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Question about gravitation

  1. Feb 22, 2007 #1
    Hello all,
    I have a simple question, that has challenged my mind for quite a while now.
    The question is: since the moon is in orbit due to the Earth's gravitational force, why doesn't the moon crash into the Earth(and the Earth crash into the sun for that matter)?
    I have the assumption that the resultant of the gravitational froces from other solar bodies keeps the moon in orbit. Please correct me if I'm wrong.
  2. jcsd
  3. Feb 22, 2007 #2

    D H

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    When you throw a ball, does it fall straight to the Earth the instant it leaves your hand? Ignoring air resistance, the ball is in free-fall (i.e., it is falling) the instant it leaves your hand. The ball also has a horizontal component to its velocity. Gravity only affects the vertical velocity. It does not affect the component of the velocity vector normal to the gravitational acceleration (i.e., the horizontal component).

    A body in orbit about another more massive object also has a component of its velocity that is normal to the gravitational acceleration. Without gravity, this normal component velocity would make the distance between the two bodies increase. With just the right velocity, the inward acceleration balances this tangential velocity to make the distance constant (a circular orbit).

    Bottom line: The moon is constantly falling toward the Earth. The reason it doesn't ever collide with the Earth is because it has a huge velocity that is nearly normal to the acceleration vector.
  4. Mar 16, 2009 #3
    Isnt it because when the moon orbits the earth it experiences a force of mv^2/r in the direction opposite to GMemm/r2..? correct me if im wrong.
  5. Mar 16, 2009 #4


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    This is true in a rotating reference frame, where earth and moon are both at rest. In this non inertial frame centrifugal forces (mv^2/r) and gravitation add up to zero. In any inertial frame, gravitation is the only force the moon experiences.
  6. Mar 16, 2009 #5


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    In other words, it's constantly missing the Earth. :smile:
  7. Mar 16, 2009 #6

    D H

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    This is true. One does not need to invoke the concept of centrifugal force to explain orbits. Invoking the concept of centrifugal force hinders the understanding of orbits. How do you explain a non-circular orbit using this concept?

    This is not true. The Moon is not in a circular orbit. I'll construct a sequence of reference frames of increasing complexity, eventually arriving at A.T.'s frame.
    • Start with the non-rotating frame with origin at the Earth-Moon barycenter. Ignoring the Sun, the other planets, and the stars, this is an inertial frame. The Moon and Earth orbit about each the origin due to gravity. No other forces are needed to explain the orbit.
    • Next in complexity is a reference frame rotating at a constant rate equal to the mean Earth-Moon orbital rate. The Moon and Earth are not stationary in this frame. In the above inertial frame, the Moon is in an elliptical orbit. The Moon's angular velocity is not constant. In this constant rate rotating frame, the Moon makes a small pseudo-orbit about it's mean position. The physics in this frame isn't too hard. You just need to account for centrifugal and Coriolis accelerations.
    • Next in complexity is a reference frame rotating with the Moon's inertial frame angular velocity. Now the Earth and Moon are constrained to lie along a line. Call this the x-axis of our reference frame. The Earth and Moon are not stationary yet; the Earth-Moon distance changes between perigee and apogee. The physics in this frame is starting to get rather messy. The Earth-Moon gravitational jerk (3rd time derivative) creeps into the equations of motion, for example. Modeling rotational states in this frame is a nightmare. But it's not as bad as A.T.'s frame ...
    • Finally, A.T.'s frame, in which the Earth and Moon have fixed positions. This is a very nice frame for graphically depicting things such as the trajectory of a space vehicle going from the Earth to the Moon (or back). This is a very un-nice frame for doing physics because it is not only rotating at a non-uniform rate, it has a time-varying distance metric.
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