Question

Suppose the gravitational force varies inversely as the nth 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

The correct option is A

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

For a planet of mass m moving with a speed v in a circular orbit of radius R around the sun of mass M, the centripetal force is given by:
$F_c=\frac{m{v}^2}{R}....\left(1\right)$

Gravitational force varies inversely as the nth power of distance. Hence,
$F_g=\frac{GmM}{R^n}....\left(2\right)$

Centripetal force is provided by the gravitational force.
$\frac{m{v}^2}{R}=\frac{GmM}{R^n}$
$v=\sqrt{\frac{GM}{R^{n-1}}}$

Time period,
$T=\frac{2\pi R}{v}$
$T=\frac{2\pi R}{\sqrt{GM/R^{n-1}}}$
$T\propto R^{\left(\frac{n+1}{2}\right)}$