Orbiting Planet Homework: Deriving Formula for Moon's Orbit Period

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The discussion focuses on deriving a formula for the period of a moon's orbit around a planet, emphasizing that the period depends solely on the planet's mass and the radius of the moon's orbit. Participants clarify that gravitational force and centripetal force must be balanced, leading to the equation Fg = Gm1m2/r^2. The conversation progresses to equate gravitational acceleration with radial acceleration, resulting in v^2 = GM/r. Ultimately, they derive the relationship involving the circumference of the orbit, leading to the formula T^2 = (4π^2r^3)/(GM). The participants express gratitude for the collaborative problem-solving process.
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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?
 
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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
 
I don't get it still. How do I derive a formula out of that?
 
F=ma right? What else does F equal? What is the formula for Force of gravity?

Set them equal to each other and let's see what happens.
 
Fg= Gm1m2/r^2...
 
ahrog said:
Fg= Gm1m2/r^2...

Yep... and? Like I said set it equal to ma and let's see what happens.

m_{moon}a=\frac{GM_{planet}m_{moon}}{r^2}

now what?
 
So I cancel out the two moons and get a=GM/r^2.
 
Yes. And when a "particle" is experiencing radial acceleration, how else can a be written?
 
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
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
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
(\frac{2\pi r}{T})^2=\frac{2^2*\pi^2*r^2}{T^2}=\frac{GM}{r}

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