Question 64: Gravitation

Question

Given below are two statements :
Statement I : A satellite is moving around earth in the orbit very close to the earth surface. The time period of revolution of satellite depends upon the density of earth.
Statement II : The time period of revolution of the satellite is $T=2 \pi \sqrt{\frac{R_e}{g}}$ (for satellite very close to the earth surface), where $R_{\mathrm{e}}$ radius of earth and $g$ acceleration due to gravity. In the light of the above statements, choose the correct answer from the options given below :

Accepted Answer

For a satellite revolving in a circular orbit of radius $r$ around the Earth (mass $M$ ), the time period ( $T$ ) is given by Kepler's Third Law
For the equilibrium along the radial direction,
$$ F_g=\frac{m v^2}{r} \Rightarrow \frac{G M_e m}{r^2}=\frac{m v^2}{r} \Rightarrow v=\sqrt{\frac{G M_e}{r}} $$
So, the time period of the revolution is,
$$ \mathrm{T}=\frac{2 \pi \mathrm{r}}{\mathrm{v}}=\frac{2 \pi \mathrm{r}}{\sqrt{\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{r}}}} $$
$$\Rightarrow $$ $$ \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{r}^3}{\mathrm{GM}_{\mathrm{e}}}} $$
where G is the universal gravitational constant and $\mathrm{M}_{\mathrm{e}}$ is the mass of Earth.
If the satellite is orbiting very close to the Earth's surface, its orbital radius $r$ is approximately equal to the Earth's radius, $\mathrm{R}_{\mathrm{e}}$.
$$ \mathrm{T} \approx 2 \pi \sqrt{\frac{\mathrm{R}_{\mathrm{e}}^3}{\mathrm{GM}_{\mathrm{e}}}} $$
Assuming the Earth is a uniform sphere, its mass M is the product of its volume and density:
$$ \mathrm{M}=\frac{4}{3} \pi \mathrm{R}_{\mathrm{e}}^3 \rho $$
Substitute this mass into our time period equation :
$$ T=2 \pi \sqrt{\frac{R_e^3}{G\left(\frac{4}{3} \pi R_e^3 \rho\right)}}=2 \pi \sqrt{\frac{3}{4 \pi G \rho}} $$
$$\Rightarrow $$ $$ T=\sqrt{\frac{3 \pi}{G \rho}} $$
Since $3, \pi$, and G are all constants, this equation proves that the time period of a satellite orbiting very close to the Earth's surface depends entirely on the density ( $\rho$ ) of the Earth.
Therefore, Statement I is true.
We also know the formula for the acceleration due to gravity (g) on the surface of the Earth :
$$ \mathrm{g}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}^2} $$
$$ \Rightarrow \mathrm{GM}=\mathrm{gR}_{\mathrm{e}}^2 $$
Substitute this expression for $\mathrm{GM}_{\mathrm{e}}$ into the time period equation $\mathrm{T} \approx 2 \pi \sqrt{\frac{\mathrm{R}_{\mathrm{e}}^3}{\mathrm{GM}_{\mathrm{e}}}}$
$$ \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{R}_{\mathrm{e}}^3}{\mathrm{gR}_{\mathrm{e}}^2}}=2 \pi \sqrt{\frac{\mathrm{R}_{\mathrm{e}}}{\mathrm{~g}}} $$
Therefore, Statement II is also true.
As both statement I and statement II are true, so the correct option is (C).

Answer

Statement I is true but Statement II is false

Answer

Statement I is false but Statement II is true

Answer

Both Statement I and Statement II are true

Answer

Both Statement I and Statement II are false

Gravitation

Choose the correct answer: