A satellite moves in a circle around the earth. The radius of this circle is equal to one half of the radius of the moon's orbit. The satellite completes one revolution is :

Options

A. `2^(-3//2)` lunar month
B. `2^(3//2)` lunar month
C. `(1)/(2)` lunar month
D. `(2)/(3)` lunar month

Answer

A

Explanation

The moon completes one revolution around the earth in one lunar month. i.e. `T_(m)= 1` lunar month and it is given that `r_(s)=(r_(m))/(2)`
From Kepler's law, `T^(2)prop R^(3)`
`therefore (T_("sat"))/(T_(m))=((r_(s))/(r_(m)))^(3//2)=((1)/(2))^(3//2)=(2^(-1))^(3//2)=2^(-3//2)`
`therefore T_(s)=2^(-3//2)T_(m)=2^(-3//2)` (lunar month)