# Ellipse Word Problem

1. Mar 23, 2006

### sportsguy3675

Having trouble with this problem.

"The mean distance from the sun to Mars is 141.7 million miles. If the eccentricity of the orbit of Mars is .093, determine the maximum distance that Mars orbits from the sun."

So basically what it is asking for is half the length of the major axis right?

And from what I could figure (a+b)/2 = 141.7 million. And that .093 = c/a. But how would I be able to solve for the a value?

Last edited: Mar 23, 2006
2. Mar 23, 2006

### daveb

The relationship between the two axes, and the relationship of the mean of the axes, gives you 2 equations in 2 unknowns.

3. Mar 23, 2006

### sportsguy3675

But there is 3 variables...right?

4. Mar 23, 2006

### daveb

No, there is a relationship between eccentricity and the two axes. Or to put it another way, do you know the relationship between a, b, and c (which give the relationship between e, a and b)

5. Mar 23, 2006

### sportsguy3675

Yeah $$c^2 = a^2 - b^2$$

6. Mar 24, 2006

### konartist

Wrong, $$a^2 = b^2 + c^2$$ is for an ellipse and $$c^2=a^2 + b^2$$ is for a hyperbola.

But please label any variable you know in terms of a, b, or c and I'll be able to help you.

7. Mar 24, 2006

### konartist

Ok, I think I understand this problem now. C tells you the distance from the center to a fixed point. Since the sun is the center of the universe and Mars is always going to be 141.7 miles from the sun we can call that C because that is the focus.

Now, to find a we know that e=c/a
now $$.093=141.7/a$$
can you figure it out from here?

8. Mar 24, 2006

### sportsguy3675

Um, $$a^2 = b^2 + c^2$$ is the same thing as the formula I posted...

But, either way, yout idea is not right. 141.7 million is the mean distance not the distance to any point on the orbit. I think the sun is supposed to be at the focus.

9. Mar 25, 2006

### daveb

Just so you know, the eccentricity is defined as sqrt(1-b^2/a^2). So by squaring this you have 2 equations in 2 unknowns: 1 of them in b^2 and a^2 and the other in b and a. Find an expresion for a from the 2nd, substitute it into the first, and solve the quadratic. This gives your value for the semimajor axis, b, which is the value you're looking for. No need to use c.

10. Mar 25, 2006

### sportsguy3675

So what you're saying is that:

$$.093 = \sqrt{1 - \frac{b^2}{a^2}}$$

and

$$.093^2 = 1 - \frac{b^2}{a^2}$$

Right?

Last edited: Mar 25, 2006
11. Mar 25, 2006

### HallsofIvy

Staff Emeritus
You were responding to the statement that c2= a2- b2 which is exactly the same as a2= b2+c2!

12. Mar 25, 2006

### sportsguy3675

Yeah, I still don't know why that person said that >_>

13. Mar 25, 2006

### daveb

14. Mar 25, 2006

### sportsguy3675

OK, so
$$.093 = 1 - \frac{b}{a}$$
and
$$.093^2 = 1 - \frac{b^2}{a^2}$$

But if I rewrite the first I get $$a = .093a+b$$ That can't be right because I still have 2 variables.

15. Mar 25, 2006

### daveb

No, 0.093^2 = 1-b^2/a^2. The other equation is the mean of a and b. I don't know where your first equation came from because .093 is not equal to 1-b/a.

16. Mar 25, 2006

### sportsguy3675

Oh, I took the square root to get that.

Anyways, what am I suppsed to do then?

17. Mar 26, 2006

### sportsguy3675

Oh, I figured it out. Turns out that 141.7 = a . So then using the eccentricity I solved for c and then for b using the relationship between the 3. Then the max is a + c and the min is b - c. :)

18. Mar 26, 2006

### konartist

Oh, sorry about that I thought that read $$c^2=a^2+b^2$$ not $$c^2=a^2-b^2$$

Hmmm, I see what you're saying about it being the distance to any point on the orbit and you're correct the sun is the focus.