You are given the following data `:g=9.81ms^(-2)`, radius of earth `=6.37xx10^(6)m` the distance the Moon from the earth `=3.84xx10^(8) m` and the time period of the Moon's revolution `=27.3days`. Obtain the mass of the earth in two different ways. `G=6.67xx10^(-11)Nm^(2)kg^(2)`.

Explanation

From Eq. (8.12) we have
`M_(g)=(gR_(E)^(2))/(G)`
Refer to information given in the Box on page 182
`=(9.81xx(6.37xx10^(6))^(2))/(6.67xx10^(-11))`
`=5.97xx10^(24)kg`.
The moon is a stallite of the Earth. From the derivation of Kepler's third law (see Eq. (8.38) ltbr. `T^(2)=(4pi^(2)R^(3))/(GM_(E))`
`M_(E)=(4pi^(2)R^(3))/(GT^(2))`
`=(4xx3.14xx3.14xx(3.84)^(3)xx10^(24))/(6.67xx10^(-11)xx(27.3xx24xx60xx60)^(2))`
`=6.02xx10^(24)kg`
Both methods yield almost the same answer. the difference between them being less than 1%