Originally posted by PiRsq
So does R3 have any relationship with the fact that the planets are orbiting in 3 dimension?
No. What we are talking about is a porportional realtionship between the Period of an orbit and its radius.
To show the math:
Force of gravity is
F_{g}= \frac {GMm}{R^2}
M is the mass of the sun in this case, and
m the mass of the planet. The
R² just means that the force falls off by the square of the distance. (double the distance and the force decreases to a quarter of its strength.)
The centripetal force needed to hold the planet in a circular path is
F_{c} = \frac{mv^2}{R}
For a circular orbit, these two forces are equal, so:
\frac {GMm}{R^2} = \frac{mv^2}{R}
m apears in both numerators, so they cancel out, and
R appears in both demoninators, canceling out one
R of
R² which gives you:
\frac {GMm}{R} = v^2
Taking the squareroot of each side gives you the orbital velocity of the Planet:
\sqrt{ \frac {GMm}{R}} = v
In one orbit the planet will travel the circumference of a circle with a radius of
R or:
C = 2\pi R
The time or period of this orbit is equal to distance/velocity or:
P = \frac{2\pi R}{\sqrt{ \frac {GMm}{R}}}
Rearranged:
P= 2\pi \sqrt{ \frac{R^3}{GM}}
Now let's say that you want to compare the periods of two different orbits at different values of
R, then:
\frac{P_{1}}{P_{2}}<br />
= \frac{ 2\pi \sqrt{ \frac{R_{1}^3}{GM}}}<br />
{ 2\pi \sqrt{ \frac{R_{2}^3}{GM}}}
2\pi cancels out:
\frac{P_{1}}{P_{2}}= \frac{ \sqrt{ \frac{R_{1}^3}{GM}}}{ \sqrt{ \frac{R_{2}^3}{GM}}}
The squareroots combine:
\frac{P_{1}}{P_{2}}=\sqrt{ \frac{ \frac{R_{1}^3}{GM}}{ \frac{R_{2}^3}{GM}}}
The GMs drop out:
\frac{P_{1}}{P_{2}}=\sqrt{ \frac{ R_{1}^3}{ R_{2}^3}}
Square both sides:
\left(\frac{P_{1}}{P_{2}}\right)^2={ \frac{ R_{1}^3}{ R_{2}^3}
or
\frac{P_{1}^2}{P_{2}^2}={ \frac{ R_{1}^3}{ R_{2}^3}
Kepler's Third law.
