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Kepler's Law

Two satellite...

Question

Two satellites A and B of masses $m_{1}$ and $m_{2} (m_{1} = 2 m_{2})$ are moving in circular orbits of radii $r_{1}$ and $r_{2} (r_{1} = 4 r_{2})$ , respectively, around the earth. If their periods are $T_{A}$ and $T_{B}$ , then the ratio $T_{A} / T_{B}$ is

A

4

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B

16

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C

2

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D

8

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Solution

The correct option is B 8
Kepler second law - square of time period is proportional to cube of radius of orbit.
$∴ T^{2} α r^{3}$
In this case, $r_{1} = 4 r_{2} \Rightarrow \frac{r_{1}}{r_{2}} = 4$
$T_{A}^{2} α r_{1}^{3}$ and $T_{B}^{2} α r_{2}^{3}$
$∴ \frac{T_{A}}{T_{B}} = (\frac{r_{1}}{r_{2}})^{\frac{3}{2}} = (4^{\frac{1}{2}})^{3}$
$\Rightarrow \frac{T_{A}}{T_{B}} = 8$

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