# Orbital momentum

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1. Oct 24, 2015

### dean barry

I know that a two body stable circular orbit has each body with equal momentum, but in the case of an elliptical orbit, if you ignore the orbit motion the two bodies reciprocate back and forth relative to each other, does this variable linear momentum rob from the circular momentum ? or is it a separate issue ?
I only attend this site for an hour or so every Saturday, and there is no hurry for an answer, thanks in advance
Dean

2. Oct 24, 2015

### Janus

Staff Emeritus
Separate. Angular momentum is conserved in an elliptical orbit also. If you think about it, the net momentum of the "in and out" component for the system is zero, as each body's momentum towards or away from the barycenter is equal in magnitude but opposite in direction to the other's. Thus it has no effect on the total momentum of the orbiting system.

3. Oct 24, 2015

### Buzz Bloom

I feel rather stupid, but I am unable to develop the equations that show that two unequal masses in circular orbits about their center of mass have the same angular momentum. Maybe I am misinterpreting what this means. I am assuming that the angular momentums being compared are with respect to the center of mass.

I looked at two Wikipedia articles, but did not find them helpful:
I would appreciate either a citation to a useful article, or even a hint or two.

4. Oct 24, 2015

### SteamKing

Staff Emeritus
The article on the Two-Body problem is the more apt here.

The section titled "Two-body motion is planar" discusses how the linear and angular momenta relate to one another and that the angular momentum is constant, or conserved.

5. Oct 24, 2015

### Buzz Bloom

Hi SteamKing:

Thanks for your post.

The following is from the Two-Body problem article, "Two-body motion is planar" section:
The motion of two bodies with respect to each other always lies in a plane (in the center of mass frame). Defining the linear momentum p and the angular momentum L by the equations (where μ is the reduced mass).
The vector r is not defined. Is it the distance between the two masses, or is it the distance from the center of mass to one or the other of the two masses?

Regards,
Buzz

6. Oct 24, 2015

### SteamKing

Staff Emeritus
If we take the center of mass of the two bodies M1 and M2 as the reference point, then the location of the c.o.m. of M1 is r1 and the location of the c.o.m. of M2 is r2. The vector r = r2 - r1.

http://scienceworld.wolfram.com/physics/Two-BodyProblem.html

7. Oct 24, 2015

### Buzz Bloom

Hi SteamKing:

Thanks you very much your prompt reply to my question, and especially for the cited science world page. I still have some confusion, but I will try to sort it out before asking more questions.

Just to make sure I am understanding the obvious correctly:
The distance between the two mutually orbiting bodies is
Since the two masses are always on a straight line that includes their c.o.m, a = |r|, and the direction of r is from M1 to M2. Is this correct?

Regards,
Buzz

8. Oct 24, 2015

### SteamKing

Staff Emeritus
I believe you have this correct.

9. Oct 24, 2015

### Buzz Bloom

Hi SteamKing:

Thanks for your quick response confirming:
a = |r|, and the direction of r is from M1 to M2.​

I learned about this stuff a very long time ago, so I may be mis-remembering what I think I know.

The reference frame in which all these vector equations are written seem to me to not be inertial. That is because both ends of the vector r are rotating, and the angular momentum is relative to an origin M1, which is revolving about the fixed c.o.m., i.e., accelerating relative the c.o.m. I seem to vaguely remember that this situation requires an adjustment in order to correct for the rotating frame of reference. Also, when I directly calculate the angular momentum relative to the c.o.m separately for each of the two masses, I get a different answer.

Can you help me find some reference that explains this?

Regards,
Buzz

10. Oct 24, 2015

### SteamKing

Staff Emeritus
I can't comment on your calculations since they are not presented.

http://web.mit.edu/8.01t/www/materials/modules/guide17.pdf

gives a more in-depth description of the two-body problem, especially its derivation. Sections 17.4 and 17.5 seem especially relevant here.

11. Oct 25, 2015

### Buzz Bloom

Hi SteamKing:

Thanks for your post, especially the citation.

Here is the relevant equation (17.3.14) for the angular momentum L from your cited paper:

This is the same as the result I calculated for the two body angular momentum using circular orbits. Note that r0
"is a constant (known as the semilatus rectum)"​
(semimajor axis in English). For a circular orbit, r0 is the radius.

Therefore, the angular momentum for each circular orbit is proportional to the square root of its radius.

From the original post #1,
"I know that a two body stable circular orbit has each body with equal momentum".​

For unequal masses, the two respective values for angular momentum of the the two masses will NOT be equal.

Regards,
Buzz

Last edited: Oct 25, 2015
12. Oct 29, 2015

### dean barry

thanks all, can i assume janus is correct ? system momentum doesnt vary, individual body angular rotation rate is related directly to its position from the barycentre at any given time ?

13. Oct 29, 2015

### Buzz Bloom

Hi dean:

I made a mistake in my post #11. I just discovered that the semilatus rectum is not the same as the semimajor axis.
The following equation relates r and ϑ for motion relative the center of mass.

The semilatus rectum is the radius r when the angle ϑ = π/2. I will have to do some work to understand what the means w/r/t the angular momentum.

Regards,
Buzz

14. Oct 31, 2015

### dean barry

Thanks for your attention all, I will chew on what I have and see if it goes anywhere
Thanks again
Dean

15. Oct 31, 2015

### Isaacsname

Amusing but perhaps off-topic, my apologies

Newton was trying to use circular orbits and couldn't get the math to work out. He was using the Sumerian measurement of one Saru { 3600 ] to estimate the earth's attraction at the moon's distance as less than the attraction at her own surface at the ratio of 1 / 3600 , this measurement was used in ancient metrology to calculate astronomical cycles

He gave up working on the problem and it wasn't until Hooke sent him a letter stating that the force that varied at the inverse of the square of the distance having an eliptic orbit with the center of force in one of the focii

Hooke sent this statement to Newton with no proof or further explanation

16. Nov 1, 2015

### Buzz Bloom

For a circular orbit, the semilatus rectum is the same as the radius. I should have seen this immediately, so I will just blame it on another senior moment.