You are given the following data `:g=9.81ms^(-2)`, radius of earth `=6.37xx10^(6)m` the distance of the Moon from the earth `=3.84xx10^(8) m` and the time period of the Moon's revolution `=27.3days`. Obtain the mass of the earth in two different ways. `G=6.67xx10^(-11)Nm^(2)kg^(-2)`.
Text Solution
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Here `g=9.81ms^(-2)` `R=6.37xx10^(6)m,r=3.84xx10^(8)m` `T=27.3` days `=27.3xx24xx60xx60s` `:. M_(e)=(gR_(e)^(2))/G=(9.81xx(6.37xx10^(6))^(2))/(6.67xx10^(-11))=5.97xx10^(24)kg` Alternative method: Since the gravitational pull provides the required centropetal force so `mr(2pi//T^(2))=GM_(e)m//r^(2)` or `M_(e)=(4pi^2r^(3))/(GT^(2))` `=(4xx(22//7)^(2)xx(3.84xx10^(8))^(3))/(6.67xx10^(-11)xx(27.3xx24xx60xx60)^(2))` `=6.02xx10^(24)kg`
You are given the following data :g=9.81ms^(-2) , radius of earth =6.37xx10^(6)m the distance the Moon from the earth =3.84xx10^(8) m and the time period of the Moon's revolution =27.3days . Obtain the mass of the earth in two different ways. G=6.67xx10^(-11)Nm^(2)kg^(2) .
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