The time period `T` of the moon of planet mars (mass `M_(m)`) is related to its orbital radius `R` as (`G`=gravitational constant)

Options

A. `T^(2)=(4pi^(2)R^(3))/(GM_(m))`
B. `T^(2)=(4pi^(2)GR^(3))/(M_(m))`
C. `T^(2)=(2piGR^(3))/(M_(m))`
D. `T^(2)=(4piM_(m)GR^(3))`

Answer

A

Explanation

Time period, `T=(2piR)/((sqrt(GM_(m))/R))=(2piR^(3//2))/(sqrt(GM_(m)))`
Where the symbols have their meanings as given. Squaring boty sides, we get
`T^(2)=(4pi^(2)R^(3))/(GM_(m))`