Question

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

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

Answer 1

The correct option is D $R^0$

Applying Newton's second law to a circular orbit, we have

$\frac{m{v}^2}{r}=\frac{4{\pi }^{2}rm}{T^2}=\frac{GMm}{r^2}$

where, $m$ is the mass of satellite, and $v$ is the orbital speed.

$T$ is the time period

$\therefore T=\frac{2\pi r^{3/2}}{\sqrt{GM}}$

For $r\approx R$

and $M=\frac{4}{3}\pi R^3\rho$ ($\rho =$ density of planet)

$\because T=\sqrt{\frac{3\pi }{\rho G}}$

i.e. $T$ is independent of $R$.