# 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.

## Answer and Explanation:

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.

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