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Eccentricity in kepler problem

  • Thread starter gulsen
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  • #1
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For a planet moving in an elliptical orbit, fraction of maximum and minumum angular is given to be n
[tex]\frac{\dot{\theta}_{max}}{\dot{\theta}_{min}} = n[/tex]
Show that
[tex]\varepsilon = \frac{\sqrt{n}-1}{\sqrt{n}+1}[/tex].

I keep finding [tex]\varepsilon = -\frac{n^2-1}{n^2+1}[/tex] :yuck: Can someone show a path to correct solution?
 

Answers and Replies

  • #2
Physics Monkey
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Hi gulsen,

I just checked myself and the answer does seem to be correct. I used the conservation of angular momentum to relate the ratio of angular velocities to the minimum and maximum radii of the orbit. These radii can be expressed in terms of the eccentricity and some other length. It just fell right out after that. If you're still having trouble, perhaps you could post some of your intermediate steps and maybe I could say more.

Hope this helps.
 
  • #3
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I've tried this:

[tex]r^2 {\dot{\theta}} = h[/tex]
[tex]\dot{\theta} = h/{r^2}[/tex]
and
[tex]r = \frac{k}{1+\varepsilon \cos(\theta)}[/tex]

so
[tex]n = \frac{ \dot{\theta}_2 }{ \dot{\theta}_1 }= (\frac{r_1}{r_2})^2 = (\frac{1+\varepsilon}{1-\varepsilon})^2 [/tex]

Thanks!
 
Last edited:
  • #4
Physics Monkey
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Ah, there is your problem. The angular momentum is [tex] \ell = m r^2 \dot{\theta} [/tex], so you shouldn't be taking that square root in your first expression. Also, since it is [tex] r^2 [/tex] that appears, you ought to have [tex] (1+\epsilon)^2 [/tex], etc instead of just [tex] (1+\epsilon) [/tex]. The incorrect square root and the missing square are just what you need to get things back on track.
 
  • #5
Physics Monkey
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You're welcome.
 
  • #6
BobG
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gulsen said:
I've tried this:

[tex]r^2 {\dot{\theta}} = h[/tex]
[tex]\dot{\theta} = h/{r^2}[/tex]
and
[tex]r = \frac{k}{1+\varepsilon \cos(\theta)}[/tex]

so
[tex]n = \frac{ \dot{\theta}_2 }{ \dot{\theta}_1 }= (\frac{r_1}{r_2})^2 = (\frac{1+\varepsilon}{1-\varepsilon})^2 [/tex]

Thanks!
Maximum angular velocity occurs at perihelion. Your equation for angular momentum pretty much proves that part.

Your true anomaly at perihelion is 0 degrees, meaning your equation for the radius should simplify pretty easily. Your true anomaly at apohelion is 180 degrees, meaning the radius should be pretty simple.

You wind up with:

[tex]r_p = a(1 - e)[/tex]
[tex]r_a = a(1 + e)[/tex]

Your a (semi-major axis) will cancel out when you compare your max angular velocity (at perihelion) to the minimum angular velocity (at apohelion).

As to Physics Monkey's comment about the angular momentum, you normally use "specific angular momentum per unit of mass" when talking about the motion of planets or satellites. When looking at how the force of gravity will accelerate an object, the mass will cancel out. There's a habit (maybe bad) of referring to the specific angular momentum as the angular momentum and relying on the reader to understand which is meant from the context.
 
Last edited:
  • #7
Physics Monkey
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Thanks for the comments, BobG, I'm sure gulsen will find them helpful though I think the problem has been resolved. Note that my comment about angular momentum was referring to a previous mistaken relation [tex] h = r \dot{\theta}^2 [/tex] that gulsen was using (which he has since edited), and has nothing to do with whether one multiplies by the mass or not.
 
  • #8
BobG
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Physics Monkey said:
Thanks for the comments, BobG, I'm sure gulsen will find them helpful though I think the problem has been resolved. Note that my comment about angular momentum was referring to a previous mistaken relation [tex] h = r \dot{\theta}^2 [/tex] that gulsen was using (which he has since edited), and has nothing to do with whether one multiplies by the mass or not.
Ah, I see. I was wondering, since it seemed like half the posts must be missing (the edit time was a minute before the explanation?).
 
  • #9
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I first tried conservation of energy, this was my original mistake. Then I tried to work out Physics Monkey on the fly, and I've produced the same mistake! Well, I just confused [tex]h = r \dot{\theta}^2[/tex] and [tex]h = r^2 \dot{\theta}[/tex] :redface:

As for the edit time, AFAIK, it's when the editing started, not when it's saved.
Again, thanks a lot for suggestion!
 

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