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

Answer: (D)

Applying Newton's second law to a circular orbit, we have
$\frac{m v^{2}}{r}=\frac{4 \pi^{2} r m}{T^{2}}=\frac{G M m}{r^{2}}$
where, $m$ is the mass of satellite, and $v$ is the orbital speed
$T$ is the time period
$\therefore T=\frac{2 \pi r^{3 / 2}}{\sqrt{G M}}$
For $r \approx R$
and $\quad M=\frac{4}{3} \pi R^{3} \rho$
$(\rho=$ density of planet $)$
$ T =\sqrt{\frac{3 \pi}{\rho G}}$
i.e. $T$ is independent of $R$