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Simple question about Gravity on Earth

  1. Mar 20, 2009 #1
    We keep getting drilled with the info in class that all objects would fall to the earth at the same speed. And it does not make sense to me

    So the equation is (Ge * m1 * m2)/r2 = F

    And F = m1 * a. There fore m1 * a = (Ge* m1*m2)/r2

    This can be simplified and the m1 eliminated by dividing it from both sides, so that a = Ge *m2(mass of earth)/r2.

    So my question is, how does this work with masses larger than the earth? Because according to this equation the only thing determining acceleration is the mass of the earth and the distance and the the Gravitational Constante Ge.

    Do you just use the mass of the larger object to calculate acceleration and disregard the smaller object's mass?
     
  2. jcsd
  3. Mar 20, 2009 #2

    mgb_phys

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    Because you also have a force on the Earth, the Earth is accelerating up to meet the falling object but normally the effect for a tennis ball is rather small so you can ignore it.

    For two stars/planets falling into each other you have to take it into account.
     
  4. Mar 20, 2009 #3

    DaveC426913

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    i.e the simpler form of the equation is only used for objects whose mass when compared to Earth's is insignificant.
     
  5. Mar 20, 2009 #4

    Integral

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    Not quite.

    Note that there are no approximations made in deriving the equation in the OP. It holds for all masses, but there is one key restriction built into it. You must use your distance from the center of the earth, that is the r in the denominator. If you use the radius of the earth you get the common value of 9.81 m/s2. So this value is only meaningful at or near the earths surface.
     
  6. Mar 20, 2009 #5
    OK. here's one for you...

    I''m sitting here on my chair veiwing PF.
    I now raise with one hand a 1/lb weight 2-feet above me.

    The earth does not move 2-feet backwards, of course, BUT, does it move at all?
     
  7. Mar 20, 2009 #6

    DaveC426913

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    Yes. It moves an amount equivalent to the Earth's mass divided your arm's mass times two feet.
     
  8. Mar 20, 2009 #7

    DaveC426913

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    Right. Forgot about the distance thing.
     
  9. Mar 20, 2009 #8
    Aah, I think I understand it now. I forgot to take into account the acceleration of the earth.

    Which would be F= m2 * a.

    m2 *a = (Ge *m1 *m2)/ r2

    cancel out the m2s and you get a = Ge *m1/r2

    So now my question is..


    Does this mean that if any object how ever large were to come close to the earth, that it would have the same acceleration as us. And that any difference in speed for the two colliding, would solely be caused by the earth also accelerating at a great speed, which it usually does not because our masses are soo insignificant?
     
  10. Mar 20, 2009 #9

    DaveC426913

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    What?
     
  11. Mar 20, 2009 #10
    I meant. Say an object like Mars or something were to come close to the earth's surface. Would it accelerate towards earth, at 9.8m/s2, just like objects on earth with really small mass?

    And would any difference in time it would take for impact by the large object, compared to how long an small object, like us would take to hit the earth be caused by the earth also accelerating. Which the earth would not usually do if small objects fall down, because their mass is so small compared to the mass of the earth.
     
  12. Mar 20, 2009 #11

    DaveC426913

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    Wellll... the part of Mars that is near Earth would feel the tug of 9.8m/s^2, yes. That there are a lot of confounding factors - mostly to do with tides (differences in pull at different points) and Roche limits - that make the question mostly academic.

    Yes, a large object would contribute its gravitational pull to the equation. (Note that you're now talking about a system whose mass is Earth PLUS Mars, so it makes sense that the attraction is stronger).
     
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