Question

A satellite having time period same as that of the earth's rotation about its own axis is orbiting the earth at a height $8$R above the surface of earth, where, R is radius of earth. What will be the time period of another satellite at a height $3.5$R from the surface of earth?

Answer 1

Time period of satellite $T^2=\left(\frac{4{\pi }^2}{GM}\right)R^3$ Where $R$ is radius of orbit.

$\Rightarrow T^2\propto R^3$

Given Time period of satellite at height $8R$ is $T_1=24$ hours,

Let $T_2$ be time period of satellite at height.

$\Rightarrow \frac{T_2^2}{T_1^2}=\frac{(3.5R+R{)}^3}{(8R+R{)}^3}$

$\Rightarrow T_2=T_1(\frac{1}{2}{)}^{\frac{3}{2}}$ $\Rightarrow T_2=\frac{24}{2\sqrt{2}}=6\sqrt{2}$ hours