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Period of a planet orbiting the Sun

  1. Mar 4, 2014 #1
    1. The problem statement, all variables and given/known data
    A planet P orbiting the Sun S is acted upon by a force F= -(mu*r)/r3 (itex won't work here for some reason :S ) per unit mass where [itex]\mu[/itex] is a positive constant and r is the vector [itex]\stackrel{\rightarrow}{SP}[/itex]
    If the orbit of the planet is a circle of radius a, show that the period of the planet is [itex] \frac{2\pi a^{3/2}}{\mu ^{1/2}} [/itex]


    The Earth and Mercury are orbiting the Sun.
    The Earth is at a mean distance of approximately [itex] 1.5 * 10^{8} [/itex] km from the Sun and Mercury at approximately [itex] 5.8 * 10^{7} [/itex] km.
    Given that the Earth's period is about 365.25 days, find the period of Mercury.
    2. Relevant equations



    3. The attempt at a solution
    Erm.... I really am not sure what to do.
    I thought It would have something to do with centripetal force, and using [itex] ω=\frac{2\pi}{t} [/itex] but I can't see how to use that to get the answer :S

    Thanks
     
  2. jcsd
  3. Mar 4, 2014 #2

    tiny-tim

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    Hi Jon.G! :wink:
    Yes.

    Write out F = ma … what do you get? :smile:
     
  4. Mar 4, 2014 #3
    [itex]\sum F=ma[/itex]

    Forces acting on P are
    [itex] F=\frac{-\mu r}{r^{3}}[/itex] and
    [itex] F=mrω^{2} [/itex] right?

    So would I go to
    [itex] ma=mrω^{2} - \frac{\mu r}{r^{3}} [/itex] ?

    I can't really see where to go the m confuses me a bit.

    Although I just realised that the question states [itex] F= - \frac{\mu r}{r^{3}} per unit mass [/itex] so would I be correct in thinking that I need to multiply this term by the mass (to get the force is N and not N kg-1?

    And then the mass could be cancelled out?
     
  5. Mar 4, 2014 #4

    tiny-tim

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    yup! :biggrin:
     
  6. Mar 4, 2014 #5
    Alright I think I'm getting somewhere now.

    If I have
    [itex] mrω^{2} = \frac{- \mu rm}{r^{3}} [/itex]
    then the m cancels, and an r
    [itex] ω^{2} = \frac{- \mu r}{r^{2}} [/itex]
    so....
    [itex] T = \frac{2\pi r}{\mu^{1/2}r^{1/2}} [/itex]



    What I'm struggling to get my head around now is the r
    I just don't know how to work with it
     
  7. Mar 4, 2014 #6
    Also, any hints as to what I'm doing wrong with itex? Then I can actually make this posts legible :S
     
  8. Mar 4, 2014 #7

    tiny-tim

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    no, you've cancelled r's wrong

    (and itex doesn't like your bold tags :wink:)
     
  9. Mar 4, 2014 #8
    Of course argh :S

    [itex] ω^{2} = \frac{\mu R}{r^{4}} [/itex]

    Still not sure how to work the vector though.
    If the orbit is a circle then wouldn't R be equal to the radius?
    But I can't just treat it as a scalar.
     
  10. Mar 4, 2014 #9

    tiny-tim

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    oh i see … you're worried about r being a vector

    it doesn't matter, just use the magnitudes of the vectors :smile:
     
  11. Mar 4, 2014 #10
    Well that makes things easier :D
    Thanks

    For the second part could I not just substitute in the value to find [itex]\mu[/itex] and then find Mercury's time period?

    (Even if this is the way, thanks Ronaldo95163. Always good to know of other ways to work through problems :smile: )
     
  12. Mar 4, 2014 #11

    tiny-tim

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    use dimensions (ie what is T proportional to?) :smile:
     
  13. Mar 4, 2014 #12
    In vector form, your two equations should read:

    [tex] \vec{F}=\frac{-m\mu \vec{r}}{r^{3}}[/tex] and
    [tex] \vec{F}=m\vec{a}=-m\vec{r}ω^{2} [/tex]
    So:[tex] \frac{-m\mu \vec{r}}{r^{3}}=-m\vec{r}ω^{2}[/tex] and
    [tex] \frac{\mu }{r^{3}}=ω^{2}[/tex]

    Chet
     
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