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Why is the moon drifting into space

  1. Mar 17, 2005 #1
    My astronomy book tried saying that the moon is spinning into outer space, away from earth,; it was blamed on tidal effects between the earth and moon and the subsequent generation of frictional forces. I dont see it.

    Also, from physics, if you look at a force diagram there should only be weight acting on the moon pulling it towards the earth, so there would be a net force towards earth. but the moon is drifting into space (confirmed by nasa experiments), so how is this possible. my physics teacher couldnt understand what i was saying.
     
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  3. Mar 17, 2005 #2

    tony873004

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    It's because the net force isn't exactly towards the Earth. The moon pulls tides on Earth. There is a high tide almost directly underneath the moon, and another high tide on the opposite side of the Earth. I say almost because as the Earth rotates, the buldge created by the moon is pulled forward slightly. This means that the Earth is not a perfect sphere, and the moon itself gets pulled mostly towards the Earth, but a tiny bit in the direction it is orbiting. This results in the moon's orbit gaining energy and that causes it to spiral out. The effect is very small, something like an inch per year, but builds up over millions of years since millions of inches is a lot of miles. The other thing that happens is that the Earth's rotation slows down a little bit. Again, a very small amount, but enough that over millions of years it adds up. Some theorize that the moon used to be a lot closer and the Earth's day was 12-18 hours long. In the distant future, after the moon spirals out more, and the Earth's rotation slows and becomes the same as the moon's orbital period, the moon will no longer spiral out. The system will be tidally locked. But other bad things (such as the Sun expandinging into a red giant and engulfing the Earth) are probably in store for Earth long before this has a chance of happening.
     
  4. Mar 17, 2005 #3

    Janus

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    The tidal interaction works like this:

    The Moon causes tidal bulges on the Earth. Ideally, these bulges would line up with the Moon. The Earth, however, turns on its axis and friction between the body of the Earth and the tidal bulges tends to drag the tidal bulges along. The Moon, meanwhile tries to keep the bulges aligned with itself. The resulting tug of war results in the tidal bulges "leading" the Moon by a bit.

    So we have an action/reaction situation here, while the Moon tries to pull "back" on the bulges, the bulges pull "forward" on the Moon. If you push or pull forward on an object in orbit, you add to its orbital energy and it climbs into a orbit with a higher average altitude.

    The other side is that while the friction between the Earth and tidal bulges pull them forward, it also pulls backward( against its rotation) on the Earth. This causes the Earth to lose rotational speed.

    In effect, the 'canted forward' tidal bulges act as a mechanism to transfer angular momentum from the Earth to the Moon.
     
  5. Mar 17, 2005 #4

    SpaceTiger

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    You need to take into consideration that the moon and earth aren't perfect spheres. They are actually kind of egg-shaped, with the long axis of the egg pointing towards the other body. Why? It has to do with the fact that each pulls more strongly on the close side of the other than the far side. The lack of sphericity means that you can no longer treat them as point masses (as I'm assuming you did in your diagram), so the problem gets a little more complicated. What ends up happening is described in more detail here.

    The short version is that the moon got tidally "locked", so that it always has the same side facing us. Given enough time, the same thing will happen to the earth (rotation period will match the moon's orbital period), so that must mean the earth is spinning down. But if the earth is spinning down, conservation of angular momentum dictates that something else must "spin up". It turns out that this something else is the moon's orbit. Over time, it gains angular momentum and moves away from the earth.
     
  6. Mar 17, 2005 #5

    tony873004

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    An interesting side note is that if the Moon's orbital period were faster than the Earth's rotational period, the exact opposite would happen. The Earth's rotation would speed up and the Moon would spiral in. This is happening to Mars' moon, Phobos.
     
  7. Mar 18, 2005 #6
    This is the easy way to think about it:

    The moon and the earth together have a certain amount of energy. The tides are like friction, they slowly drain the Earth-Moon energy. This causes the moon to slow down, and so it drifts further out.
     
  8. Mar 18, 2005 #7

    tony873004

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    No, it causes the moon to speed up. If the moon slowed down it would drift inward. Friction only applies to the Earth part of it. It makes the Earth's rotation slow down.
     
  9. Mar 18, 2005 #8

    Janus

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    Easy, but wrong. If the Moon were losing energy it would fall into a lower orbit, not climb away.

    While it is true that higher orbits mean lower orbital speeds and thus lower kinetic energies, The increased altitude of the orbits involve an increase of gravitational potential energy that is greater than the decrease in kinetic energy.
     
  10. Mar 18, 2005 #9

    SpaceTiger

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    An interesting way of looking at it is that gravitational systems have a negative specific heat. That is, if you add energy, the particles/bodies in the system will slow down (have a lower temperature).
     
  11. Mar 20, 2005 #10
    When we move a satellite to a higher orbit we must fire the rocket. Unless the moon is somehow already past the point of no return, there must be a force acting on it. Tidal patterns on Earth would, I think, slow the moon down-sort of like spinning a raw egg compared to a hard boiled one. This is a very good question. I think the moon is receding from the Earth at about 2.4 cm/year, now I wonder, is it accelerating?
     
  12. Mar 20, 2005 #11

    pervect

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    No, tidal forces act to speed the moon up, not slow it down, as several posters have already mentioned. However, the tidal forces act as a whole to dissipate energy, while conserving angular momentum. The Earth loses more energy by spinning down than the moon gains by climibing to a higher orbit - the difference powers the Earth's tides.
     
