I am completely aware that theories and calculations don't do a very good job at predicting reality.
I think you will find that Sir Issac Newton (1642-1727)came up with some perfectly good, predictive maths almost 400 years ago concerning the moons, and most other solar bodies orbital behaviour. Also, more recent calculations show that the moon is moving
I think you will find that Sir Issac Newton (1642-1727)came up with some perfectly good, predictive maths almost 400 years ago concerning the moons, and most other solar bodies orbital behaviour. Also, more recent calculations show that the moon is moving away
from the earth "the Moon slowly receding from Earth at the rate of approximately 38 millimetres per year.
but is not expected to ever escape our gravitational influence.
28th May 2006 - 01:13 AM
There are really four separate questions here.
1: Why does the earth's gravity not cause the moon to come crashing down immediately?
The centrifugal force idea is just a fiction. If you want to do physics in an accelerated coordinate system attached to the moon (so tht you can ignore the acceleration) then you need a fictitious "centrifugal force" to balance the equations. Basically, this approach takes the equation F=ma and moves the ma over to the left side to get F-ma=0. If you do this, and pretend that -ma is a force (the centrifugal force) and that there is no acceleration, then since the new equation is algebratically the same as the old one, you will get the right answers to your calculations. It doesn't explain anything at all really; it just allows you to pretend there is no acceleration if you feel more comfortable doing that.
What is really going on is that the moon is travelling in a circular path around the earth, and all circular motion has an acceleration. This is because acceleration is defined as rate of change of velocity, velocity is a vector, vectors have direction, and the direction of the velocity is continuously changing as the moon goes around its circular path. Since the velocity vector continuously turns to stay tangent to the orbital path, the tip of the velocity vector is always being nudged to one side, in the direction toward the earth. Therefore the motion has an acceleration, and that acceleration points towards the earth.
Force is required to produce this acceleration, and this force is called the centripetal force. It is not fictitions like the centrifugal force, but a perfectly real force. However, it is emphatically not some new force that you get by going around in a circle (a common misconception); rather it is just a convenient name for the amount of force you need in order to go around in the circle. You must have some real identifiable force to supply this need. In this case, the major force acting on the moon (and the only one consistently pulling in the right direction, toward the earth) is the earth's gravity. So the centripetal force is what you need, and the earth's gravity is what you have. Now apply a basic principle of logic: if you don't have what you need, then you can't do what you do. Since the moon is actually going in a circle around the earth, it must have what it needs to do so, and so the earth's gravity must be exactly the right size to fill the requirement called centripetal force. If this seems like a coincidence, see point 4 below.
Here is an alternate way to look at the idea of something being in orbit: Imagine you stood on top of a tall mountain (try Kilimanjaro, on the equator) and threw a ball very hard in a horizontal direction. Gravity would cause the path of the ball to curve downwards and hit the ground. If you threw the ball harder, it would travel farther before hitting the ground. Now suppose you could throw it extremely fast, so that it could travel for say, 1000km before hitting the ground. If it travels that far, the fact that the earth is not flat will come into play: the ground will curve downward under the ball, so it will travel farther than expected before hitting the ground. In fact, if you threw it a little faster still, the ground would curve away under the ball at the same speed that the ball was falling due to gravity, and it would then be no closer to the ground later than when you threw it. Since that is so, the ball must appear to be travelling horizontally over whatever region of the earth it is above, and so it is acting exactly the same way as when it was first thrown. Therefore this will continue forever, with the ball always falling, but the ground always curving out from under it, so that it never comes down. This is an orbit. Of course such a low orbit would encounter air friction, and the ball's forward velocity would be reduced so much that it would spiral down and hit the ground, but a higher orbit, above the atmosphere, would continue indefinitely.
2: Why does the energy loss from the tides not cause the moon to spiral down into the earth?
