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Drop the Moon 1m From the Earth, what's the acceleration?

  1. Mar 31, 2009 #1
    We know that free fall acceleration is independent of mass.

    9.8 m/s^2 on the Earth Surface


    1.63 m/s^2 on the Surface of the Moon.

    But what would happen if you dropped the Moon on the Earth or, the other way around, dropped the Earth on the Moon? Which acceleration would it be? Is the acceleration indeed have some dependency on mass?


  2. jcsd
  3. Mar 31, 2009 #2
    You can compute this yourself by equating Newton's second law with his third law.
  4. Mar 31, 2009 #3


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    The general formula that describes the gravitational attraction between two objects is
    [tex]F = \frac{G M_1 M_2}{r^2}[/tex]
    where G is the gravitational constant and r is the distance between the centers of gravity. When you take M1 = mass of the earth or moon, and r approximately equal to the radius of the earth or moon (the radius of the earth is about 6000 kilometers, so dropping an object at 1 meter above its surface would make it a negligible 6000,001 kilometers) you can write this as
    [tex]F = M_2 g[/tex]
    [tex]g = \frac{G M_1}{R^2}[/tex]

    If you use this formula, you always get the right answer. In the case of earth/moon attraction I would suggest using it, because - unlike an apple falling towards the earth - the attraction of one to the other is not negligible for either.
  5. Mar 31, 2009 #4
    Consider two masses M1 and M2 (one could be Earth, the other the Moon). The force that M1 exerts on M2 is given by:

    F = - M1 M2 G/r^2 r-hat

    Here r is the distance between the two masses, G the gravitational constant and r-hat is the unit vector that points from M1 to M2. By Newton't third law the foce that M2 exerts on M1 should be the same in magnitude, but the opposite in sign. You can see this from the above formula. If we had taken M1 to denote the other mass, then everything would be reversed, but the formula is symmetric, except that r-hat would point in the opposite direction.

    From Newton's second law:

    F = M a

    it follows that the acceleration of M2 due to the gravitational force of M1 is given by:

    a2 = - M1 G/r^2 r-hat

    This does not depend on the mass M2. In case of the Moon, you see that the Moon will not accelerate at 1 g, because the Moon is very far from the Earth (note that r should be taken from the center of the Eart, so the value of g at the Earth's surface follows by putting r = 6378 km, the radius of the Earth).

    Similarly, you also see that M1 accelerates toward M2. To find the relative acceleration, you need to add up the two accelerations.

    Note that the Moon is always accelerating toward the Earth. As the Moon revolves around the Earth, it's velocity is constantly changing direction in the direction of Earth. This is called centripetal acceleration. This explains why the Moon isn't falling down on the Earth, despite it accelerating toward the Earth.
  6. Mar 31, 2009 #5
    So are you all saying that the relative acceleration does depend on both objects, however, when we talk about free fall being independent of mass, as it is generally between Earth and an object with negligible mass it can be ignored?


  7. Mar 31, 2009 #6


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    Yes. Both objects attract each other.
    But yes, usually you can ignore one.

    If you wonder why, try using the "full" law to calculate the acceleration of the earth due to a 100 g object at 1 meter above its surface (using F = m a).
    [Note that the force is always the same by the way, it's because of the m in F = m a that the same force leads to different accelerations hence motions.]
  8. Mar 31, 2009 #7

    D H

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    While the relative acceleration of two objects toward each other does indeed depend on both objects masses, the acceleration of either object from the perspective of an inertial observer only depends on the mass of the other object. In this sense, free fall acceleration truly is mass-independent.
  9. Mar 31, 2009 #8
    There was a discussion some time ago on another forum if a feather and a hammer would take precisely the same time to fall when dropped from rest at some height above the Moon's surface.

    Soon in that discussion the Moon had to be replaced by an ideal Moon that had no atmosphere at all, otherwise that would be a trivial dominant effect.

    Some said that the Moon will accelerate to the hammer more than to the feather. But I objected to this simplistic reasoning, because you cannot pretend that the entire Moon will react as a rigid body. When you are standing on the Moon your feet are depressing the Moon's surface a bit. When the hammer is dropped the surface rebounds. This perturbation will eventually travel to all parts of the Moon's interior.
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