Question

A spaceship orbits around a planet at a height of $20\text{km}$ 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 $24\text{hours}$ around the planet ?
[Given: Mass of planet $=8\times {10}^{22}\text{kg}$, Radius of planet$=2\times {10}^6\text{m}$, Gravitational constant $G=6.67\times {10}^{-11}{\text{Nm}}^2/{\text{kg}}^2$]

• A. $13$
• B. $17$
• C. $11$
• D. $9$

Answer 1

The correct option is C $11$

Given:
Height of spaceship $=20\text{km}$
Mass of planet $8\times {10}^{22}\text{kg}$
Radius of planet$=2\times {10}^6\text{m}$,
Gravitational constant $G=6.67\times {10}^{-11}{\text{Nm}}^2/{\text{kg}}^2$

Time period of revolution,

$T=\frac{2\pi r}{v}$

Velocity of the spaceship,

$v=\sqrt{\frac{GM}{r}}$

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

Substituting the value,
we get

$T=2\pi \sqrt{\frac{(202\times {10}^4{)}^3}{6.67\times {10}^{-11}\times 8\times {10}^{22}}}\text{sec}$

$T=7812.2\text{s}$

$T\approx 2.17\text{hr}$

$\Rightarrow \text{Number of revolutions}=\frac{24}{2.17}\approx 11$.

Hence option $\left(D\right)$ is correct.