Respuesta :
We can solve the problem by using Kepler's third law, which states:
[tex]\frac{4 \pi^2}{T^2}=\frac{GM}{r^3}[/tex]
where T is the period of revolution of the Moon around the Earth, G is the gravitational constant, M the Earth's mass and r the average distance between Earth and Moon.
Using the data of the problem:
[tex]T=27.3 d \cdot 24 \cdot 60 \cdot 60 = 2358720 s=2.36 \cdot 10^6 s[/tex]
[tex]r=384000 km=3.84 \cdot 10^8 m[/tex]
We can re-arrange the equation and find the Earth's mass:
[tex]M=\frac{4 \pi^2 r^3}{GT^2}=\frac{4 \pi^2 (3.84 \cdot 10^8 m)^3}{(6.67 \cdot 10^{-11})(2.36 \cdot 10^6 s)^2}=6.0 \cdot 10^{24} kg[/tex]
The approximate mass of the earth is : 6 * 10²⁴ kg
Given data :
Average time moon orbits ( T ) = 27.3 days
Average distance of moon from earth ( r ) = 384,000 Km = 3.84 * 10⁸ m
Determine the approximate mass of the earth
Applying kepler's third law
[tex]\frac{4\pi ^2}{T^2} = \frac{GM}{r^3}[/tex] ----- ( 1 )
where : T = 27.3 days = 2.36 * 10⁶ secs, r = 3.84 * 10⁸ m , G = constant
Therefore :
M ( mass of earth ) = [tex]\frac{4\pi ^2r^3}{GT^2}[/tex] ---- ( 2 )
insert values into equation ( 2 )
M = 6 * 10²⁴ kg
Hence we can conclude that The approximate mass of the earth is : 6 * 10²⁴ kg.
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