Period of a planet orbiting the Sun

[Jon.G]

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 \mu is a positive constant and r is the vector \stackrel{\rightarrow}{SP}
If the orbit of the planet is a circle of radius a, show that the period of the planet is \frac{2\pi a^{3/2}}{\mu ^{1/2}}


The Earth and Mercury are orbiting the Sun.
The Earth is at a mean distance of approximately 1.5 * 10^{8} km from the Sun and Mercury at approximately 5.8 * 10^{7} 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 ω=\frac{2\pi}{t} but I can't see how to use that to get the answer :S

Thanks

 

[tiny-tim]

Hi Jon.G!

I thought It would have something to do with centripetal force, and using ω=\frac{2\pi}{t} …

Yes.

Write out F = ma … what do you get? ​

 

[Jon.G]

\sum F=ma

Forces acting on P are
F=\frac{-\mu <b>r</b>}{r^{3}} and
F=mrω^{2} right?

So would I go to
ma=mrω^{2} - \frac{\mu <b>r</b>}{r^{3}} ?

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

Although I just realized that the question states F= - \frac{\mu <b>r</b>}{r^{3}} <b>per unit mass</b> 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 canceled out?

 

[tiny-tim]

Although I just realized that the question states F= - \frac{\mu <b>r</b>}{r^{3}} <b>per unit mass</b> 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 canceled out?

yup!

 

[Jon.G]

Alright I think I'm getting somewhere now.

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



What I'm struggling to get my head around now is the r
I just don't know how to work with it

 

[Jon.G]

Also, any hints as to what I'm doing wrong with itex? Then I can actually make this posts legible :S

 

[tiny-tim]

Alright I think I'm getting somewhere now.

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

no, you've canceled r's wrong

(and itex doesn't like your bold tags )

 

[Jon.G]

Of course argh :S

ω^{2} = \frac{\mu R}{r^{4}}

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.

 

[tiny-tim]

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

 

[Jon.G]

Well that makes things easier :D
Thanks

For the second part could I not just substitute in the value to find \mu 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 )

 

[tiny-tim]

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

use dimensions (ie what is T proportional to?)

 

[Chestermiller]

Of course argh :S

ω^{2} = \frac{\mu R}{r^{4}}

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:

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

Chet