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

Answer

• There are two satellites
• Satellite¹ takes 4 hours to orbit around a circular orbit of radius R
• Satellite² orbits on a circular path of radius 3R
• Time taken = ?

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$\quad$ ● Here we use Kepler's Law which states that the square of the time taken will be equal to the cube of the radius

$\quad$ ● T² ∝ R³

$\sf:\implies {\bigg\lgroup \dfrac{T_1}{T_2} \bigg\rgroup}^{2} = {\bigg\lgroup \dfrac{R_2}{R_1}\bigg\rgroup}^{3}$

$\sf:\implies \bigg\lgroup \dfrac{T_1}{T_2} \bigg\rgroup = \sqrt{{\bigg\lgroup \dfrac{R_2}{R_1}\bigg\rgroup}^{3}}$

$\sf:\implies \bigg\lgroup \dfrac{T_1}{T_2} \bigg\rgroup = {\bigg\lgroup \dfrac{R_2}{R_1}\bigg\rgroup}^{\frac{1}{2}\times 3}$

$\sf:\implies \bigg\lgroup \dfrac{T_1}{T_2} \bigg\rgroup = {\bigg\lgroup \dfrac{R_2}{R_1}\bigg\rgroup}^{\frac{3}{2}}$

$\sf:\implies T_2 = T_1 \times {\bigg\lgroup \dfrac{R_2}{R_1}\bigg\rgroup}^{\frac{3}{2}}$

$\quad$ ● Here the R₁ will be R, then T₁ is 4 Hours and R₂ will be 3R

$\sf:\implies T_2 = 4\times {\bigg\lgroup \dfrac{3R}{R}\bigg\rgroup}^{\frac{3}{2}}$

$\sf:\implies T_2 = 4\times 3^{\frac{3}{2}}$

$:\implies\underline{\boxed{\pink{\mathfrak{ t_2 = 4\sqrt{27} \ hours}}}}$

Answer 2

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

• Satellite in a circular orbit of radius R has a period of 4 hours
• Another satellite with orbital radius 3R

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

• Time of second satellite???

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

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

• T² ∝ R³

• (T1/T2) ² = (R1/R2) ³

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

$\: \: \:$

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

$\: \: \: \:$

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

$\: \: \: \:$

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

Therefore,

• Time of second satellite = 4√27 hours