Question

For a given density of the planet, the orbital period of a satellite near the surface of planet of radius $R$ is proportional to:

• A. $R^{\frac{1}{2}}$
• B. $R^{\frac{3}{2}}$
• C. $R^{\frac{-1}{2}}$
• D. $R^0$

Answer 1

The correct option is D $R^0$

From Kepler's third law, $\frac{T^2}{r^3}=\frac{4{\pi }^2}{GM}$
where $r$ is radius, semi-major axis of the elliptical orbital of satellite.

Near the planet surface $r=R$ ; $M=\frac{4}{3}\pi R^3\rho$
So, $T=\sqrt{\frac{3\pi }{G\rho }}\propto R^0$