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

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SUMMARY

When applying Kepler's Second Law, the radius used should be the radius of the orbit, not the radius of the object itself. For example, Earth's orbital radius is approximately 1.49e11 m, while its radius is 6.38e6 m. The distinction is crucial as Kepler's Laws specifically pertain to orbital mechanics, and the radius of the orbit determines the dynamics of the celestial body in motion. Understanding this concept is essential for correctly solving problems related to orbital dynamics.

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  • Understanding of Kepler's Laws of planetary motion
  • Familiarity with orbital mechanics
  • Basic knowledge of gravitational forces and equations
  • Ability to interpret mathematical equations related to orbits
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  • Study the derivation and implications of Kepler's Laws
  • Learn about the gravitational constant (G) and its applications in celestial mechanics
  • Explore the concept of orbital radius and its effect on orbital period
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Astronomy students, physics enthusiasts, and anyone studying celestial mechanics will benefit from this discussion, particularly those looking to deepen their understanding of orbital dynamics and Kepler's Laws.

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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.
 
Last edited:
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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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