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

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

The centripetal force $\left(F_c\right)$ necessary to keep the planet in its circular orbit is provided by the gravitational force $\left(F_g\right)$ between the planet and the sun, i.e.
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
$F_c=\frac{m{v}^2}{R}....\left(1\right)$
If the gravitational force were to vary inversely as the nth power of distance, the gravitational force would be equal to
$F_g=\frac{GmM}{R^n}....\left(2\right)$
Equating equations (1) and (2), we get
$v=\sqrt{\frac{GM}{R^{n-1}}}$
Now, time period, $T=\frac{2\pi R}{v}=\frac{2\pi }{\sqrt{GM}}{R}^{\left(\frac{n+1}{2}\right)}\Rightarrow T\propto R^{\left(\frac{n+1}{2}\right)}$
Hence, the correct choice is (a).