# Period of a planet orbiting the Sun

• Jon.G
In summary: The force of gravity on the planet is F= -(mu*r)/r3. 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.
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.

## 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

Hi Jon.G!
Jon.G said:
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 realized 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 canceled out?

Jon.G said:
Although I just realized 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 canceled 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

Jon.G said:
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 )

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.

Jon.G said:
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 )

Jon.G said:
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?)

Jon.G said:
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.

$$\vec{F}=\frac{-m\mu \vec{r}}{r^{3}}$$ 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

## 1. What is the period of a planet orbiting the Sun?

The period of a planet orbiting the Sun is the time it takes for the planet to complete one full orbit around the Sun. This is also known as the planet's orbital period or year.

## 2. How is the period of a planet's orbit calculated?

The period of a planet's orbit is calculated using Kepler's third law, which states that the square of the orbital period is proportional to the cube of the semi-major axis of the orbit. This means that the period is directly related to the size of the planet's orbit.

## 3. What factors can affect the period of a planet's orbit?

The period of a planet's orbit can be affected by its distance from the Sun, as well as the mass of the planet and the mass of the Sun. Other factors such as gravitational forces from other planets and the shape of the planet's orbit can also have an impact.

## 4. What is the difference between a planet's sidereal period and its synodic period?

A planet's sidereal period is the time it takes for the planet to complete one full orbit relative to the stars. On the other hand, a planet's synodic period is the time it takes for the planet to return to the same position relative to Earth and the Sun. This is the period used for determining the planet's apparent motion in the sky.

## 5. Can the period of a planet's orbit change over time?

Yes, the period of a planet's orbit can change over time due to various factors such as gravitational interactions with other objects, or changes in the planet's orbit caused by external forces. However, these changes are typically very small and may take millions of years to have a noticeable impact on the planet's orbital period.

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