Although the rate of energy loss is very large by our standards, it is very, very small compared to the amount of kinetic and potential energy available in the earth-moon system. The kinetic energy of just the moon's orbital motion, for example, is 1/2*(7.5E22 kg)·[(2·pi·3.84E8 m)/(29 day * 86400 sec/day)]^2 = 3.5E28 Joules. At this rate, the tides could remove energy at a rate of 1000GW for 1.1 billion years before using up this energy. Of course there are other energies in the system as well; this is just an example calculation to show the general magnitudes involved. (As a matter of fact, the moon's kinetic energy would increase, not decrease, as it spiralled into the earth; however, the amount of gravitational potential energy released by lowering its altitude would be exactly twice the amount by which the kinetic energy increased, since the orbit is circular. Therefore the actual energy released would be equal to the moon's initial orbital kinetic energy.)
3: Why does the moon actually move away from the earth rather than toward it?
As mentioned in point 2, there are several energies in the earth moon system, including the moon's and earth's rotations and the earth's orbital kinetic energy around the center of mass of the system. First, let's deal with this center of mass idea. The main point here is that forces within the system (such as the earth and moon pulling on each other) cannot have any effect on the motion of the system's center of mass. If this were possible, conservation of momentum would be violated, and you could do crazy stuff like sit inside a box and hammer so hard on the top lid that you and the box rose up in the air and flew. This was the principle behind the infamous "Dean drive" which absorbed the attention of a lot of people back in the 1950's. However, things that are too good to be true aren't true, and so we get the conclusion that the center of mass of the earth moon system moves around the sun in a nice ellipse just as if the earth and moon weren't pulling on each other. This means that, since the center of mass of the system is the average position of all the mass, then the earth and moon stay on opposite sides of it as they orbit, which means that really the earth and moon travel in circles around this point, rather than the moon directly orbiting the earth. The earth 80 times more massive than the moon, so the center is correspondingly closer to the earth than the moon. In fact, it is about 4800 km from the center of the earth, which is actually beneath the earth's surface, so the earth is more like wobbling than orbiting. Still, it has kinetic energy, about 1/80 as much as the moon (bigger mass but smaller velocity, velocity is squared, so the velocity factor dominates).
Now as mentioned in point 1, the moon's gravity must be exactly strong enough to cause the earth's centripetal acceleration toward the center of mass point. However, since the whole earth accelerates together, while parts of the earth are nearer the moon than others, the moon's gravity must average out to the right strength across the whole earth. Of course that means that it is too strong on the side of the earth closest to the moon, and too weak on the side away from the moon, causing tidal bulges at both these points. On the side toward the moon, the extra force causes the water to accelerate toward the moon too much, so it gets a little bit ahead of the rest of the earth and forms a bulge, while on the opposite side, the lack of force cause the water to lag behind the bulk of the earth, so it forms a trailing bulge. These bulges always stay pointed toward and away from the moon, so as the earth turns, the oceans and continents sweep through the bulges, which means that from the point of view of someone on the earth, it appears that the bulges sweep around the earth. Since there are two of them, you get two tides a day. (Almost: since the moon is going around its orbit, when the earth has turned once around, the moon is not in the same position as it was 24 hours before, so the earth has to turn a little further to catch up with the new positon. This makes the tides take about 25 hours not 24 to go through one complete cycle.)
This sloshing motion of the tides clearly uses up some energy, which is ultimately converted to heat at the earth's surface. The interesting question is, how exactly does this energy get taken from the orbits and spins and turned into heat? In order to understand this, you have to take into account the fact that there is a time lag in the tidal bulges. Since the tidal bulges are just shapes filled with water, and that water is participating in the earth's rotation, the tidal bulges tend to get carried along as the earth turns. Since the earth's rotation is considerably faster than the moon's revolving (1 day versus 29 days), the tidal bulges are carried forward somewhat. (If the earth turned slower than the moon orbited, they would be carried backward.) This means that they don't exactly point toward and away from the moon, but are tilted at a slight angle away from the line from earth to moon. Now think of the effect of the mass in these tidal bulges: each bulge is going to exert a small amount of gravitational attraction on the moon, and the bulge near the moon will pull in a direction slightly forward along the moon's orbit, because earth's rotation has carried it to a forward position. Similarly, the bulge far from the moon will pull in a direction slightly backward along the moons orbit. You might think that these two effects would cancel, and mostly they do, but the near bulge has a stronger effect on the moon than the far bulge, simply because it is nearer. Therefore, there is a slight net force pulling forward on the moon as it orbits. This does work on the moon, giving it additional orbital kinetic energy, so the moon gradually spirals away from th earth, not toward it.
