How do I Calculate GMm with Given a and T for Gravitation Problem?

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Homework Help Overview

The discussion revolves around a gravitation problem involving the calculation of GMm using given values for the semi-major axis (a) and the orbital period (T). The context includes references to Kepler's laws and gravitational equations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of Kepler's third law, particularly the relationship a^3/T^2. There are attempts to relate the semi-major axis to the radius and questions about how to derive GM from the provided values. Some participants consider the implications of ignoring eccentricity in their calculations.

Discussion Status

The discussion is active, with participants exploring different interpretations of Kepler's third law and its application to the problem. Some guidance has been offered regarding assumptions about eccentricity, and there are indications of progress in understanding parts of the problem, though no consensus has been reached on the overall approach.

Contextual Notes

Participants note that the problem specifically asks for the mass of Jupiter, which is significantly larger than the mass of its moons, suggesting a focus on simplifying assumptions in the calculations.

Clara Chung
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Homework Statement


The question is attached

Homework Equations


Kepler's laws and gravitational equations

The Attempt at a Solution


the question gave me a and T, I related it with Kepler's third law a^3/T^2, then I don't know what to do next. The answer of part a is 1.27 x10^17 /r^2 , part b = 1.9 x 10^27, How can I find GMm as the question only provided a and T?
 

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What did you get out of Keplers third law ?
 
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BvU said:
What did you get out of Keplers third law ?
If I assume a=r, I can get GM?
 
I think you can ignore the excentricity, yes.
 
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BvU said:
I think you can ignore the excentricity, yes.
How can I get m after that?
 
Kepler's third law says a^3/T^2 is 'constant', namely ...
(They only ask for M of Jupiter and that's >> m of moons)
 
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BvU said:
Kepler's third law says a^3/T^2 is 'constant', namely ...
(They only ask for M of Jupiter and that's >> m of moons)
Thanks I got part A and B
 
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