step solution

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

\begin{aligned} & \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}} .\end{aligned}

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