Question

Find earth's approximate mass from the fact that the moon orbits earth in an average time of 27.3 days at an average distance of 384,000 kilometers. (hint: the moon's mass is only about 180 of earth's.)

Answer 1

We can solve the problem by using Kepler's third law, which states:

$\frac{4 \pi^2}{T^2}=\frac{GM}{r^3}$

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:

$T=27.3 d \cdot 24 \cdot 60 \cdot 60 = 2358720 s=2.36 \cdot 10^6 s$

$r=384000 km=3.84 \cdot 10^8 m$

We can re-arrange the equation and find the Earth's mass:

$M=\frac{4 \pi^2 r^3}{GT^2}=\frac{4 \pi^2 (3.84 \cdot 10^8 m)^3}{(6.67 \cdot 10^{-11})(2.36 \cdot 10^6 s)^2}=6.0 \cdot 10^{24} kg$

Answer 2

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

$\frac{4\pi ^2}{T^2} = \frac{GM}{r^3}$ ----- ( 1 )

where : T = 27.3 days = 2.36 * 10⁶ secs,  r = 3.84 * 10⁸ m , G = constant

Therefore :

M ( mass of earth ) = $\frac{4\pi ^2r^3}{GT^2}$ ---- ( 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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