Question

A geostationary satellite is orbiting around an arbitrary planet $P$ at a height of $11R$ above the surface of $P$, $R$ being the radius of $P$. The time period of another satellite in hours at a height of $2R$ from the surface of $P$ is ___. $P$ has the time period of rotation of $24$ hours.

• A. $\frac{6}{\sqrt{2}}$
  
• B. $3$
• C. $6\sqrt{2}$
• D. $5$

Answer 1

The correct option is B $3$

From Kepler's law: $$T^2\propto R^3$$where $R$ is distance of satellite from the centre of the planet.

Since time period of a geostationary satellite is equal to the time period of rotation of planet. In this case that is equal to $24$ hours.

Using Kepler's law
${\left(\frac{24}{T}\right)}^2={\left(\frac{12R}{3R}\right)}^3$

$T=3\text{hour}$