Weighing the Earth : You are given the following data: g = 9.81 `ms^(-2)`, `R_(E) = 6.37 ×x 10^(6)` m, the distance to the moon R = ` 3.84 ×x 10^(8)` m and the time period of the moon’s revolution is 27.3 days. Obtain the mass of the Earth `M_(E)` in two different ways.

Explanation

From Eq. (8.12) we have
`M_(E) = (g R_(E)^(2))/(G)`
= `(9.81 xx (6.37 xx 10^(6))^(2))/(6.67 xx 10^(11))`
= 5.97`xx 10^(24)` kg.
The moon is a satellite of the Earth. From the derivation of Kepler’s third law
` T^(2) = (4 pi^(2) R^(3))/(GM_(E))`
`M_(E) = (4 pi^(2) R^(3))/(GT^(2))`
`= (4 xx 3.14 xx 3.14 xx (3.84)^(3) xx 10^(24))/(6.67 xx 10^(-11) xx (27.3 xx 24 xx 60 xx 60 )^(2))`
= 6.02 `xx 10^(24)` kg
`Both methods yled almost the same answer. the difference between them being less than 1%.