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 a time period of rotation of $24hours$.

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

Answer 1

The correct option is B

$3h$

Step 1. Given data:

For geostationary satellite radius of orbit, $\begin{array}{rcl}{r}_1& =& R+11R\end{array}$

$\begin{array}{rcl}\therefore r_1& =& 12R\end{array}$

For the another satellite radius of orbit, $r_2=R+2R$

$\therefore r_2=3R$

As the time period of a geostationary satellite is same as the time period of rotation of the planet so,

Time period of geostationary satellite, $T_1=24hours$

Time period of another satellite, $T_2=?$

Step 2. Calculating time period of satellite:

According to Kepler's Third law of periods:

$T\hspace{0.17em}=2\pi \sqrt{\frac{r^3}{GM}}$

where, $r=$radius of orbit

$G\&M=$Constant

Putting values in the above equation:

$T{\hspace{0.17em}}_1=2\pi \sqrt{\frac{{\left(12R\right)}^3}{GM}}$ and, ____$\left(1\right)$

$T{\hspace{0.17em}}_2=2\pi \sqrt{\frac{{\left(3R\right)}^3}{GM}}$ ____$\left(2\right)$

Dividing equation $\left(2\right)$ by $\left(1\right)$.

$\Rightarrow$$\frac{T{\hspace{0.17em}}_2}{T_1}=\sqrt{\frac{{\left(3R\right)}^3}{{\left(12R\right)}^3}}$

$\Rightarrow$$\frac{T{\hspace{0.17em}}_2}{24}=\sqrt{\frac{1}{64}}$ $\left[T_1=24h\right]$

$\Rightarrow$ $T_2=\frac{1}{8}\times 24$

$\therefore T_2=3h$

Hence, correct option is (B).