Question

A geostationary satellite is orbiting the Earth at a height of $6R$ above the surface of the earth, $R$ being the radius of the Earth. The time period of another satellite, at a height of $\text{2.5}R$ from the surface of the Earth, is

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

Answer 1

Answer: (A)

A geostationary satellite has a time period of $T=24hr$ . From Kepler's law,
$T^2\propto R^3\Rightarrow \frac{T_2^2}{T_1^2}=\frac{R_2^3}{R_1^3}\Rightarrow \frac{T_2^2}{{\left(24\right)}^2}=\frac{{\left(R+\text{2.5}R\right)}^3}{{\left(R+6R\right)}^3}$
$T_2=\sqrt{\frac{24\times 24}{8}}=6\sqrt{2}hr$ .