Question

A spaceship orbits around a planet at a height of $20\, km$ from its surface. Assuming that only gravitational field of the planet acts on the spaceship, what will be the number of complete revolutions made by the spaceship in $24\, hours$ around the planet ?
[Given : Mass of planet = $8 \times 10^{22} \; kg $;
Radius of planet = $2 \times 10^6 \; m$,
Gravitational constant $G = 6.67 \times 10^{-11} \; Nm^2/kg^2$]

• A. 9
• B. 11
• C. 13
• D. 17

Answer 1

Answer: (B)

$F_{g} = \frac{mv^{2}}{r} $
$ \frac{GMm}{r^{2}} = \frac{mv^{2}}{r} $
$ V = \sqrt{\frac{GM}{r}} = \sqrt{\frac{\left(6.67 \times10^{-11}\right)\left(8\times10^{22}\right)}{2.02 \times10^{6}}} $
$V= 1.625 \times10^{3}$
$ T = \frac{2\pi r}{V} $
$ n \times T =24 \times60 \times60 $
$ n \left[\frac{2\pi\left(2.02\times10^{6}\right)}{1.625 \times10^{3}}\right] = 24 \times3600 $
$n = \frac{24\times3600 \times1.625 \times10^{3}}{2\pi\left(2.02\times10^{6}\right)}$
$ n = 11$