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Question

Physics Question on Keplers Laws

The rotation period of an earth satellite close to the surface of the earth is 83 minutes. The satellite in an orbit at a distance of three times earth radii from 2 jts surface will be

A

83 minutes

B

83×883\times \sqrt{8} minutes

C

664 minutes

D

249 minutes.

Answer

664 minutes

Explanation

Solution

: T2=T1(r2r1)3/2=83(R+3RR)3/2{{T}_{2}}={{T}_{1}}{{\left( \frac{{{r}_{2}}}{{{r}_{1}}} \right)}^{3/2}}=83{{\left( \frac{R+3R}{R} \right)}^{3/2}} =83×8=664=83\times 8=664 minutes.