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Kepler's 3rd Law

  • Thread starter Reverie29
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  • #1
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Homework Statement


A satellite is in a circular orbit very close to the surface of a spherical planet. The period of the orbit is 2.49 hours.
What is density of the planet? Assume that the planet has a uniform density.


Homework Equations



T^2 = (4(pi)^2*r^3) / GM

The Attempt at a Solution


Okay, so I coverted the period into seconds and got 8964 seconds.
Then I rearranged the equation to get
M/r^3 = 4(pi)^2 / GT^2, assuming that M/r^3 would get me density.
So then according to that Density = 4(pi)^2 / (6.67e-11 N*m^2/kg^2)(8964 s)^2 which gives 7366 kg/m^3 which is not correct.

Or am I missing something about density? Density is mass divided by area. Should I be finding a radius to find the area and then find the mass somehow... I don't know. I'm confused on what to do.
 

Answers and Replies

  • #2
rock.freak667
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[tex]\rho = \frac{M}{V}[/tex]

Assuming the planet is a perfect sphere, [itex]V=\frac{4}{3} \pi r^3[/itex]

So

[tex] \rho = \frac{M}{\frac{4}{3} \pi r^3} = \frac{3M}{4\pi r^3}[/tex]
 
  • #3
Janus
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Your problem is in assuming that M/r^2 gives you density.

Density is mass divided by volume. So what is the formula for the volume of a sphere?
 
  • #4
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Okay.
The density of a sphere is = 3M / 4(pi)r^3. And I have already solved for M/r^3. I tried multiplying by 3 and dividing by 4pi, but still got an incorrect answer. I got 17,356 kg/m^3.

Should I be looking at another equation?
 
  • #5
Dick
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3/(4pi)<1. Why did your answer increase?
 
  • #6
rock.freak667
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[tex]T^2=\frac{4\pi r^3}{GM}[/tex]

[tex]\frac{1}{T^2}=\frac{GM}{4\pi r^3}[/tex]

[tex]\frac{1}{T^2}=\frac{G}{3} \frac{3M}{4\pi r^3}[/tex]

[tex]\frac{1}{T^2}=\frac{G}{3} \rho[/tex]

and then you got [itex]\rho [/itex] to be that value? If so and you calculated correctly...that should be the answer.
 
  • #7
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I have no idea why it increased! I guess I must be calculator retarded. I've got it now, thanks!!!
 

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