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Questions about Gravity and Acceleration of masses.

  1. Aug 14, 2010 #1
    So as we all know the acceleration of an object is independent of it's mass (neglecting friction)...so if you dropped a rock and a car they should technically accelerate downward at the same rate. But what about if the object we were dropping was really massive, like so massive it would have its own gravitational field, wouldn't that field then affect the earth's gravitational field and pull the earth a bit towards it, and so this massive object wouldn't accelerate downward at the same rate as say a rock? Which would mean that acceleration due to gravity isn't actually completely independent of mass?
    Where am I going wrong?
  2. jcsd
  3. Aug 14, 2010 #2


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    Even small objects pull the earth toward them.
    You have to be careful to measure acceleration as change in distance from some fixed point - not as distance to the earth.
  4. Aug 14, 2010 #3


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    Well normally you will have g=GM/r2 where 'r' is the radius of the Earth and 'M' is the mass of the Earth (we are assuming you are dropping things from a small height above the Earth).

    Now you see 'g' is independent of any other mass except for the Earth.

    g=acceleration due to gravity = force exerted per unit mass on an object in a gravitational field = F/m = GMm/r2/m=GM/r2

    If you are dropping another mass which is comparable to the mass of Earth (5.9742 × 1024 kg), then both masses would matter and so would the distance between them.

    How you are dropping such a mass is another matter.
  5. Aug 14, 2010 #4
    Right, I just used massive objects for a better example...but say you "dropped" Jupiter right above the earth, jupiter has a much stronger gravitational field, so wouldn't Jupiter remain stationary and the earth would accelerate towards it...so the acceleration of Jupiter would be zero as oppose to 9.8m/s^2? I realize this example is ridiculous but it's the best I could think of to get my point across.
  6. Aug 14, 2010 #5


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    The force between them would be F=GMm/r2 where 'r' is the distance between the masses 'M' and 'm'. So essentially they'd both be under the same force by Newton's 3rd Law.
  7. Aug 15, 2010 #6
    Hey Josh :)

    If you "dropped" Jupiter at some point relative to the Earth, both would feel the same force [tex]F=\frac{GMm}{r^2}[/tex], where M is the mass of Jupiter and m the mass of Earth. If you imagine some arbitrary point between Jupiter and the Earth, Jupiter would accelerate towards it with [tex]a=\frac{Gm}{r^2}[/tex] and the Earth would accelerate towards it with [tex]a=\frac{GM}{r^2}[/tex].
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