Question

A geostationary satellite is orbiting the earth at a height of $5R$ above the surface of the earth has a time period of $24$ hours. $R$ being the radius of the earth.The time period of another satellite in hours at a height $2R$ from the surface of the earth is

• A. $6\sqrt{2}$
• B. $\frac{6}{\sqrt{2}}$
• C. $5$
• D. $10$

Answer 1

$\text{Step 1: Kepler's Third Law}$

From Kepler's Third Law "The square of the time period $\left(T\right)$ of revolution of a planet around the sun in an elliptical orbit is directly proportional to the cube of its semi-major axis $\left(r\right)$"

$T^2\propto r^3$

So, $T_1^2\propto r_1^3$ and $T_2^2\propto r_2^3$

$\therefore T_2=T_1{\left(\frac{r_2}{r_1}\right)}^{\frac{3}{2}}$ $....\left(1\right)$

$\text{Step 2: Time period Calculation}$

Given, Time period of first satellite $T_1=24hr$

Orbiting radius of first satellite

$r_1=5R+R=6R$

Orbiting radius of second satellite

$r_2=2R+R=3R$

$\therefore$ From Equation $\left(1\right)$, Time period of second satellite:

$T_2=24hr{\left(\frac{3R}{6R}\right)}^{\frac{3}{2}}=\frac{24}{2\sqrt{2}}hours$

$T_2=6\sqrt{2}hours$

Hence time period of another satellite in hours at height $2R$ from the surface of earth is $6\sqrt{2}hours$. Option $\left(A\right)$