Find Density Given Period of Orbit

  • #1

Homework Statement


A satellite is in a circular orbit very close to the surface of a spherical planet. The period of the orbit is 2.53 hours. What is the density of the planet? Assume that the planet has a uniform density.

Homework Equations


T^2=(4pi^2r^3)/GM
V=4/3piR^3
Density= Mass/ Volume


The Attempt at a Solution


So far, I have calculated the amount of seconds in the orbit, but that is where my understanding of this comes to an abrupt halt.
How could I get the radius without the speed of the satellite? Is it not true that the satellite could orbit ANY planet in 2.53 hours if it had enough speed?
Some equations call for the gravity of planet, which is not given.
I guess you could say that I need a little "Barney-style" walkthrough on this one.
 

Answers and Replies

  • #2
Charles Link
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Try writing ## M=density \cdot V ## and put it into your T^2 equation. Also ## R ## and ## r ## are the same in this problem. Do you see how the ## R's ## will cancel? You can look up the value for ## G ##. (It should be in your textbook). When ## G ## is M.K.S., the density will be in kilograms/meter^3. (assuming T is in seconds).
 
  • #3
Keith_McClary
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Eliminate R from the equations:
##T^2=(4\pi^2R^3)/GM##
##V=\frac{4}{3}\pi R^3##
 
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  • #4
Eliminate R from the equations:
##T^2=(4\pi^2R^3)/GM##
##V=\frac{4}{3}\pi R^3##
Ok so,
I've got:
9108^2=G((4pi^2)/(D*(4/3)pi))
 
  • #5
Charles Link
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Ok so,
I've got:
9108^2=G((4pi^2)/(D*(4/3)pi))
G=6.67 E-11 in M.K.S. units if I remember it correctly. You can solve for D, but your ## G ## belongs in the denominator.
 
  • #6
G=6.67 E-11 in M.K.S. units if I remember it correctly. You can solve for D.
Where the heck did you find that?
 
  • #7
G=6.67 E-11 in M.K.S. units if I remember it correctly. You can solve for D, but your ## G ## belongs in the denominator.
Gettin there:
9108^2=(4pi^2)/(6.67e11(D)(4/3)pi)
 
  • #8
Pardon my slow advance on the problem, this is my first time taking physics, and i've been out of school for 8 years.
I'm not too savvy on how to isolate the D out of the denominator.
 
  • #9
Charles Link
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Pardon my slow advance on the problem, this is my first time taking physics, and i've been out of school for 8 years.
I'm not too savvy on how to isolate the D out of the denominator.
You multiply both sides of the equation by ## D ##. This will put the ## D ## on the left side and the ## D's ## on the right side (upstairs and downstairs) will cancel. Meanwhile your ## \pi^2/\pi=\pi ## and ## \pi=3.14159 ##. Looks like your algebra may be a little rusty, but this is a good calculation to get some practice with it...Notice also how the 4's cancel on the right side (4/4=1), and the "3" (in the 4/3) will wind up in the numerator(upstairs) on the right side. (The ## G ## that you asked about is the universal gravitational constant and it is ## G=6.67 E-11 ##).
 
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  • #10
I got 1.70333034e-19
Doesn't sound right.
 
  • #11
jbriggs444
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Ok so,
I've got:
9108^2=G((4pi^2)/(D*(4/3)pi))
Keep everything symbolic until you have a formula for the quantity you are after. Measured numeric values in a formula make the algebra more difficult.
 
  • #12
Charles Link
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I got 1.70333034e-19
Doesn't sound right.
The G is a 6.67 E-11 in the denominator. (I think you might have incorrectly used 6.67 E+11. G=6.67 E-11). It will get multiplied by the 9108^2 and the result is approximately 5.2 E-3 in the denominator. There is a ## 3 \pi ## in the numerator so the result will be ((3)(3.14))/(5.2 E-3). Next step is to convert this to a single number. Please try to verify these estimates. The result is in kilograms per cubic meter. (To compare to the density of water, you need to convert to grams per cubic centimeter. Water has density 1 gram per cubic centimeter. That should tell you if your answer makes sense.)
 
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  • #13
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Where the heck did you find that?
It is a fundamental constant, you have to either know it or look it up. Your textbook, Wikipedia, WolframAlpha, Google, ... all have the value.
Keep everything symbolic until you have a formula for the quantity you are after. Measured numeric values in a formula make the algebra more difficult.
This really helps. Also, keep the units once you plug in values, that gives an additional cross-check that everything is right.
 

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