Question

A Geo-stationary satellite is orbitting the earth at a height of $4R$ above the earth's surface. The time period of another satellite at a height of $2R$ $(\text{in}\text{hours}$) is:
($R$ is the radius of earth)

• A. $$24\times {\left(\sqrt{\frac{3}{5}}\right)}^3$$
• B. $$12\times {\left(\sqrt{\frac{3}{5}}\right)}^3$$
  
• C. $$6\times {\left(\sqrt{\frac{3}{5}}\right)}^3$$
• D. $$72\times {\left(\sqrt{\frac{3}{5}}\right)}^3$$
  

Answer 1

The correct option is A $$24\times {\left(\sqrt{\frac{3}{5}}\right)}^3$$

The time period of a satellite revolving around a planet of radius $R$ and mass $M$ at height $h$ above the surface of planet is given by,

$$T=\frac{{(R+h)}^{\frac{3}{2}}}{\sqrt{GM}}$$

For a given planet, $\frac{1}{\sqrt{GM}}$ term is a constant.

For a Geo-stationary satellite, $T=24\text{hours}$

At a height $h=4R$, we have

$\Rightarrow 24=\frac{1}{\sqrt{GM}}\times {(R+4R)}^{\frac{3}{2}}$

$\Rightarrow 24=\frac{1}{\sqrt{GM}}\times {\left(5R\right)}^{\frac{3}{2}}........\left(1\right)$

Time period of another satellite at height $2R$ is,

$T=\frac{1}{\sqrt{GM}}\times {(2R+R)}^{\frac{3}{2}}$

$\Rightarrow T=\frac{1}{\sqrt{GM}}\times {\left(3R\right)}^{\frac{3}{2}}.......\left(2\right)$

From equation $\left(1\right)$ and $\left(2\right)$ we have,

$\frac{T}{24}=\frac{{\left(3R\right)}^{\frac{3}{2}}}{{\left(5R\right)}^{\frac{3}{2}}}={\left(\frac{3}{5}\right)}^{\frac{3}{2}}$

$\therefore T=24{\left(\sqrt{\frac{3}{5}}\right)}^3$

Hence, option $\left(A\right)$ is the correct answer.