Kepler's Law

By Anonymous on Thursday, April 19, 2001 - 11:59 am :

Has anyone got any idea how Kepler's third Law can be proved?

It's the one that says:
The square of the time a planet takes to complete an orbit is directly proportional to the cube of the planet's average distance from the sun.

Thanks!!!

By Sean Hartnoll (Sah40) on Friday, April 20, 2001 - 11:36 am :

It is easiest to prove for a circular orbit. You use Newton's Law for the gravitational force

F = G M m / r^{2}

and the centrifugal force

F = m v^{2} / r

r

is the distance,

v

is the velocity,

m

is the mass of the planet,

M

is the mass of the sun and

G

is the gravitation constant.

Okay. Now the velocity is related to the PERIOD (time taken to go round) by $v = 2 π r / T$ , because the total distance of the orbit is $2 π r$ (length of a circle radius $r$ ). So the centrifugal bit becomes $F = 4 m π^{2} r / T^{2}$ . Set the two forces to be equal because the planet is in equilibrium:

$4 m π^{2} r / T^{2} = G M m / r^{2}$

Which when you rearrange gives

$T^{2}$ proportional to $r^{3}$ , which I think is Kepler's third law.

This result was historically very important, as it meant that Newton's new law of gravitation was able to reproduce the experimental result of Kepler's law.

The derivation for general elliptic orbits is a bit more complicated.

Sean