# Using Kepler's Second Law - which radius to use?

1. Jul 20, 2011

### 5.98e24

1. The problem statement, all variables and given/known data
This is just a general question.

When using Kepler's second law, which radius am I supposed to use to sub into r? Is it the radius of the object (ex. Earth's is 6.38e6 m) or the radius of orbit (ex. Earth's is 1.49e11 m)?

2. Relevant equations
C = (GM)/(4pi^2) = (r^3)/(T^2)

3. The attempt at a solution
I don't know the answer. I've seen solutions that involved using the radius of the object AND the radius of the orbit but I don't know when to use which.

I understand that the answer to this may be painstakingly obvious... but I really would appreciate any responses.

Last edited: Jul 20, 2011
2. Jul 20, 2011

### bacon

Keplers Laws deal with orbits, therefore you would use the radius of the orbit. The reason you have seen the radius of orbit and the radius of the object is most likely because the problem is asking how far from the surface of the object the orbit is.

3. Jul 20, 2011

Many thanks!