A satellite in a circular orbit of radius R has a period of 4 hours. Another satellite with orbital radius 3R around the same planet will have a period (in hours)

Options

A. 16
B. 4
C. `4sqrt(27)`
D. `4sqrt(8)`

Answer

C

Explanation

From Kepler's third law, `T^(2)prop R^(3)`
`therefore (T_(1))/(T_(2))=((R )/(3R))^(3//2)` or, `(T_(1))/(T_(2))=((1)/(3))^(3//2)`
or, `T_(2)=T_(1)(3)^(3//2)=4sqrt(27)` hours.