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The gravitational acceleration g

  1. Sep 5, 2013 #1
    Suppose a particle of mass M is under gravitational attraction. The Newton's law of gravitation says that F=GMm/r^2, and the part Gm/r^2 is g (acceleration due to gravity how?)
  2. jcsd
  3. Sep 5, 2013 #2


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    Why don't you look up the mass of the earth, the radius of the earth, and G and plug them into Gm/r^2 and see what you get? Is it close to the usual value of g?
  4. Sep 5, 2013 #3


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    M = mass of planet
    m = mass of object/particle

    but Newton also says..

    F = mg

    You can do the rest.
  5. Sep 5, 2013 #4


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    At the surface of the earth, Gm/r^2= g using m= mass of the earth, r= radius of the earth.
  6. Sep 6, 2013 #5
    At only one point is GM/r^2 exactly equal to 9.8. However, since r only changes slightly with respect to its value at heights we experience, for all intents and purposes, g=9.8m/s^2.
  7. Sep 6, 2013 #6


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    Like noted above, you get it by equating Newton's gravity law with Newton's second law.
    You can plug in some numbers here: Earth's Gravity.
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