Question

to prove t^2 directly proportional to r^3

Answer 1

Kepler's Third Law states that the square of the time period of orbit is directly proportional to the cuber of the semi-major axis of that respective orbit. (the semi-major axis for a circular orbit is of course the radius) Mathematically this can be represented as: T2 / r3 = k where k is a constant. The value k is related to physical constants such that k = 4pi2/GM where G is the gravitational constant and M the mass of the object at the centre of the orbit (NOT the object doing the orbiting!)

 

How did Kepler arrive at this result? Unfortunately, through experiment, which is not particularly convenient for us, but, thankfully we have knowledge Kepler had not! 

 

The result can be obtained surprisingly easily, assuming we have the necessary tools. 

We will need the following four equations:

Circular Motion: a = v2/r; v = wr = 2pi/T

Gravitational attraction: F = GMm/r2 

Newton's Second Law: F = ma

 

Substituting circular motion and gravitational attraction into the above formula yields:

mv2/r = Gmm/r2 

Cancelling the m's multiplying by r and by GM gives:

v2/GM = 1/r

This is very close to the result we want, one more substitution should give us the desired equation. Notice that v = wr = 2rpi/T from circular motion equations.

And so we have:

4pi2/GMT2 = 1/r3

Multiplying by T2:

T2/r3 = 4pi2/GM as required!