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Newton's law of gravitation

  1. Apr 9, 2015 #1
    1. The problem statement, all variables and given/known
    according
    to my calculus III textbook, the gravitational force acting on an object at ##x = <x,y,z>## is ##F(x)=-\frac{mMG}{|x|^3} x ##. What's the point of having a cube on the bottom. Why shouldn't I memorize it as ##F(x)=-\frac{mMG}{|x|^2} ##

    2. Relevant equations
    See above

    3. The attempt at a solution
     
  2. jcsd
  3. Apr 9, 2015 #2
    IMG_5063.JPG
     
  4. Apr 9, 2015 #3

    gneill

    User Avatar

    Staff: Mentor

    x is a vector. ##G\frac{M m}{|\mathbf{x}|^2}## is the magnitude of the force, while ##\frac{\mathbf{x}}{|\mathbf{x}|}## is a unit vector in the direction of the force.
     
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