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. `28`
B. `2.8`
C. `27.3`
D. `2.73`

Answer

C

Explanation

Given `k=10^(-13)x1/((24xx60xx60)^(2))d^(2)1/((1//1000)^(3))km^(-3)`
`=1.33xx10^(-14)d^(2)km^(-3)`.
For the moon, `r=3.84xx10^(5)km`.
`:. T^(2)=kr^(3)=1.33xx10^(-14)xx(3.84xx10^(5))^(3)=27.3 days`