# Calculate in days the expected period of rotation of the moon around the earth. Mass of earth is...

## Question:

Calculate in days the expected period of rotation of the moon around the earth. Mass of earth is 5.977 x 10{eq}^{24} {/eq} kg, and the mean radius of the moon orbiting around the earth is 3.844 x 10{eq}^8 {/eq} m. Use {eq}\displaystyle T = \frac{4 \pi^2 r^3}{GM} {/eq}.

## Kepler's Third Law

Kepler's third law of planetary motion states that the period of an object orbiting a star/planet is related to the semi-major axis of the elliptical orbit. The relationship of the third law is:

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

Here:

{eq}\displaystyle T {/eq} is the period of the object's orbit in seconds, {eq}\displaystyle r {/eq} is the semi-major axis (radius for a circular orbit) of the elliptical orbit in meters, {eq}\displaystyle G \approx 6.67 \times 10^{-11} \: \text{m}^3/\text{kg s}^2 {/eq} is the gravitational constant, and {eq}\displaystyle M {/eq} is the mass of the star/planet that is being orbited (assuming it to be much larger than the orbiting object's mass) in kilograms.

To calculate the expected period of the Moon's orbit around the Earth, we will use Kepler's third law.

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

We know the mean radius of the Moon's (nearly) circular orbit is {eq}\displaystyle 3.844 \times 10^8 {/eq} m, and that the Earth's mass is {eq}\displaystyle 5.977 \times 10^{24} {/eq} kg. Plugging these into the equation gives a period of:

{eq}\displaystyle T = \sqrt{\frac{4 \pi^2 (3.844 \times 10^8)^3}{(6.67 \times 10^{-11})(5.977 \times 10^{24})}} \approx 2.37 \times 10^6 \: \text{s} {/eq}

Using the fact that there are 86400 seconds in a day allows us to convert this period to days as:

{eq}\displaystyle T \approx \frac{2.37 \times 10^6}{86400} = 27.43 \: \text{days} {/eq}

Thus, the expected period of the Moon's orbit around the Earth is approximately 27.43 days. 