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Question

Question: One project after deviation from its path, starts moving round the earth in a circular path at radiu...

One project after deviation from its path, starts moving round the earth in a circular path at radius equal to nine times the radius at earth R, its time period will be

A

2πRg2 \pi \sqrt { \frac { R } { g } }

B

27×2πRg27 \times 2 \pi \sqrt { \frac { R } { g } }

C

πRg\pi \sqrt { \frac { R } { g } }

D

8×2πRg8 \times 2 \pi \sqrt { \frac { R } { g } }

Answer

27×2πRg27 \times 2 \pi \sqrt { \frac { R } { g } }

Explanation

Solution

=2π(9)3/2Rg= 2 \pi ( 9 ) ^ { 3 / 2 } \sqrt { \frac { R } { g } } r=9Rr = 9 R (given)].