# Intersecting Orbits

1. Apr 8, 2010

### atarr3

1. The problem statement, all variables and given/known data
Two masses, m and 2m, orbit around their CM. If the orbits are circular, they don't intersect. But if they are very elliptical, they do. What is the smallest value of eccentricity for which they intersect?

2. Relevant equations

$$c=\sqrt{a^{2}+b^{2}}=\frac{k\epsilon}{1-\epsilon^{2}}$$ where a and b are the lengths of the semi-major and semi-minor axis, respectively. And $$k=\frac{L^{2}}{m\alpha}$$

3. The attempt at a solution
I'm having trouble visualizing this problem with elliptical orbits. When they're circular it's like they are concentric circles. Mass 2m has an orbital radius of 1/3 distance of separation and mass m has an orbital radius of 2/3 distance of separation. I have equations that relate eccentricity to the lengths of the semi-major and semi-minor axis, but I'm confused as to how I can actually find those values. I'm also confused as to when the orbits intersect. Does this occur when the semi-major/semi-minor axis of one orbit equals that of the other? I'm just looking for a little guidance in this problem. The only thing I know is that the CM is a focus for both elliptical orbits. Thanks guys!

2. Apr 8, 2010

### nickjer

To picture it better, check out:

http://en.wikipedia.org/wiki/Gravitational_two-body_problem

The first image shows 2 bodies orbiting in an elliptical orbit. The 2nd image only proves what you were thinking about when you talked about concentric circles.

I am not quite sure the equations you put up. But using the equations on the wikipedia page you can solve your problem. To understand when the orbits intersect, think of the concentric circles and picture them squeezing to form slight ellipses with the CM now a focal point and not the center. Soon the smaller internal ellipse formed by the heavier mass will touch the edge of the larger ellipse from the smaller mass. If things keep getting more elliptical then it will appear similar (but not exactly) like the first image on that page.

3. Apr 8, 2010

### nickjer

Here is a better image of what is happening. The vector $$\vec{r}$$ is just the distance between both masses.

[PLAIN]http://img534.imageshack.us/img534/5048/ellipse.png [Broken]

If you drew out only $$\vec{r}$$ it would appear as an ellipse with eccentricity $$\epsilon$$ which you need to find that causes those 2 ellipses you see in the figure above (which share the same focal point) to intersect. The formula for $$r(\theta)$$ can be seen on the wikipedia page I gave in the previous post.

EDIT: I don't pride myself on my artistic skills :(

Last edited by a moderator: May 4, 2017
4. Apr 8, 2010

### atarr3

I'm still a little confused by what you mean by "If you draw it out using only r". Other than that, I think I understand what's going on here. Thanks again for your help!

5. Apr 8, 2010

### atarr3

I can assume that the 2m mass always orbits 1/3 the distance of separation away from the CM and the m always orbits 2/3 the distance of separation away from the CM, right?

6. Apr 9, 2010

### D H

Staff Emeritus
No.

The answer is "yes" if by that you mean that given a separation r(t) between the two objects, the distances between the CM and the 2m and 1m objects are respectively r(t)/3 and 2r(t)/3.

I get the feeling you are thinking this question is asking for the bodies to collide. They won't. At some point in time, body1 will be exactly where body2 was some time before that.

Look at nickjer's diagram. It is pretty good. In particular, it shows where this intersection occurs.

7. Apr 9, 2010

### nickjer

Yes, just as D H said if you r(t) is the separation distance. And this is the key to solving the problem. Now pay close attention to the extrema's, when the separation distance is a maxima and minima in that figure. They will be related somehow when the orbits intersect in the way I drew them.

8. Apr 9, 2010

### atarr3

Well at the extremas one of the masses will be intersecting the other orbit. At a minimum mass m will intersect and at a maximum mass 2m will intersect. I can also see the the maximum radius of orbit for mass 2m equals the minimum radius of orbit for mass m, and this occurs at the intersection as well.

Last edited: Apr 9, 2010
9. Apr 9, 2010

### atarr3

The only thing I'm confused about in the equation for r(theta) is the angular momentum term. I don't know how to relate the angular momentum for mass 2m to mass m, but I'm guessing it has to do with the intersection point. I'm not sure if I can make this assumption, but at the intersection is the angular momentum for mass 2m twice that of mass m? Looking at the equation for angular momentum, $$L=r\timesp=mvr$$ I can see that at the intersection $$r$$ is equal for both cases, the smaller mass, $$m$$ is 1/2 the larger mass. But that leaves $$v$$, which I don't know how to relate them to one another. Unless I can somehow find it from taking a derivative of r(theta). I feel like I'm overcomplicating this for myself.

10. Apr 9, 2010

### atarr3

I can't get it to show the cross product for angular momentum, but it's supposed to say L = r X p

11. Apr 9, 2010

### atarr3

Ok I think I just got the answer. I'll post it once I'm done with my classes today just make sure. Thanks for all of your help guys!

12. Apr 9, 2010

### atarr3

Ok so given $$\frac{1}{r(\theta)}=\frac{\mu}{h^{2}}\left(1+\epsilon cos(\theta)\right)$$, this is the equation for the finding the distance of separation as a function of theta. Going back to what I said about $$r_{2max}=r_{1max}$$ I can find a way to relate these to one another. Also looking at the distance of separation, these radii also equal $$\frac{1}{3}$$ the maximum distance of separation and $$\frac{2}{3}$$ the minimum distance of separation. The maximum distance of separation is given by $$r = \frac{h^{2}}{\mu(1-\epsilon)}$$ The minimum distance of separation is given by $$r =\frac{h^{2}}{\mu(1+\epsilon)}$$. Substituting these into $$r_{2max}=r_{1min}$$ and canceling out common terms I get $$\frac{1}{1-\epsilon}=\frac{2}{1+\epsilon}$$, which yields $$2-2\epsilon=1+\epsilon$$, so $$3\epsilon=1$$ and therefore $$\epsilon=\frac{1}{3}$$

Last edited: Apr 9, 2010
13. Apr 9, 2010

### D H

Staff Emeritus
Or you could have just used

\aligned r_{\text{min}} &= a(1-e) \\ r_{\text{max}} &= a(1+e) \endaligned

for a given orbit. From this,

$$e = \frac{r_{\text{max}}-r_{\text{min}}}{r_{\text{max}}+r_{\text{min}}}$$

For all of the orbits of the two objects about their center of mass, the apoapsis of the smaller object will be twice the apoapsis of the larger object. The orbits will just intersect when the periapsis of the smaller object is equal to the apoapsis of the larger object. Thus, for the smaller object, $r_{\text{max}} = 2r_{\text{min}}$ and thus

$$e = \frac{2r_{\text{min}}-r_{\text{min}}}{2r_{\text{min}}+r_{\text{min}}} = \frac 1 3$$

14. Apr 9, 2010

### nickjer

Sorry about my delay in responding. But this looks right. And I didn't think of D H's way, but both get the same result :)