# Period of a planet orbiting the Sun

## 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.

## 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

## Answers and Replies

tiny-tim
Science Advisor
Homework Helper
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? $\sum F=ma$

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

So would I go to
$ma=mrω^{2} - \frac{\mu r}{r^{3}}$ ?

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

Although I just realised that the question states $F= - \frac{\mu r}{r^{3}} per unit mass$ 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?

tiny-tim
Science Advisor
Homework Helper
Although I just realised that the question states $F= - \frac{\mu r}{r^{3}} per unit mass$ 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! Alright I think I'm getting somewhere now.

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

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

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

tiny-tim
Science Advisor
Homework Helper
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 cancelled r's wrong

(and itex doesn't like your bold tags )

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
Science Advisor
Homework Helper
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 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
Science Advisor
Homework Helper
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
Mentor
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