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

Ist method:
`g = (GM_(E))/(R_(E)^(2))` or `M_(E) = (gR_(E)^(2))/(G)`
`:. M_(E) = (9.81 xx (6.37 xx 10^(6))^(2))/(6.67 xx 10^(-11))`
`= 5.97 xx 10^(24) kg`
2nd method : The Moon is the satellite of the Earth. Then, time period of revolution of Moon is
`T = (2 pi R)/(upsilon) = (2pi R)/(sqrt(GM_(E)//R))`
Squaring both the sides, we have
`T^(2) = (4 pi^(2)R^(3))/(GM_(E))` or `M_(E) (4 pi^(2) R^(2))/(GT^(2))`
`:. M_(E) = (4 xx (22//7)^(2) xx (3.84 xx 10^(8))^(3))/((6.67 xx 10^(-11))xx (27.3 xx 24 xx 60 xx 60)^(2))`
`= 6.02 xx 10^(24) kg`