Question

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

• A. $6.45h$
• B. $5.39h$
• C. $6.23h$
• D. $5.48h$

Answer 1

The correct option is A $6.45h$

According to Kepler's law,
$\frac{T_1^2}{T_2^2}=\frac{R_1^3}{R_2^3}$

$\Rightarrow \frac{24\times 24}{T_2^2}=\frac{6\times 6\times 6\times R^3}{2.5\times 2.5\times 2.5\times R^3}$

$\Rightarrow T_2^2=\frac{24\times 24\times 2.5\times 2.5\times 2.5}{6\times 6\times 6}=6.45h$