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**Kepler's Laws of Planetary Motion**: Back in the late 16th century, Johannes Kepler studied the motion of the planets around the sun. Amid his writings, three laws were formed. These laws are known as Kepler’s Laws of Planetary Motion, or simply, Kepler’s Laws.

The first law stated that the orbit of a planet regarding the Sun was an ellipse, having the Sun’s center of mass as the focus.

The second law stated that a line joining a planet with the Sun sweeps out equal areas in equal intervals of time.

The third law stated that the mean distance cubed over the period squared was the same for each planet. Therefore

(r^{3}_{A}/T^{2}_{A}) = (r^{3}_{B}/T^{2}_{B})

[where r is the distance from the sun the planet and T is the amount of time relative to earth, for the planet to make one orbit around the sun].

Kepler’s third law in particular is useful for finding out tons of data. For example, we can find out the distance from the sun to a given planet, star, or comet. Also, we may calculate the period of the planet, star, or comet if we already have its distance from the sun.

Given that earth is 1.5 x 10^{11} m away from the sun and a certain comet orbits the sun in 923 earth days, find the average distance from the sun to the comet.

*Solution:*

Use Kepler’s Third Law:

(r^{3}_{A}/T^{2}_{A}) = (r^{3}_{B}/T^{2}_{B}) [Earth will be A and the comet will be B.]

Plug in known values.

(1.5 x 10^{11} m)^{3} /365d = r^{3}_{B}/923d

Get r^{3}_{B} by itself.

r^{3}_{B} = [(1.5 x 10^{11} m)^{3}(923d)] / (365d)

r^{3}_{B} = 8.5 x 1033 m^{3}

Take the cube root to solve for just r.

r _{B} = 2.04 x 10^{11} m

This is the average distance between the sun and the comet, according to Kepler's Laws of Planetary Motion.

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