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Ellipse Word Problem 
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#1
Mar2306, 11:35 AM

P: 45

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? 


#2
Mar2306, 12:01 PM

P: 925

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



#3
Mar2306, 12:07 PM

P: 45

But there is 3 variables...right?



#4
Mar2306, 12:31 PM

P: 925

Ellipse Word Problem
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
Mar2306, 12:37 PM

P: 45

Yeah [tex]c^2 = a^2  b^2[/tex]



#6
Mar2406, 12:12 AM

P: 48

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


#7
Mar2406, 12:16 AM

P: 48

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 [tex].093=141.7/a[/tex] can you figure it out from here? 


#8
Mar2406, 07:51 PM

P: 45

Um, [tex]a^2 = b^2 + c^2[/tex] 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
Mar2506, 09:28 AM

P: 925

Just so you know, the eccentricity is defined as sqrt(1b^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
Mar2506, 11:12 AM

P: 45

So what you're saying is that:
[tex] .093 = \sqrt{1  \frac{b^2}{a^2}} [/tex] and [tex] .093^2 = 1  \frac{b^2}{a^2} [/tex] Right? 


#11
Mar2506, 01:31 PM

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#12
Mar2506, 01:59 PM

P: 45

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



#13
Mar2506, 04:33 PM

P: 925




#14
Mar2506, 05:55 PM

P: 45

OK, so
[tex] .093 = 1  \frac{b}{a} [/tex] and [tex] .093^2 = 1  \frac{b^2}{a^2} [/tex] But if I rewrite the first I get [tex] a = .093a+b [/tex] That can't be right because I still have 2 variables. 


#15
Mar2506, 08:49 PM

P: 925

No, 0.093^2 = 1b^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 1b/a.



#16
Mar2506, 09:11 PM

P: 45

Oh, I took the square root to get that.
Anyways, what am I suppsed to do then? 


#17
Mar2606, 09:42 PM

P: 45

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
Mar2606, 11:19 PM

P: 48

Oh, sorry about that I thought that read [tex]c^2=a^2+b^2[/tex] not [tex]c^2=a^2b^2[/tex] 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. 


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