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Laws of Newton -
Kepler's Third Law

Kepler's Third Law states that for all objects moving around a fixed source of gravitational field along elliptical orbits the ratio of a square of their period of rotation to a cube of a semi-major axis is the same.

As in the case of the Kepler's First Law, this Third Law has been based on numerous experiments and years of observation.

Based on all the knowledge conveyed in previous lectures on Kepler's Laws, we will derive this Third Law theoretically.

Let's make a simple derivation of Kepler's Third Law in case of a circular orbit.

In this case the velocity vector of an object circulating around a central point is always perpendicular to a position vector from a center to an object.
Since the gravitational force is collinear with a position vector, it is also perpendicular to velocity, which is tangential to a circular orbit. Therefore, gravitational force makes no action along a velocity vector which makes the magnitude of the velocity vector constant.

Let's introduce the following characteristics of motion:
t - absolute time,
r - radius of a circular orbit of a moving object,
F - vector of gravity,
M - mass of the source of gravitational field,
m - mass of object moving in the gravitational field,
r - position vector from the source of gravitational field to a moving object,
r'=v - velocity vector of a moving object,
r"=v'=a - acceleration vector of a moving object,
T - period of circulation,
ω=2π/T - scalar value of angular velocity,
Here bold letters signify vectors, regular letters signify scalars and magnitudes of corresponding vectors, single and double apostrophes signify first and second derivative by time.

Constant magnitude v of velocity vector means constant angular velocity ω and obvious equality v=r·ω.

Magnitude a of an acceleration vector can be simply found by representing a position vector as a pair of Cartesian coordinates (x,y):
x = r·cos(ωt)
y = r·sin(ωt)
x' = −r·ω·sin(ωt)
y' = r·ω·cos(ωt)
x" = −r·ω²·cos(ωt)=−ω²·x
y" = −r·ω²·sin(ωt)=−ω²·y
and, therefore,
r" = −ω²·r
(collinear with r and F)
from which follows
a = |a| = |−ω²·r| = ω²·r

According to the Newton's Second Law,
F = m·a
According to the Universal Law of Gravitation,
F = G·M·m/r²
Therefore,
a = ω²·r = G·M/r²
from which follows
ω² = G·M/r³

Since ω=2π/T,
4π²/T² = G·M/r³
T²/r³ = 4π²/(G·M) - constant
End of proof for circular orbit.