  13. Mar 24, 2005 #12
    Then, why are satellites crashing down to earth unlike the moon which is getting flung out to space? Are they too light? Too close?
     
  14. Mar 24, 2005 #13

    selfAdjoint

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    Too close. Their orbits cut through the upper parts of the atmosphere, so they experience air drag, which slows them down.
     
  15. Mar 24, 2005 #14
    I have always remembered this explanaition :

    the earth slows down due to the friction with the oceans (the tidal bulges). Part of this energy goes to a rise in potential energy of the earth-moon system because of the extra forces between the bulges and the moon (a torque is exterted between the bulges and the moon because both are NOT aligned). The other (less in magnitude) amount of energy goes to the rise in kinetic energy of the moon (due to conservation of angular momentum).

    Now, normally the radius of the moon would lower because the kinetic energy is bigger, but in the same time the potential energy has become much greater and this can only happen if the radius of the moon-orbit has become larger.

    the net effect yields a lunar orbit of bigger radius. It is a bit like the radius rises 5 km (because the potential energy rises) and lowers 2 meters (because the kientic energy rises). In the end both energy conservation and angular momentum conservation are respected.

    I have one question though : how do we know for sure that more energy from the 'slowing down' of the earth goes to the rise in potential energy (coming from the force between the moon and the tidal bulges) and less energy goes to the rise in kinetic energy of the moon ? Does anyone have the calculations to prove this ?

    regards
    marlon
     
  16. Mar 24, 2005 #15

    Phobos

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    Remember that an orbit is a balacing of tangential velocity and gravity between the two objects. If you increase the velocity of the moon (as Janus described), then it can overcome the gravitational attraction more and move further away from the Earth (move to a higher orbit).

    Of course, this is not free energy...increasing the moon's tangential velocity is at the expense of decreasing the Earth's rotational speed.
     
  17. Mar 24, 2005 #16

    tony873004

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    On a kinda related subject, how strong is the Earth's grip on the Moon's tidal buldge? I don't imagine its very strong, and that if the Moon did have a period other than sychronous, that it would take a long time for Earth to slow it down to sychronous again.


    If the Moon had a slow rotation as viewed from Earth so that from Earth we witnessed it rotate once a year (ie in 6 months from now the far side of the moon was facing us, and 6 months later, the familiar side was facing us again), how long would it take for the Earth to put the brakes on the Moon's rotation and make it sychronous again?

    How much energy would it take to make the moon rotate, as viewed from Earth, no matter how slowly, so that the far side was facing us at times? IE: If an astronaut drove a moon buggy in a westward direction on the Moon's equator and travelled around the Moon once, how massive would the moonbuggy have to be to impart a rotation on the moon that exceeded the strength of Earth's grip on the tidal buldge and break the sychronous rotation for at least one rotation?
     
  18. Mar 24, 2005 #17
    Yes, but total energy needs to be conserved so increasing the kinetic energy of the moon will indeed result in a higher orbit. However if the radius has increased then so has the potential energy between the moon and the earth. So the increase in both kinetic and potential energy must come from the decrease in rotational energy of the earth. What i am wondering about is what calculation shows us how much of the earth's rotational energy goes to the increase in kinetic energy and potential energy of the moon ?


    marlon
     
  19. Mar 24, 2005 #18

    BobG

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    You're counting the same increase twice. You add energy to the Moon's orbit, period.

    Theoretically, there's two ways you could do that. You could change the potential energy by instantly teleporting the Moon to a new location (okay, that doesn't happen). Or, you could add kinetic energy.

    Once you've added the energy, the balance between kinetic and potential energy is automatic. In other words, if you only accelerated the Moon once, the Moon would have to return back through that same point (the one where you accelerated it) every orbit. You're affecting the opposite side of your orbit - how far away the Moon will reach on the opposite side. As that distance increases, the balance shifts from kinetic energy towards potential energy.

    In reality, the Moon is accelerated at every point in its orbit (not just one point). You see the end result of an infinite number of little accelerations, which is why it appears you added speed to wind up with a slower speed (for any given point in an orbit, you're looking at the opposite side of the orbit than where the acceleration took place).
     
  20. Mar 24, 2005 #19
    Thanks BobG

    marlon
     
  21. Mar 25, 2005 #20

    pervect

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    There's a theorem called the virial theorem which says that the kinetic energy of an orbiting body with an inverse square law force law is -1/2 the potential energy.

    (The potential energy is negative, the kinetic energy is positive, that's why the factor is negative).

    So when the moon goes to a higher orbit, if you increase the potential energy by 2 units, making it less negative, the kinetic energy decreases by 1 unit, and the total energy increases by 1 unit.

    See for instance Goldstein, "Classsical mechanics", pg 83-85.

    For a sanity check, consider a circular orbit.

    The potential energy is -GM/R

    Setting the centripetal acceleration equal to the force of gravity gives
    GMm/R^2 = mv^2/R, so mv^2=GM/R, thus .

    5*m*v^2 = kinetic energy = GM/2R

    which confirms the relationsip

    kinetic energy = -1/2 potential energy

    Note that the factor of -1/2 is specific to the inverse square force law.
     
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