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.

Explanation

Given
`k=10^(-13)s^(2)m^(-3)`
`=10^(-13)[(1)/((24xx60xx60)^(2))d^(2)][(1)/((1//1000)^(3)km^(3))]`
`=1.33xx10^(-14)d^(2)km^(-3)`
Using Eq. (8.38) and the given value of k.
the time period of the moon is
`T^(2)=(1.33xx10^(-14))(3.84xx10^(5))^(3)`
T=27.3k
Note that Eq. (8.38) also holds for elliptical orbits if we replace `(R_(E)+h)` by the semi-major axis of the ellipse. the earth will then be at one of the foci of this elipse.