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ORBIT: change in orbital distance

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  1. Mar 20, 2016 #1
    1. The problem statement, all variables and given/known data
    we know the mass of the moon, Mm, and the earths, Me, and also the initial distance between their centers as the moon orbits the earth, Rem.

    Now if the earth’s angular velocity about its own axis is slowing down from a initial given angular velocity, ωi to a final angular velocity (due to tidal friction), ωf

    Find the final orbital distance between the earth and moon as a consequence. Ignore the rotation of the moon about its own axis and treat it as a point object in circular orbit about the center of a fixed (but spinning) earth.

    2. Relevant equations
    t=Iα torque

    L=Iω or r x p angular momentum

    T=2π √ r3/GM orbital period

    3. The attempt at a solution

    I do not conceptually understand why a change in the earths rotation would even change the radius of the moon's orbit around the earth. From the 3rd equation I wrote we see that r does not depend on the earth's rotation. Why would it even make a difference how fast earth is rotating?
     
  2. jcsd
  3. Mar 20, 2016 #2

    gneill

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    Friction. The tidal bulge that's induced on the Earth due to the Moon's gravitational influence (and to a lesser extent the Sun's, too) is dragged slightly by the friction between the ocean water and the rotating Earth. So the bulge is always being pulled ahead of the Earth-Moon line by friction. (The bulge reforms continuously, but the dragging is continuous, too). This slightly off-centerline mass produces a torque via gravitational pull between it and the Moon. The torque tends to slow the Earth's rotation and speed up the Moon in its orbit. In effect, the bulge couples the Earth's rotation to the Moon so that angular momentum can be "moved" from the Earth's rotation to the Moon's orbit.
     
  4. Mar 20, 2016 #3
    Oh I understand how the earth slows down now. But why does this affect the moon's orbit? Because the force of gravity doesn't change because the earth does not become more massive or something
     
  5. Mar 20, 2016 #4

    gneill

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    The bulge pulls the Moon forward in its orbit, speeding it up.
     
  6. Mar 20, 2016 #5
    Because the earth would not be a perfect sphere anymore right?

    So would I use momentum conservation in this one with the moon being r x p?
     
  7. Mar 20, 2016 #6

    gneill

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    Right.
    Yes, conservation of angular momentum is the right approach.
     
  8. Mar 20, 2016 #7
    Ok, I came up with this: but I don't know the initial/final velocities of the moon. I could use 2 pi r / T if I was given T but I am not


    r x p = rp (since right angle)

    Iω + r x p = Iω + r x p
    IEarth ωearth initial + dinitial vinitial mmoon = IEarth ωearth final + dinitial vfinal mmoon
     
  9. Mar 20, 2016 #8

    gneill

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    If you assume that the Moon's orbit is always approximately circular (that is, assume that the change in the Moon's orbital radius is very slow and remains essentially circular over time), then you should be able to find an expression for the velocity of an object in a circular orbit that you can apply. Hint: the velocity depends upon the mass of the primary and the orbital radius.
     
  10. Mar 20, 2016 #9
    Yes, doing f=ma I got v^2 =GM/r. Now I have v in terms of r, G, & m.

    This must be it. So now I would be able to use conservation of angular momentum
     
  11. Mar 20, 2016 #10

    gneill

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    Yes.
     
  12. Mar 20, 2016 #11
    Thank you for the help gneill!
    I'll try that when I get the chance.
     
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