I am reading why does E=mc[sup]2[/sup] and I am not sure about one part where they talk about the Earth, the Moon, and conservation of angular momentum.
The argument goes 1) Gravitational pull from the moon creates a bulge in the oceans which causes the tides - OK I knew that. 2) Friction between the Earth’s surface and the bulge causes the planet’s rotation to slow down - knew that bit too. 3) The principle of conservation of angular momentum - where does the lost angular momentum go? To the moon causing it to speed up, that’s where!
The bit that puzzles me, and is not alluded to in the text, is not that angular momentum is preserved, nor that increasing the speed of the Moon will preserve it, but HOW does angular momentum get transferred from Earth rotation to the Moon. What acts on the Moon to make it orbit faster?
I suspect that the answer here has to be ‘gravity’ - it’s the only force by which the Earth and the Moon act on each other, and it makes some sense to me that angular momentum could be transferred via such a force, just like regular momentum could.
However, I can’t think at the moment of any other example of angular momentum being transferred via gravity - or for that matter an analogy of angular momentum being transferred from one body to another via some other force - to make this sound more plausible to either of us. I hope some smart Doper will come along to help you out there.
It still doesn’t make sense, though. Based on this theory, as I understand it, they claim (a) the moon’s rotation has slowed down to the point where one rotation = one revolution (i.e., the frozen moon theory) AND for whatever reason the moon cannot turn any slower, so then it’s orbit must begin slipping away from earth which results in a longer (slower) orbital period…all in the name of conservation of angular momentum.
The moon creates a tidal bulge on the earth. The earth is spinning faster than the orbital period of the moon, so the earth keeps trying to move the bulge away from the moon, leading the moon. The moon pulls back on that bulge, slowing the earth down.
Note that the earth also creates a tidal bulge on the moon. The moon currently spins at the same rate as its orbital period (i.e. once every ~30 days). If it spun more rapidly, it would be in the same situation the earth is currently in, i.e. it would be moving its tidal bulge away from facing the earth, and the earth would be pulling back on it, slowing the rotation of the moon.
Suppose instead that the moon were spinning more slowly. Now the tidal bulge starts lagging, and the earth pulls on it, increasing the spin rate of the moon.
As far as changing orbital speed of the moon is concerned, consider the earth as two seperate centers of mass: one for the main part of the earth, the other for the tidal bulge that the earth keeps rotating away from the moon. The COM of that bulge exerts a gravitational force on the moon; that force has a component which is tangential to the moon’s orbital path around the earth’s main COM, thereby increasing the moon’s orbital speed.
These effects - tidal locking of the earth’s/moon’s rotations, as well as the change in orbital velocity/distance - are happening simultaneously, not sequentially.
Theory predicts that the same thing will happen between the earth and the sun, i.e. the earth’s 24-hour day will lengthen until at some point it is tidally locked with the sun, and the earth will rotate only once every 365 days.
It is not so much the angular momentum that gets transferred, but rather the energy required to move the moon into a larger orbit, which requires transfer of energy to increase the gravitational potential energy. The angular momentum is that move into a higher orbit. The fact that the moon is rotating more slowly is secondary.
The moon is about tens times as far away as when it formed. Can you imagine tides 1000 times as high as currently? Also every 2 1/2 hours since a day took only around five hours.
An object in a more distant orbit has a longer orbital period but the object is actually moving faster, it just takes longer for it to complete an orbit. Read Space by James A. Michener for a nice explanation of how, when in orbit, you must go faster to go slower and you must go slower to go faster. (Actually you’d probably find much better explanations elsewhere, but I enjoyed Space.)
“To speed up, you slow down; to slow down, you speed up.”
As far as the transfer of angular momentum from one body to another, consider the gravitational swing-by maneuvers like that performed by Cassini-Hyugens; as it swings around the planet, it is pulled by the gravity of that body, and acquires both a small amount of orbital momentum (from the orbit of the planet around the sun) and rotational momentum (from the rotation of the planet) as it is pulled along into its new trajectory. The amount of momentum transfer is negligible in terms of its effect on the planet, but can be large as evidenced by the velocity change of the probe. This causes the planet to both slow its progression in its orbit (and fall very slightly sunward) and retard its rotation. If you performed this maneuver hundreds of times a day for thousands of years, you’d eventually end up with a measurable amount of momentum change.
The orbital speed of Mercury is about 107,000 MPH. Venus is about 78,000 MPH. Earth is about 67,000 MPH. So, an object more distant is (A) has a longer orbital period, and (B) is moving slower. For a circular orbit, it appears that the orbital speed is inversely proportional to the square root of the distance.
The general formula is Kepler’s Third Law, which states that P^2 is proportional to a^3, where P is the orbital period, and a is the semimajor axis (the average between the closest distance and the furthest distance).
chrisk- read newton’s comments on his law of gravity. He considered his law as a problem in the sense that the law of gravity does not propose the existence of gravty as material stuff pulling on each other in mutual attaction. For Newton the ‘law’ is one of 'action at a distance. Presuming this go directly to the moon just after lunar capture from a linear motion straightline. From observation there is a rotation and orbit induced - two modes of angular momentum. If applying a mutual attraction force, the developed and balanced angular momentum will be corrupted.
Basically, there ain’t no gravity field, forces of attraction etc,
The action at a distance occurs instantaneously, thus the reason for no force
transfers between earth and moon for instance.
Consider the following. The moon pulls on the earth’s ocean waters creating the nearside bulge. The moon is also pulling on the far side of the earth which would also pull water to the moon’s direction. However, observation has two, more or less equivalent high-tide bulges. The only resolution of this contradiction is that mutual forces of gravity ain’t - conservation of angular momentum is.
youy asked for a Doper’s evaluation, so put this your pipe and smoke it if you will- -inhaling helps.
