# Units for Kepler's 3rd Law

1. Oct 10, 2015

### Calpalned

1. The problem statement, all variables and given/known data
Kepler's 3rd law can be written as $p^2=a^3$
If p, the period, is given in years, what are the units for a?

2. Relevant equations
n/a

3. The attempt at a solution
The answer is AU. Is there a proof for this or is this merely a definition? Thank you.

2. Oct 10, 2015

### Daeho Ro

Let's consider the Earth.

If we choose the "year" for period, then the value of $a$ also have to be $1$ with some unit and the only thing is $1 \text{AU}.$ You may think that this is the definition for $\text{AU}.$

BTW, be careful about dimensions of each side. $(\text{Year})^2 \neq (\text{AU})^3,$ and so usually we write the law as $p^2 \propto a^3.$

3. Oct 10, 2015

### Krylov

Please pardon my ignorance, but I would feel that, strictly speaking, we cannot know the units of $a$ as long as we do not know the units of the proportionality constant $C$ in the relation $\frac{a^3}{p^2} = C$? And vice versa, of course.

4. Oct 10, 2015

### Daeho Ro

Otherwise, you may fix first the units of $a$ and $p$, then you can get the constant $C$ with some unit.