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.

Weighing the Earth : You are given the following data: g = 9.81 ms^(-2

1.	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.
Answer» SOLUTION :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%.