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Equation of Gravity's influence relative to objects

  1. Apr 17, 2013 #1
    Hello, I'm trying to figure out the following formula/question. If I have an object and I wanted to reduce the effective outside gravity to as minimal as possible how would I calculate the results. I would like to note I have just lately become sparked with the passion of higher math and science and have not taken any of the normal classes that would normally help me know how to procced accordingly.

    I have found the following formual in two forms (not sure which is correct or if one had typo).

    At a point between the earth and moon, the magnitude of the earth's gravitational acceleration equals the magnitude of the moons gravitational acceleration. What is the distance of that point from the center of the earth? The distance from the center of the earth to the center of the moon is 383000km and the radius of the earth is 6370km. The radius of the moon is 1738km and the acceleration of gravity at it's surface is 1.62m/s^2.

    F=ma F=mMG/R^2(Tau) => a =GM/R^2(Tau)

    GMsub1/R^2 = GMsub2/(3.383E^8-R^2)
    GMsub1/R2 = GMsub2/(3.383*108- R)^2 **(I think the power outside was a typo because it wouldn't balance right otherwise)

    I know this is finding the "zero" point or equal between the moon and Earth's center of mass pull on something the same. I want to know if I added and 3 point to one side:

    GMsub1/R^2 = (GMsub2/(3.383E^8-R^2))+(GMsub3/(3.383E^8-R^2))

    Would this then tell you where the zero point is between the first object and the other two or could someone point me in the right direction for text on these type of questions?

    Thank You for any help,
    A Curious Monkey's cousin
  2. jcsd
  3. Apr 17, 2013 #2


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    Welcome to PF!

    Hello Galric! Welcome to PF! :smile:

    (try using the X2 and X2 buttons just above the Reply box :wink:)
    No, you're adding forces, and forces are vectors, so you must add them like vectors. :wink:

    (if that means nothing to you, say so, and i'll explain)
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