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Circular motion

1,375 bytes added, 14:02, 19 December 2016
Added an example
<math>a = \frac{v^2}{r}</math>
Kepler's third law of planetary motion states that the square of the time period of a planet is proportional to the cube of the orbital distance. This can be seen as the [[earth]] undergoes circular motion around the [[sun]]. The force, <math>F</math>, acting on the earth is just [[gravity]] so
<math>F = \frac{GMm}{r^2}</math>
:<math>G</math> is Newton's gravitational constant
:<math>M</math> is the [[mass (science)|mass]] of the sun
:<math>m</math> is the mass of the earth
:<math>r</math> is the distance between the earth and the sun
From Newton's second law, this is can be related to the acceleration as:
<math>\frac{GMm}{r^2} = m \omega^2 r</math>
Since the angular speed, <math>\omega</math>, is equal to <math>2 \pi / T</math> where <math>T</math> is the time period, we can rewrite the previous equation as
<math>T^2 = \frac{4 \pi^2}{GM} r^3</math>
This is Kepler's third law.
In reality, the orbits of the planets are not circular, instead they are elliptical. This derivation also assumes the mass of the earth is small compared to that of the sun. This is reasonable in this case but not for two stars that orbit as part of a binary system. When this more general case is accounted for, Kepler's third law becomes
<math>T^2 = \frac{4 \pi^2}{G(M+m)} a^3</math>
Where <math>a</math> is the [[semi-major axis]] of the [[ellipse]].