Weighing the earth : You are given the following data: g=9.81 ms^(-2). R_E= 6.37xx10^6 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 : 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= 6.37xx10^6 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=(9.81xx(6.37xx10^6)^2)/(6.67xx10^(-11))=5.97xx10^24` kg The moon isa satellite of the Earth. According toKepler.s THIRD law `T^2=(4pi^2R^3)/(GM_E) M_E=(4pi^2R^3)/(GT^2)` `=(4xx3.14xx3.14xx(3.84)^3xx10^24)/(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%.