Weighing the earth : Youare given the following data : g =9.81 ms^(-2).R_(E)= 6.37 xx10^(6) m, the distance to the moon R = 3.84xx10^(8)m and the time period of the moons revolution is 27 .3 days . Obtain the mass of the earth M_E in two different ways.

Weighing the earth : Youare given the following data : g =9.81 ms^(-2)

1.	Weighing the earth : Youare given the following data : g =9.81 ms^(-2).R_(E)= 6.37 xx10^(6) m, the distance to the moon R = 3.84xx10^(8)m and the time period of the moons revolution is 27 .3 days . Obtain the mass of the earth M_E in two different ways.
Answer» Solution :We know that `M_E=(gR_E^2)/G=(9.81 xx(6.37xx10^(6))^2)/(6.67 xx10^(-11))=5.97xx10^(24)kg` The moon is a SATELLITE of the EARTH. According to Kepler.s Third law `T^(2) = (4PI^2R^3)/(GM_E) M_(E) =(4pi^(2)R^3)/(GT^2)` `= (4xx3.14 xx3.14 xx(3.84)^3xx10^(24))/(6.67xx10^(-11) xx(27.3xx24xx60xx60)^(2))=6.02xx10^(24)` kg Both methods yields almost the same answer , the difference between them being less than 1% .