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Is the Moon's linear momentum p constant?

  1. Mar 11, 2004 #1
    Question:

    "The moon revolves around the Earth. Is the Moon's linear momentum p constant?(the p is supposed to be in italics and with an arrow over it pointing to the right. It's kinetic energy constant? Assume the orbit is circular."

    Being this a concept question and not a math question im having some trouble visualizing this problem. What exacly should I consider in order to formulate a good explanation? Any help will be STRONGLY appreciated since I have no physics background whatsoever.
     
  2. jcsd
  3. Mar 11, 2004 #2
    Linear momentum is a vector function, i.e. it has both direction and magnitude. It depends on the magnitude of the mass of the object moving, the magnitude of the velocity with which the object is moving, and the direction in which the object is moving. If the linear momentum is to be constant, then all three of these must be constant. Here's a question for you, then: is the moon always moving in the same direction?

    As for kinetic energy, kinetic energy is a scalar function, i.e. it has only magnitude. It depends on the magnitude of the mass and the magnitude of the velocity, but not the direction of the velocity. That said, let us consider the "circular" orbit of our super-conveniently perfect physics world moon. A circle is radially symmetric, i.e. it looks the same regardless of the angle from which you view it from the center. Because of this, there cannot be any differences in the magnitude of velocity, or else we would be able to distinguish which part of the circular orbit it is in. Since we can't do this, the magnitude of the velocity must be constant. The moon's mass is obviously constant, so the moon's kinetic energy will also be constant.

    Hope this helps.

    cookiemonster
     
  4. Mar 11, 2004 #3

    turin

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    Homework Helper

    What basic laws of physics can you reasonably consider with the given information/situation (hint: what is conserved?)? Can the moon by itself be considered as a closed system (hint: why not?)?
     
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