Solving Orbital Mechanics: 24 Hours to Seconds Conversion

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Homework Help Overview

The discussion revolves around a problem in orbital mechanics, specifically focusing on converting time from hours to seconds and applying gravitational formulas related to circular motion. Participants are exploring the relationships between angular velocity, gravitational force, and radius in the context of orbital dynamics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss converting 24 hours to seconds and consider using gravitational equations. There is an exploration of the relationship between angular velocity and gravitational force, with some questioning the correct application of formulas and the role of the Earth's radius in calculations.

Discussion Status

Some participants express confidence in their approaches, while others suggest reconsidering the radius used in calculations. There is acknowledgment of different interpretations of the problem, and guidance is offered regarding the application of gravitational formulas.

Contextual Notes

Participants mention the need to account for the Earth's radius in their calculations, indicating a potential constraint in the problem setup. There is also a reference to homework rules that may influence the discussion.

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I can't decide what to do with this question. I am studying for a final. I converted 24 hours to seconds and tried using the Fg = -gmm/r but I don't think that's going to work...

any leads? Thanks guys
 
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You're on the right track. Think about circular motion, too.
 
theFuture said:
You're on the right track. Think about circular motion, too.

ok I think I am getting it.

r*(omega)^2 = GM/r^2

then i can cancel the R's to get

Omega^2 = GM/r ?
 
triden said:
ok I think I am getting it.

r*(omega)^2 = GM/r^2

then i can cancel the R's to get

Omega^2 = GM/r ?

Yes,it's correct.To check your answer,though,u should be gettin round about 35000km.
 
you mean omega^2 = GM/r^3
 
That's one of my favorite problems.

I think the answer is about 24,000mi. Don't forget to subtract the radius of the earth. I always forget to do that!
 
i Think radius should be (R+r) and use F=GMm/(R+r)^2
 
saltrock said:
i Think radius should be (R+r) and use F=GMm/(R+r)^2

Why complicate?Use "r" as your length variable (the radius of the trajectory)and then,once u got the result,subtract the mean radius of the Earth which is round about 6371km.
 

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