# Where Does the Angular Momentum Go?

When the earth, for example, moves slower in its orbit when it is furthest from the sun…where does the decrease in angular momentum go? Is it magically transferred to the sun, or something? Maybe it causes all CD players to play backwards so we can find all the hidden messages…?

• Jinx

The earth moves slower when it’s further from the sun. Angular momentum is constant. In fact, that’s the meaning of Kepler’s 2nd Law. It was originally phrased as: “a line connecting a planet to the sun will sweep out equal areas in equal times”. That just means angular momentum (r*v, where r: distance from the sun, v: velocity) is conserved.

It causes a variety of organisms to divert momentarily from their natural paths. Walter Becker once tuned his guitar at that moment, and all the strings were in tune. He couldn’t play right for a week. Pat Boone recorded a heavy metal album. It became an annual thing with Niel Young. He went rockabilly, electronic, and retro-blues. Rush Limbaugh turned left. It hurt his neck, and we know where that led.

There is no decrease in angular momentum – it is conserved and constant. This is because the area swept out by the earth is the same for any given period of time, regardless of its distance and velocity.

IOW, the changes in the distance from the sun and the orbital velocity exactly offset each other at the earth moves in its eliptical orbit.

Angular momentum is I*w. I is the moment of inertia (which in this case equals the distance from the sun to the center of the earth multiplied by the mass of the earth) and w is the angular velocity in radians/sec.

The angular velocity slows down by the same amount as the moment of inertia increases as the earth recedes from the sun, as several others mentioned.

You may find this counterintuitive. Consider two planets of equal mass in circular orbit about the same star. One is closer in, and it has higher speed and higher angular velocity. The other is further out and has lower speed. Guess which one has more angular momentum? If you said the slower one, you’re right.

Does the net angular momentum of the Cosmos equal zero?

>Does the net angular momentum of the Cosmos equal zero?

Wow, good question. Let me speculate.

It must be constant.

It must be small, in the sense that the mass-weighted mean angular rotation rate must be very slow.

I can see where net zero angular momentum is consistent with universes that expand or contract or have quintessence etc etc, and it’s not at all obvious whether nonzero angular momentum could be consistent likewise.

Certainly there must be very big regions - clusters of clusters of galaxies - where the angular momentum isn’t zero. Over these distances relativistic effects would be important and I am fuzzy on the question of what constraints appear if we say angular momentum must appear conserved to all observers, but I bet we must say it.

Wow…

Yes, almost certianly. Any net angular momnetum would be observable as a certain kind of anistropy in the CMBR and would almost certainly affect the observed abundance of different elements. These place a very low upper limt on the angular momentum of the universe, of course if the universe did have angular momentum time-like loops (i.e. time travel) might be possible (see Godel universes).

Angular momentum = M = I*w

I = m*d and w ~ 1/t ; m = planet mass, d = distance to planet center, t = time for one orbit.

Time, t, for one period is related to the distance, d, from the primary by:

t[sup]2[/sup] ~ d[sup]3[/sup] or t ~ d[sup]3/2[/sup]

so M ~ (md)/t ~ md/d[sup]3/2[/sup] ~ m/d[sup]1/2[/sup]

If planet one is at distance = 1 unit and planet 2 is a distance 2 units then:

M[sub]2[/sub]/M[sub]1[/sub] ~ (d[sub]1[/sub]/d[sub]2[/sub])[sup]1/2[/sup] = 1/2[sup]1/2[/sup]

and outer planet 2 has 70% of the angular momentum of planet 1

I = md[sup]2[/sup].

Achernar, while I appreciate the conciseness of your post, could you elaborate on what the ratio would be?

I can’t even reproduce David Simmon’s result. Isn’t 1/2[sup]1/2[/sup] 84+% reather than 70%? And trying to apply your correction, I get an unlikely result.

Help for the math challenged, please.

David Simmons’ math was totally correct, but for that initial typo. Here’s the post, with changes made in bold.

To be honest, I didn’t think this all the way through when I first posted, but it’s something I knew from quantum mechanics. In the Bohr atom, each circular electron orbital corresponds to a certain (quantized) angular momentum. The innermost one has angular momentum L = h-bar, and the angular momentum increases as you go out, 2h-bar, 3h-bar, etc.

And I’m not sure what the matter is here, but 1/2[sup]1/2[/sup] is indeed equal to 0.7071068…

Thanks very much.

I used the Google calculator with .5 exp .5 (and variations thereof). Plugging in .5^.5 gives the correct result, as does the Windows calculator.

Correct. I left out the second d multiplier and carefully derived the wrong result. Oh well.

.5^.5 = (1/2)^.5 which is not equal to 1/2[sup]1/2[/sup] or 1/sqrt(2)

Technically, yes, but the parentheses were implicit. What the Google calculator was doing was using exp as short hand for e^, so .5 exp .5 evaluates to 0.5(e^.5), which is indeed 0.82

At the risk of being overly pedantic the ground state orbital angular momentum of the hydrogen atom is actually zero. But Achernar is absolutely correct in saying that in the Bohr model it equals hbar. Which of course is one of the problems with Bohr’s theory.

Like hell it doesn’t! Two blunders on the same thread and in the same day.

Don’t stop me now, I’m on a roll!