Question

find the mass of the earth if it's radius is 6.37×10^6m . the distance to the moon is 3.84×10^8m and the time period of Moon's revolution is 27.3 days.​

Answer 1

Answer:

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Explanation:

$\sf \: Here, g=9.81m s−2 \\ \sf\:R=6.37×106m, r=3.84×108m \\ \sf \: T=27.3days =27.3×24×60×60s \\ \sf ∴M_{e} = \frac{ {gR _{e}}^{2} }{G} = \frac{9.81 \times {6.37 \times 10^{6} }^{2} }{6.67 \times 10^{ - 11} } \\ \sf = 5.97 \times {10}^{24} kg$

Alternative method: Since the gravitational pull provides the required centripetal force, so

$\sf \: mr(2\pi/ {T}^{2} ) = GM_{e} {m/ r}^{2} or \: M_{e} = \frac{ {4\pi}^{2} {r}^{3} }{GT^{2} } \\ \sf \: = \frac{4 \times (22 \times 7)^{2} \times {(3.84 \times {10}^{8} )}^{3} }{6.67 \times {10}^{ - 11 } \times {(27.3 \times 24 \times 60 \times 60)}^{2} } \\ \sf = \: 6.02 \times {10}^{24} kg$

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