Orbiting of Planet

  • Thread starter ahrog
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  • #1
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Homework Statement


When a moon orbits a planet, it can be shown that the period of the moon's orbit depends only on the mass of the planet and the radius of the moon's orbit.
a) Draw and label a diagram to illustrate the variables.
b) Derive a formula for the period of a moon's orbit. The only variables in the final answer should be the mass of the planet and the radius of the orbit. All other values should be constants.


Homework Equations


Fg= Gm1m2/r^2
g=Gm1/r^2


The Attempt at a Solution


Okay, I don't really understand how to make formula's to get an answer like this. I figure that the variables in the formula would be the period of orbit, mass and radius. The constants would probably be pi and G.

Could someone give me a hint at how to do this?
 

Answers and Replies

  • #2
LowlyPion
Homework Helper
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If it's orbiting then you know that centrifugal is balanced by gravitational attraction.

You also know that v = ωr = 2πf*r = 2πr/T
 
  • #3
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I don't get it still. How do I derive a formula out of that?
 
  • #4
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F=ma right? What else does F equal? What is the formula for Force of gravity?

Set them equal to each other and lets see what happens.
 
  • #5
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Fg= Gm1m2/r^2...
 
  • #6
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Fg= Gm1m2/r^2...
Yep... and? Like I said set it equal to ma and let's see what happens.

[tex]m_{moon}a=\frac{GM_{planet}m_{moon}}{r^2}[/tex]

now what?
 
  • #7
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So I cancel out the two moons and get a=GM/r^2.
 
  • #8
3,003
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Yes. And when a "particle" is experiencing radial acceleration, how else can a be written?
 
  • #9
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v^2/r=GM/r^2 And then if I cancel out the 2 r's I would get v^2=Gm/r...

Would I use v=d/t here? If so, then for (d/t)^2=GM/r^2 what would cancel out the d?
 
  • #10
3,003
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You are very close, but not exactly. Instead of distance we should use something else.... what is the distance called when we are talking about a circle?
 
  • #11
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Circumference, which is 2pir?

Then (2pir/t)^2=GM/r^2
Which would end up as

2pir^3/Gm=T^2

Would that be right?
 
  • #12
3,003
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[tex](\frac{2\pi r}{T})^2=\frac{2^2*\pi^2*r^2}{T^2}=\frac{GM}{r}[/tex]

Don't forget to square everything in the parentheses. And then you've got it.
 
  • #13
46
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Thank you sooo much for going through this with me C: I really appreciate it
 
  • #14
3,003
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No probs :wink:
 

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