Given that `T^(2) = kR^(3)`, express the constant `k` of the above relation in days and kilometres. Given, `k= 10^(-13)s^(2) m^(-3)`. The Moon is at a distance of `3.84 xx 10^(5) km` from the earth. Obtain its time period of revolution in days.

Options

A. `16d`
B. `24d`
C. `27.3 d`
D. `19d`

Answer

C

Explanation

Given, `k = 10^(-13) s^(2) m^(-3)`
`= 10^(-13) [(1)/((24 xx 60 xx 60)^(2))d^(2)] [(1)/((1//1000)^(3)km^(3))]`
`= 1.33 xx 10^(-14) d^(2) km^(-3)`
`T^(2) = k R^(3)` or `T = (k R^(3))^(1//2)`
`:. T = [1.33 xx 10^(-14) xx (3.84 xx 10^(5))^(3)]^(1//2)`
`27.3 d`