Question

A satellite in a circular orbit of radius $R$ has a period of $4\,hours$. Another satellite with orbital radius $3 \, R$ around the,same planet will have a period (in hours)

• A. $16$
• B. $4$
• C. $4\sqrt{27}$
• D. $4\sqrt{8}$

Answer 1

Answer: (C)

$4\sqrt{27}$

Answer 2

According to Kepler's third law
$T^{2} \propto R^{3}$
$\Rightarrow \frac{T_{2}}{T_{1}}=\left(\frac{R_{2}}{R_{1}}\right)^{3 / 2}$
$\therefore \frac{T_{2}}{T_{1}}=\left(\frac{3 R}{R}\right)^{3 / 2}$
$\Rightarrow \frac{T_{2}}{T_{1}}=\sqrt{27}$
$\therefore T_{2}=\sqrt{27} T_{1}=\sqrt{27} \times 4=4 \sqrt{27} h$