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The rotation period of an earth satellite close to the surface of the earth is 83 minutes. The time period of another earth satellite in an orbit at a distance of three earth radii from its surface will be

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83 minutes

249 minutes

$83 \times \sqrt{8}$ minutes

664 minutes

answer is C.

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Detailed Solution

For first satellite $r_{1} = R$ and $T_{1} = 83 min ute$
For second satellite $r_{2} = 4 R$
$T_{2} = T_{1} {(\frac{r_{2}}{r_{1}})}^{3 / 2} = T_{1} {(4)}^{3 / 2} = 8 T_{1} = 8 \times 83 = 664 minutes$

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