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Weighing the Earth : You are given he fo...

Weighing the Earth : You are given he following data: g=9.81 `ms^(-2) R_(E)=6.37xx10^(6)m`, the distance to the moon `R=3.84xx10^(6)m`, and the time period of the moon's revolution is 27.3 days. Obtain the mass of the Earth `M_(g)` in two different ways.

Text Solution

Verified by Experts

We know that
`M_E=(gR_E^2)/G=(9.81xx(6.37xx10^6)^2)/(6.67xx10^(-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^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%.
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