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Homework Help: Gravitational potential of a sphere

  1. Jun 3, 2014 #1
    1. The problem statement, all variables and given/known data
    So in this video: https://www.youtube.com/watch?v=rm3x2X0X_Sc&t=210 Why does g.out and g.in have values as shown on the video? I can not for life of my understand it.

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Jun 3, 2014 #2


    Staff: Mentor

    Its a solid sphere so as you pass through the surface and approach the center, the gravitational effects are from the core of the sphere not the shell above you.

    In a hollow sphere, the effect of gravity is zero.

    You can think of a solid sphere as a collection of shells fitting inside each other.
  4. Jun 3, 2014 #3
    yeah but I still don't understand why it's [tex]\frac{GM}{x^{2}}[/tex] in x>R case and [tex]\frac{GMx}{R^{3}}[/tex] in the other one. I have no idea where the x^2 go and r^3 and x came

    Please just help me understand it, I'll never have physics again in my life after tommorows test.
  5. Jun 3, 2014 #4
    Okay nvm I get it, first one was just multiplied by a ratio of what is generating the gravitational effect for our current position, so [tex]\frac{x}{R}[/tex], am I right?
  6. Jun 3, 2014 #5


    User Avatar
    Gold Member

    If you consider the sphere as being composed of many smaller spheres of radius x, then each one has volume (4/3)πx3. The total mass of the solid sphere is M. If we suppose a uniform sphere, the mass of a sphere of radius x, M(x), is given by M x3/R3.

    Sub this into the expression g(x) = G M(x)/x2. In this way you can derive the field outside too using continuity. If you've ever done electrostatics, this is basically an analogous version of Gauss' Law for gravity.
    Last edited: Jun 3, 2014
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