# Thread: Help with elliptical path of Jupiter around the sun.

1. ## Help with elliptical path of Jupiter around the sun.

Hi.

I'm trying to figure out the path of Jupiter around the sun.

It is an ellipse, so I'm trying to get it into the x^2/a^2+y^2/b^2=1 form.

I have the minimum (460,000,000 miles) and maximum(508,000,000 miles) distances.

My teacher has given me three equations to work with, to help me figure the problem out:

Max distance+Min distance=length of Major Axis

A=Length of major axis/2

Major axis-Minor axis/2=C

With these three equations, I have the following information:

Major Axis=968,000,000 miles
A=484,000,000 miles
B=?
C=?

I was wondering if someone could explain how to figure out the minor axis?

Also, the sun is at a point of (C,0) ...so does that mean that my C value would be the C in that ordered pair?

2. Originally Posted by lpaige2004
Hi.

I'm trying to figure out the path of Jupiter around the sun.

It is an ellipse, so I'm trying to get it into the x^2/a^2+y^2/b^2=1 form.

I have the minimum (460,000,000 miles) and maximum(508,000,000 miles) distances.

My teacher has given me three equations to work with, to help me figure the problem out:

Max distance+Min distance=length of Major Axis

A=Length of major axis/2

Major axis-Minor axis/2=C

With these three equations, I have the following information:

Major Axis=968,000,000 miles
A=484,000,000 miles
B=?
C=?

I was wondering if someone could explain how to figure out the minor axis?

Also, the sun is at a point of (C,0) ...so does that mean that my C value would be the C in that ordered pair?
The sun is at a focus of the ellipse. The distance from the centre of the ellipse (the midpoint of the major axis) to the focus is ae, where e is the eccentricity. So $e = \frac{484,000,000 - 460,000,000}{484,000,000}$. Once you know e, you can find the minor axis b from the formula $b^2 = a^2(1-e^2)$.

3. Originally Posted by Opalg
The sun is at a focus of the ellipse. The distance from the centre of the ellipse (the midpoint of the major axis) to the focus is ae, where e is the eccentricity. So $e = \frac{484,000,000 - 460,000,000}{484,000,000}$. Once you know e, you can find the minor axis b from the formula $b^2 = a^2(1-e^2)$.

Hi. Thank you for the reply.

I have a few questions though, if you don't mind. Eccentricity is measured from c/a, correct? How did you get: $e = \frac{484,000,000 - 460,000,000}{484,000,000}$?

Also, am I supposed to take the square root of everything, once I plug in the numbers, or am I supposed to square the numbers I plugged in?

Thank you so much for your help. It's really appreciated.

4. Hello lpaige2004

In your original post, I'm not sure what you mean by this:
Originally Posted by lpaige2004
...Major axis-Minor axis/2=C ...
so let me see if I can clarify things.

The equation of the ellipse, as you have said, is:
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
and you have correctly calculated $a$ to be $4.84 \times 10^8$.

These are the other things you need to know:

• The centre of the ellipse is at $(0,0)$.

• The foci of the ellipse are at $(\pm ae,0)$.

• The relation between $a, b$ and $e$ is:

$a^2e^2 = a^2-b^2$ (1)
• In the case of the elliptical orbit of a planet around the sun, the sun is at one of the foci of the ellipse.

• The greatest distance from a focus to a point on the ellipse is $a(1+e)$; the shortest distance is $a(1 - e)$.

Opalg has used this last fact, together will the values $a = 4.84 \times 10^8$ and $a(1 - e)=4.6\times10^8$, to write down an expression for $e$.

What you now need to do:

• Evaluate $e$.

• Use equation (1) above to find the value of $b$.

• Use your values of $a$ and $b$ to write down the equation of the ellipse.

• If you also need to know the distance of the sun from the centre of the orbit - the distance I think you have called $c$ - then it's $ae$ (as stated above).