Question

A spherical planet has uniform density $\frac{\pi}{2}\times 10^4 \,kg/m^3$ Find out the minimum period for a satellite in a circular orbit around it in seconds. (Use $G = \frac{20}{3} \times 10^{-11} \frac{N-m^2}{kg^2})$

• A. $7500$
• B. $3000$
• C. $4500$
• D. $6000$

Answer 1

Answer: (B)

Time period is minimum for the satellites with minimum radius of the orbit i.e. equal to the radius of the planet. Therefore.
$\frac{GMm}{R^{2}} = \frac{mv^{2}}{R} $
$V = \sqrt{\frac{GM}{R}} $
$T_{min} = \frac{2\pi R}{\sqrt{\frac{GM}{R}}} = \frac{2\pi R\sqrt{R}}{\sqrt{GM}}$
using $M =\frac{4}{3}pR^{3} \cdot\rho \,\,T_{min} = \sqrt{\frac{3\pi}{G\rho}}$
Using values $T_{min} = 3000 \,s$