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Find the Period

  1. Sep 15, 2007 #1
    Of an asteroid whose Mass is 2.0*10^-2 times that of earth's and whose distance from the sun is twice the earth's distance from the sun. Find the period in years.

    I am supposed to use the concept of gravitational F=centripital force[tex]=m\frac{v^2}{r}[/tex] and the fact that [tex]v=\frac{2\pi r}{T}[/tex]

    so this is my attempt:

    [tex]F_g=\frac{GMm}{r^2}=m(\frac{v^2}{r})[/tex]

    implies [tex]\frac{GMm}{r^2}=\frac{m4\pi^2r}{T^2}[/tex]

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

    This M though is the mass of the sun correct?

    Casey
     
  2. jcsd
  3. Sep 15, 2007 #2
    Maybe just the acting of posting it helps, cause I got this one too. 2.8 years.

    Thanks,
    Casey
     
  4. Sep 15, 2007 #3

    G01

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    Yes, M is the mass of the sun and m is the asteroid in this case.
     
  5. Sep 15, 2007 #4

    G01

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    Glad to hear it!:smile:
     
  6. Sep 16, 2007 #5

    dynamicsolo

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    Properly speaking, you will probably learn in your course about Newton's generalization of Kepler's Third Law, in which case M is actually the total mass of the "system". But since the asteroid's mass is negligibly small compared to the Sun's, you get entirely satisfactory precision using the Sun's mass alone. (You'd see a slight discrepancy if we used Jupiter as the second mass; for binary stars, you *must* use the sum of the stars' masses.)

    You can avoid the annoyance of needing to know G and M (or even pi!) or even worrying about units by solving a comparison ratio. Both the Earth and this asteroid orbit the Sun, so we have

    T^2 for asteroid 4(pi^2)(r for asteroid ^3)/GM
    ______________ = _________________________

    T^2 for Earth 4(pi^2)(r for Earth ^3)/GM

    (this doesn't show as neatly as I'd like and I don't have a TeX manual)

    or just

    (T for ast. / T for Earth)^2 = (r for ast. / r for Earth)^3 .

    The asteroid orbits twice as far from the Sun as the Earth does, so

    (T for ast. / T for Earth)^2 = 2^3 = 8 ;

    thus the asteroid's orbital period is sqrt(8) times the Earth's (which is one year) or
    about 2.83 years.

    The comparison ratio approach is very tidy when the problem provides the right sorts of information. The only extent to which you even deal with units is in making sure that quantities to be compared in ratios are in the same units.
     
  7. Sep 16, 2007 #6

    Nice one dynamicsolo. I like how you saw a proportion in this; I have a hard time seeing thise right off the bat. When I drew a diagram of this I had a feeling I could do something to this affect, but I have never been good at proportions.

    I need to go ahead and find some more problem like this so I can learn how to spot ratios better.

    I always like it when somebody reads through a post; sees that the OP has been answered and has the solution; and still takes the time to show the OP another way it could have been solved. PF rocks!:smile:

    Thanks again,
    Casey
     
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