On the other hand, the equal and opposite reaction forces act on the earth, where they pull backwards on the tidal bulges, causing the earth's rotation to slow down. Therefore, the actual source of energy in the tides is the earth's rotational kinetic energy, not the moon's orbital kinetic energy; in fact, the moon is gaining, not losing energy. There is evidence that roughly 200 million years ago, the earth was spinning significantly faster, and a day was about 20 hours long. (I don't recall the details of this evidence. It came from paleontology, and agrees nicely with the predicted amount of slowing we would expect from the tidal effects mentioned above.)
4: Why is the moon's orbit circular in the first place? In fact, why are all the planets' orbits circular and why does the whole planetary system lie in a plane instead of planets orbiting the sun in just any old tilted orbits?
The answer to this question depends on conservation of angular momentum and a fact about orbital mechanics. First the fact, which I will not prove: out of all orbits with the same angular momentum, the circular orbit has the lowest energy. Now about conservation of angular momentum: The angular momentum of a system can only change when a torque acts on it from outside. Just as with forces and linear momentum, torques acting between objects within a system do not change the system's total angular momentum, but just redistribute it within the system. In order to exert a torque on something, you must have a force acting on it, and this force must be pushing off-center, not passing through the axis the system is turning around. The tidal bulges mentioned above in 3 exerted a torque on the moon, speeding it up in its orbit and therefore increasing both its kinetic energy and its angular momentum.
Now consider a system where the central object is spinning fast, and there are one or more other objects orbiting it more slowly. We saw in 3 that tidal bulges will always steal kinetic energy from the fast spinning object, deliver kinetic energy to the slow orbiting object, and wasts some energy in the process. Therefore the overall energy of the system will decrease. However, the angular momentum of the system will not decrease, because these tidal torques are internal to the system. Therefore, the ratio of kinetic energy to angular momentum will decrease as time goes on. Since the smallest kinetic energy is achieved when the orbit is circular, all orbits will become more circular over time. Therefore it is no coincidence that most of the planetary orbits are circular (they were circularized by the tides raised in the planets by the sun) or that the orbits of most of the moons are circular (they were circularized gy the tides raised in the moons by the planets). However, this process takes time, and so newly captured moons could not be expected to be in circular orbits yet, and also this process is slower where forces are weaker, so that planets farther from the sun (such as Pluto) might not have had time yet for their orbits to circularize. (Or Pluto may not have been there very long, or the pulls of other planets may have affected it and prevented its orbit from becoming circular, or all three.)
If you now look at the solar system as a whole, the total angular momentum of the solar system is constant because there are no significant outside torques acting on it (the other stars are much, much too far away), but because of tidal friction, it has been losing orbital kinetic energy for a long time. Therefore, it must be at nearly the lowest possible energy level consistent with its total angular momentum. Since there are several planets, lowest energy is achieved for each one if its orbit is circular, as discussed above, but what about the relations of these orbits to each other? Since angular momentum is a vector, which for each orbit points perpendicular to the plane of the orbit, these vectors must be added together to get the total angular momentum. The largest result is obtained when adding vectors that are all pointing in the same direction, so to get the greatest possible ratio of angular momentum to kinetic energy (which is the same as the lowest possible ratio of kinetic energy to angular momentum), all the angular momentum vectors should point the same direction. Therefore, all the orbits of the planets should lie in the same plane, and the solar system will evolve toward this configuration over time. By now the planetary orbits lie nearly perfectly in a single plane (again, except for Pluto).
