Question

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

Answer 1

We know that Kepler's Third law states that, The square of it's period of revolution around Sun is directly proportional to cube of mean distance of a planet from the Sun. Then,

$\longrightarrow\sf{ {T}^{2} ∝ {r}^{3} }$

i.e,

$\longrightarrow\sf{ \frac{ {T}^{2} }{ {R}^{3} } = Constant \: K \: .... \: (1)}$

$\longrightarrow\sf{ ( { \frac{T_1}{T_2} )}^{2} = ( { \frac{R_2}{R_1} )}^{3} }$

$\longrightarrow\sf{ \frac{T_2}{T_1} = (\frac{R_2}{R_1} ) {}^{\frac{3}{2} } }$

Thus, we get,

$\star \: \boxed{\sf\green{T_2 = t_1( \frac{R_2}{R_1} ) {}^{ \frac{3}{2} } }}$

Here,

$\sf{R_1 =R}$

$\sf{T_1 = 4}$

$\sf{R_2 = 3R}$

Thus, Period of second satellite will be:

$\longrightarrow\sf{t_2 = 4( \frac{3R}{R} ) {}^{ \frac{3}{2} } }$

$\longrightarrow\sf{t_2 = 4 \times 3 {}^{ \frac{3}{2} } }$

$\longrightarrow{\underline{\boxed{\sf{T_2 = 4 \sqrt{27} \: hr}}}}$

Thus,

• The same planet will have a period of 4 √27 hours.