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