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Kepler Orbits and ellipses

  • Thread starter Emspak
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Homework Statement



I am trying to see if I am on the right track with this.

The problem: A kepler orbit (an ellipse) in Cartesian coordinates is: $$(1−\epsilon^2)x^2 + 2\alpha \epsilon x + y^2 = \alpha^2$$.
The task is to show that the major and minor axes are: $$a = \frac{\alpha}{\sqrt{1-\epsilon^2}} = \frac{k}{2|E|} \text{ and } b = \frac{\alpha}{\sqrt{1-\epsilon^2}} = \frac{L}{\sqrt{2m|E|}}$$

Well and good, I noticed that the general equation for an ellipse has that middle term [itex]2\alpha \epsilon x [/itex] but I could get rid of it if I just assume the ellipse's center is at the origin. Then I can say the general form for the ellipse is [itex]\frac{x^2}{a}+\frac{y^2}{b}=1[/itex] and go from there. When I do that I can redue the original equation to: $$\frac{(1−\epsilon^2)x^2}{\alpha^2} + \frac{y^2}{\alpha^2} = 1$$

Plugging in fo a (that is, noting that under the x is the value for a) I see I can make the denominator under x equal to [itex]\frac{\alpha^2}{(1-\epsilon^2})[/itex] which would make [itex]a=\frac{\alpha}{\sqrt{1-\epsilon}}[/itex] and I can do the same thing for b, getting me [itex]b=\frac{\alpha}{\sqrt{1-\epsilon}}[/itex] as well.

It's the next step I am a bit shaky on. Assuming [itex]\alpha = \frac{L^2}{mk}[/itex] I am not entirely sure how to get the last step. I was thinking that to get total energy (E) I would just add the vectors of radial and tangental velocity, and plug that into [itex]KE= \frac{1}{2} mv^2[/itex]. But I am trying to determine if I am in the right ballpark. It occurred to me I have to account for potential energy as well, though.
 

Answers and Replies

  • #2
D H
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The task is to show that the major and minor axes are: $$a = \frac{\alpha}{\sqrt{1-\epsilon^2}} = \frac{k}{2|E|} \text{ and } b = \frac{\alpha}{\sqrt{1-\epsilon^2}} = \frac{L}{\sqrt{2m|E|}}$$
There's something wrong here. I suspect you are supposed to show that the *semi* major axis (not major axis) is ##a=\frac{\alpha}{1-\epsilon^2}## (note the lack of a square root) and that the *semi* minor axis is ##b=\frac{\alpha}{\sqrt{1-\epsilon^2}}##.

Well and good, I noticed that the general equation for an ellipse has that middle term [itex]2\alpha \epsilon x [/itex] but I could get rid of it if I just assume the ellipse's center is at the origin.
The ellipse's center is *not* at the origin. Set y to zero in the original equation. This is not of the form (x-c)^2 = 0.
 

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