I think this has answered probably more questions than you actually had, so I'll stop now!
29th May 2006 - 04:02 AM
The point I was making in raising this question was about the reductionist view of science teacher that we get at schools these days. They would explain the whole thing by a simple formula showing how to calculate centripetal accelerration and how centrifugal force balance out the gravity pull of the Earth to keep the Moon in a stable orbit with perfect balance of forces. Then this is generalised so that we have "For all havenly bodies out there, this is how it works ...". But when asked about the loss of energy in the system, many people are at a loss.
A more simplistic view of the whole idea about a satelite orbiting a bigger havenly body could be produced by example we see in textbooks. If you shoot a cannon ball out of the Earth at a correct speed and angle, it will reach the outer space of the Earth and where the forces are balanced, it will start to go around the Earth in a stable manner. And a simple formula will demonstrate teh relationship betweenthe distance from the Earth and the speed of the cannon ball ... so that the forces are in balance. It's so important to note here that the cannon ball takes the initial injection of energy to move into orbit. And its orbit is determined by the amount of the initial injection of energy. Therefore it loses energy until the balance is achieved. So the balance is not simply there by design but it is achieved due to the dissipation of energy. If you choose a rocket instead of a cannon ball, then you would achieve the same result by firing the engine to add more kinetic energy into the rocket to help it move into a desired orbit. Either case balanced is achieved by removing or adding energy.
Now that the cannon ball is in a stable orbit. The question arises is "Will it stay there forever or will it fall down or breaks away?" Common sense suggests that if it has enough enery to break away, it would have done so when it is shot up there. So it loses energy and settled into an orbit. It can only lose more energy and fall back down or stay stable in orbit forever. The train of thinking goes further. If it does not lose energy then it should stay there forever. However if it loses more energy, it will have to spiral down. And at this point of thought, one would say the Moon will have to down to Earth.
Now it starts to get tricky. No one know how the Earth-Moon system was formed. It's crazy to suggest that the Moon shot off from Earth into orbit. If it did, perhaps it will eventually lose enough energy to spiral back down. Therefore the cannon ball analogy does not hold very well here. Another conjecture is that the Moon was captured as it was a havenly body flying through space into the influence of the Earth's gravity field. After the initial great disturbance, the Earth and Moon formed a motion system where two bodies rotate around each another in a stable fashion. In this case some of the energy were converted into work which brought the two bodies to a state of balanced forces.
I find that it does not really matter if the system was formed in either of the two conjectures above, there is a common logic
They all start with the initial state where the forces were imbalanced. Then they moved into the state of balance where some energy were transformed into work and that defined their relationship spatially in space. Then the logic goes on saying that this state of balance can only change with loss or gain of energy to one or both of the objects. The affect of the change on the Moon as the smaller object is much bigger. Therefore with a loss of energy, one would naturally conclude that the Moon would spiral into the Earth. In this case the loss of energy is clear in the evidence of tides. This loss of energy is probably on both the spin of the Earth around its own axis and on the Moon when it travels in orbit.
But if it is proven that the Moon is drifting away from Earth, then there are many things here not explained at all. One possibility is that the Moon - Earth system is still in the stablisation phase after the formation. And that means balance has not been achieved at all. And the loss of energy to the tides is happening and helping to bring in the balance. My original point is that the observation window of humans in time is just too small for us to fully appreciate this whole thing. The only thing I feel pretty sure is that leaking energy ultimately must lead to the satelite spiraling to Earth. But this may never happen. The Sun might become a red giant and engulfed all the planets by that time.
But between now and eternity, school science teachers will continue to tell the kids that all sattelites will just happy stay in theihr orbits to eternity because of a little equation that they write on the board. And the textbook writers will continue to support this reductionist view of how this works.
