The mean radius of the earth's orbit of mercury is `6xx10^(10)m`. The mercury will revolve around the sun in nearly

Options

A. `sqrt(2/5)` years
B. `215` years
C. `(2/5)^(2)` years
D. `(2/4)^(3//2)` years

Answer

D

Explanation

`T prop r^(3//2)`
`(T_(m))/(T_(e))=((r_(m))/(r_(e)))^(3//2)=((6xx10^(10))/(1.5xx10^(11)))^(3//2)=(2/5)^(3//2)`
`T_(m)=(2/5)^(3//2) T_(e)=(2/5)^(3//2)xx1=(2/5)^(3//2) years`