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.

Answer 2

$\huge{ \underline { \rm{ \large{ \purple{Given:}}}}}$

Satellite in a circular orbit of radius R has a period of 4 hours

Another satellite with orbital radius 3R

$\huge{ \underline{ \rm{ \large{ \green{Find:}}}}}$

Time of second satellite???

$\huge{ \underline{ \rm{ \red{ \large{Solution:}}}}}$

From Kepler's 3rd law T² is proportional to r³, the orbital period squared is proportional to the distance cubes

T² ∝ R³

T² ∝ R³(T1/T2) ² = (R1/R2) ³

${ \implies{ \sf{ \frac{ {(4)}^{2} }{ {(T _{2})}^{2} } = \frac{ {(R)}^{3} }{ {(3R)}^{3} } }}}$

${ \implies{ \sf{ {( T_{2} )}^{2} = 27 \times 16}}}$

${ \implies{ \sf{ { T_{2} } = \sqrt{27 \times 16} }}}$

${ \implies{ \sf{ \red{ T_{2} = 4 \sqrt{27} hours }}}}$

Therefore,

Time of second satellite = 4√27 hours.

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