# Given that the Moon's period about the Earth is 27.32 days and the distance from the Earth to the...

## Question:

Given that the Moon's period about the Earth is 27.32 days and the distance from the Earth to the Moon is {eq}3.84\times10^8 \;m {/eq} estimate the mass of the Earth. Assume the orbit is circular.

## Kepler's Laws:

Kepler's Law is one of the first mathematical attempts to understand the motion of planets and the celestial objects. It is devised by the astronomer Johannes Kepler from the astronomical observations of Tycho Brahe. Kepler's Laws are a precursor to the Newtonian Theory of Universal Gravitation which explains the motion of celestial objects.

The Kepler's Laws are:

1.)The orbit of an object follows a shape of an ellipse

2.)An orbiting object covers equal areas during equal times or as the object comes closer its speed increases or as it is farther away the speed decreases.

3.)The third is law is written mathematically as:

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

Where:

{eq}T {/eq}=Period

{eq}G {/eq}=Gravitational constant

{eq}M {/eq}=Mass

The mass of the Earth is approximately {eq}6.02\times 10^{24} kg {/eq}

Solution:

First Converting 27.32 days to seconds

{eq}27.32 days \times \frac{24 hrs}{1 days} \times \frac{60 mins}{1 hr} \times \frac{60 s}{ 1 min} {/eq}

={eq}2,360,448 s {/eq}

Kepler's Third Law is written as:

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

and rearranging to find M

{eq}M=\frac{4\pi^{2}}{GT^{2}}r^{3} {/eq}

{eq}=\frac{4\pi^{2}}{(6.67\times10^{-11} \frac{m^{3}}{kg s^{2}})(2,360,448 s)^{2}}(3.84\times10^{8}m)^{3} {/eq}

= {eq}6.02\times 10^{24} kg {/eq}