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Gravitational force inside the earth

  1. Oct 15, 2012 #1
    I read the proof that for a thin spherical shell with mass M, the potential energy U for a point mass m inside the shell is constant (depending only on the radius R of the shell) and thus the gravitational force on m is 0 at any point inside the shell.

    Then they talk about a point mass inside the earth: they say that thanks to what we discovered about the energy and force on m inside a shell - we can conclude that the gravitational force on m inside the earth (A uniform sphere) is only due to the mass that's "locked-up" inside the sphere with radius r, that represents the distance between the center of the earth and m, and all the earth's mass "outside" of r doens't contribute to the gravitational force at all. Also - the earth's mass that's "inside" exerts gravitational force as though it was all concetrated at the center of the earth.

    I don't understand - what's the difference between a shell and the earth? why in one we have 0 force and on the other it's not 0? and how did they conclude all of the above (mass "outside" m exerting no force, mass "inside" m exerting force as if it was all in the center of the earth) from the spherical shell proof?
     
  2. jcsd
  3. Oct 15, 2012 #2
    At any point inside the Earth, you can regard the Earth as the sum of the spherical shell "above" the point and the spherical ball "below" the point. The force due to the shell is zero as proved. So the entire force is only due to the ball. Is this clear so far?
     
  4. Oct 15, 2012 #3
    No, how can you treat the "above" as a shell if you have a ball inside the shell itself?
     
  5. Oct 15, 2012 #4
    This is the principle of superposition. The total force of gravity if the sum of individual point forces. They are independent of each other. So you can consider the points inside the shell independently of the points inside the ball.
     
  6. Oct 15, 2012 #5
    but doesn't the calculation in the proof for the shell relies on the fact that the mass is distributed in a certain way in the shell? doesn't this assumption change when you have a ball inside the shell?
     
  7. Oct 15, 2012 #6
    The proof relies on how the mass is distributed INSIDE the shell. It makes no assumptions on the distribution OUTSIDE the shell. Simply put, the force due to a point in the shell gets cancelled by the forces due to the other points in the shell, REGARDLESS of any masses present elsewhere.
     
  8. Oct 15, 2012 #7
    Ok so what you're saying is basiclly this (I hpoe:)): Outside the radius r (the distance between center of earth to body) you can treat all of this mass as a collection of concentric spherical shells that strech all the way to the surface of the earth. These shells exert no net gravitational force on the body.
    And for The remaining mass that makes up the solid sphere with radius r?
     
  9. Oct 15, 2012 #8
    The rest is the ball. The ball can also be considered as a collection of concentric spherical shells. What is the force due to a shell outside it?
     
  10. Oct 15, 2012 #9
    F=[itex]\frac{GM(mass of shell)*m(mass of body)}{r^2}[/itex]
     
  11. Oct 15, 2012 #10
    So the force is the same as that of a point with the mass of the shell, the point situated in the centre of the shell. What then can be said of the force due to a ball?
     
  12. Oct 15, 2012 #11
    That I do not know :) I guess that a ball still behaves in this way since it's also a collection of concentric shells, so the force it exerts on the body at r is as the same if all the ball's mass was concetrated at its center... but what's the ball's mass?
     
  13. Oct 15, 2012 #12
    Correct.

    You need to model its density as a function of radius. For the Earth, it is quite complex. The only general statement that can be made is that the gravity increases up to a certain radius, then decreases down to zero in the centre.
     
  14. Oct 15, 2012 #13
    ok thanks!
     
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