# Kepler's 3rd law - which object in numerator?

1. Mar 6, 2013

### aphysicsmanduh

Hi folks! I have been searching and searching and cannot come up with a clear answer. When applying Kepler's 3rd law to 2 objects to solve for an unknown period or radius, which orbiting object (satellite) is in the numerator and which is in the denominator?

Here is the formula I am referring to:

( TA / TB )^2 = (RA / RB)^3

so if you're solving for orbiting period you would have TA = √(RA/RB)^3 * TB^2

I get different answers if I switch them so there must be some sort of "rule" that tells me which object is "A" and which is "B". From my searching, it looks like period and radius "A" are for the less massive object. I also wasn't sure if it had something to do with which object was closest to original object being orbited. An example is the moons of Jupiter. If I was trying to find the period of Ganymede given its radius and the info. for Io, which object would be the numerator and which would be the denominator in Kepler's 3rd law equation.

Any help would be great so I can sleep tonight! :)

2. Mar 6, 2013

### TurtleMeister

Is this what you're looking for?

$$T=2\pi\sqrt{\frac{a^3}{G(m_1+m_2)}}$$

"a" is the sum of the semi major axis of the two bodies.

3. Mar 6, 2013

### Bandersnatch

It doesn't matter which is which.

The erroneous results might be due to:

Perchance you've lost a square root there?

TA = √(RA/RB)^3 * √TB^2