A spaceship orbits around a planet at a height of $20\ \text{km}$ from its surface. Assuming that only the 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 $= 8\times10^{22}\ \text{kg}$, radius of planet $= 2\times10^6\ \text{m}$, gravitational constant $G = 6.67\times10^{-11}\ \text{N m}^2/\text{kg}^2$]
Answer: (D) $11$
Orbit radius $r = 2.02\times10^6\ \text{m}$.
$$T = 2\pi\sqrt{\frac{r^3}{GM}} = 2\pi\sqrt{\frac{8.24\times10^{18}}{5.34\times10^{12}}} = 2\pi\times1243 \approx 7.81\times10^3\ \text{s}$$
Revolutions in a day: $\dfrac{86400}{7810} \approx 11.06$, so $11$ complete revolutions.
Solution by Sreeraj P, M.Sc Physics