The period of a satellite in a circular orbit | vedclass

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Question

(C) According to Kepler's third law of planetary motion,the square of the time period $T$ is proportional to the cube of the orbital radius $R$,i.e.,$

Vedclass · Accepted Answer

$8T$

Vedclass · Answer

(C) According to Kepler's third law of planetary motion,the square of the time period $T$ is proportional to the cube of the orbital radius $R$,i.e.,$T^2 \propto R^3$.Given for the first satellite: $T_1 = T$ and $R_1 = R$.For the second satellite: $R_2 = 4R$ and we need to find $T_2$.Using the ratio formula:$\frac{T_2}{T_1} = \left( \frac{R_2}{R_1} \right)^{3/2}$Substituting the values:$\frac{T_2}{T} = \left( \frac{4R}{R} \right)^{3/2}$$\frac{T_2}{T} = (4)^{3/2}$$\frac{T_2}{T} = (2^2)^{3/2} = 2^3 = 8$Therefore,$T_2 = 8T$.

The period of a satellite in a circular orbit of radius $R$ is $T$. The period of another satellite in a circular orbit of radius $4R$ is:

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