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If the tides on earth one meter high how far away is the moon?

  1. May 11, 2005 #1
    Hi all, I am reviewing for my final and cannot seem to find how to solve the following

    If the tides on earth one meter high how far away is the moon?

    What equations should I use to calulate how the moon effects the tides on earth?
     
  2. jcsd
  3. May 12, 2005 #2

    HallsofIvy

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    The point of tides is that the gravitational force acting on the water is different from the gravitational force acting on the earth. That's because we can treat the gravitational force on the earth as acting on the center of the earth while the force acting on the water is acting at the surface of the earth. The fundamental equation you need is F= GmM/r2, used twice. G is the gravitational constant, m the mass of the water (or earth), M the mass of the moon and r is (1) the distance from the center of the earth to the center of the moon (2) the distance from the surface of the earth to the center of the moon (the figure in (1) minus the radius of the earth). The G and M terms should cancel out and if you set the difference in force to the weight of the water, the mass of the water should cancel as well.
     
  4. May 14, 2005 #3
    Thanks for the help but I am still having a hard time figuring out how to do this. I know the density of water, but volume should I use?

    Once I know this I have to figure out how much torque the moon is exerting on the earth. Form the I have to fingure out how much torque the moon will be exerting on the earth in one year. I know the moon is moving a way at 4cm/year from the earth.

    Because the toruqe will have decreased the day should be slightly longer.
     
  5. May 15, 2005 #4

    Andrew Mason

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    This is an unanswerable question based on the information you have given. The tidal forces cause the earth to rise as well as the water. The difference in height of the water will depend on the depth of the water as well as local geography. Some of the highest tides in the world are in the Bay of Fundy between Nova Scotia and New Brunswick (50 feet). It has less to do with the strength of the moon's gravity at that point and much more to do with the frequency of the tidal oscillation and natural period of the bay itself and the shape of the Minas Basin.

    AM
     
  6. May 15, 2005 #5
    OK here is the exact question he gave in class,

    If the tides are a meterhigh, how long will you birthday be next year?
     
  7. May 15, 2005 #6

    Andrew Mason

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    I calculate the answer to be 23 hours 56 minutes.

    AM
     
  8. May 16, 2005 #7
    Thanks for the help Andrew. Could you please show me how got the anwser 23 hours 56 minutes.

    I know as the moon moves away from the earth the earth is slowing down. Some of the angular momnet of the earth is "lost" on the moon.

    Thank again for your help.
     
  9. May 16, 2005 #8

    Andrew Mason

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    Well how accurate do you want it to be? 86164.091 seconds or 23 hours 56 minutes 4.091 seconds: that is the length of an earth day. It increases a few microseconds per century due to the earth losing angular momentum to the moon but that is all.

    AM
     
  10. May 16, 2005 #9
    thanks for the quick reply. I understand that the change will be minimal but how can i mathmatically show how much angular momentum is lost on the moon each year.
     
  11. May 16, 2005 #10

    Andrew Mason

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    You can't. You have to measure it. It is way too complicated to predict mathematcally.

    AM
     
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