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Weighing the Earth : You are given the f...

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.

Text Solution

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`implies g = 9.81 ms^(-2)`
`G = 6.67 xx 10^(-11) Nm^(2) kg^(-2)`
`R_(E) = 6.37 xx 10^(6) m`
(i) Acceleration due to gravity ,
`g = (GM_E)/(R_E^2)`
`M_E = (gR_E^2)/G`
`= (9.81 xx (6.37 xx 10^6)^2)/(6.67xx10^(-11))`
`=(398. 059 xx 10^(12))/(6.67xx 10^(-11))`
`= 59.679 xx 10^(23)`
`M_E ~~ 5.97 xx 10^(24) kg`
(ii) `R = 3.84 xx 10^(8) m`
`T = 27.3 ` days
`= 27.3xx 24 xx 3600 s`
`= 23587 xx 10^(2) s`
From Kepler.s third law.,
`T^(2) = (4pi^(2) R^3)/(GM_E)`
`= (4xx(3.14)^(2)xx(3.84xx10^(8))^3)/(6.67xx10^(-11)xx(23587xx10^2)^2)`
`= 0.00000060178 xx 10^(31)`
`~~ 6.02 xx 10^(24)` kg
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