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

AI Thread Summary
When applying Kepler's Second Law, the radius to use is the radius of the orbit, not the radius of the object itself. This is because Kepler's Laws pertain specifically to orbital mechanics. In some cases, both the radius of the orbit and the radius of the object may be referenced to determine the distance from the object's surface to the orbit. The distinction is crucial for accurately solving problems related to celestial orbits. Understanding this will clarify when to use each radius in calculations.
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Homework Statement


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)?

Homework Equations


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

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.
 
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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.
 
bacon said:
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.

Many thanks! :smile:
 
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