29th May 2006 - 05:23 AM
They arnt lying to you for the reason that they are too lasy to explain it all. If a physics teacher explained all the variables about something, all the assumptions and so on, im sure most of the class wouldnt have a clue whats going on. Call it lies-to-children if you will, there is no point knowing all the little details when very few will get the little details.
29th May 2006 - 05:49 AM
Yes, I agree that many science teachers at the high school level give very shallow and unsatisfying (and sometimes just plain wrong) explanations to questions such as this one. In my home state (Washington state, USA) high school teachers are only required to have a college degree in education, but are not required to have a college degree in the subject they are teaching. Most high school science teachers majored in education and minored in their chosen subject. They know barely more about it than is required to teach the high school class, and a bright student can easily poke holes in their knowledge. A friend of mine informs me that in his home state (Montana, USA) high school teachers are required to have a college degree in the subject they teach, so perhaps the situation is better in some places than others. There is absolutely no excuse for a teacher at the college level to give a shallow answer like the ones you have received, but it is much less common there.
As to your other points:
I mentioned in point 4 that the tidal friction takes energy from the faster rotating body (faster in terms of angular frequency, not velocity) and passes some of it to the slower body, while wasting some of it as heat. If you think about this, you will see that the system can only stop changing when all the rotational speeds are the same in the system. This shows that you are correct when you say that the earth moon system has not yet reached its final stable situation. In fact, the moon has already lost all of its excess rotational speed, so that it rotates on its axis in the exact same time it takes to orbit the earth (which is why we always see the same face of the moon), but the earth is still spinning considerably faster than once per month, of course. Since the earth is much heavier than the moon, it has much more rotational energy to lose, and also loses it slower, since the tides caused by the moon on the earth are not as strong as the tides caused by the earth on the moon. Therefore, the earth has not completed the process of shedding its excess rotational speed. The moon will continue to gain orbital speed as the earth's rotation will continue to slow down, until far in the future (several billion more years, if the earth lasts that long before being incinerated by the sun) the earth will eventually keep the same face turned to the moon, and the whole system will appear to spin together like a single solid object. The moon will hang in the earth's sky in the same place, and vice versa, and they will keep the same face toward each other as they orbit their common center of gravity.
Of course, the sun also exerts tides on the earth (about half as strong as the tides of the moon, which is why on full moon or new moon days, the tides are especially strong, as the two effects work together, and on half moon days, the tides are weak because the effects partly cancel), so that on an even longer time scale, the earth moon system as a whole will gradually lose energy due to the tides of the sun, which will push the earth farther from the sun while stealing energy from the earth moon system's orbital kinetic energy and from the sun's rotational energy. At that point the sun's rotation will slow down, the moon will spiral into the earth, and the composite earth moon will rotate so slowly that it will keep one face towards the sun. This well take a VERY long time of course, because the amount of orbital angular momentum in the earth moon system is much, much larger than what is in the earth's rotation, while the sun's tides, being weaker, will remove the energy at a lower rate. It is also possible that the process could halt at some point due to a phenomenon called resonance, which might keep the moon's orbital period at some fixed fraction of the earth's orbital period. I would have to calculate that to see what happens.
Anyway, the basic process is one of synchronization and circularization, with the faster spinning motion losing energy, the slower spin gaining energy, and some energy wasted as heat. Because this process is occuring both between the moon and earth and also, at a slower rate, between the sun and the earth moon system as a whole, the moon will first move away from the earth until the earth's rotational kinetic energy is used up, and then the effect of the sun will dominate and the moon will begin to spiral in as energy is lost from the system as a whole. This also means that you are correct in saying that it does not matter much how the earth moon system formed, because the process will bring them to a circular orbit eventually. This process is mostly complete today, with a nice circular orbit, moon rotating at the same speed as it is orbiting, but the earth still turning too fast.