Period of a planet orbiting the Sun

  • Thread starter Jon.G
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  • #1
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Homework Statement


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.

Homework Equations





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
 

Answers and Replies

  • #2
tiny-tim
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Hi Jon.G! :wink:
I thought It would have something to do with centripetal force, and using [itex] ω=\frac{2\pi}{t} [/itex] …

Yes.

Write out F = ma … what do you get? :smile:
 
  • #3
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[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?
 
  • #4
tiny-tim
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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?

yup! :biggrin:
 
  • #5
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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
 
  • #6
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Also, any hints as to what I'm doing wrong with itex? Then I can actually make this posts legible :S
 
  • #7
tiny-tim
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Alright I think I'm getting somewhere now.

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

no, you've cancelled r's wrong

(and itex doesn't like your bold tags :wink:)
 
  • #8
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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.
 
  • #9
tiny-tim
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Still not sure how to work the vector though.

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

it doesn't matter, just use the magnitudes of the vectors :smile:
 
  • #10
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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: )
 
  • #11
tiny-tim
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For the second part could I not just substitute in the value to find [itex]\mu[/itex] and then find Mercury's time period?

use dimensions (ie what is T proportional to?) :smile:
 
  • #12
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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.

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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