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Why does gravity cancel out for all points inside a sphere?

  1. May 17, 2006 #1

    I've heard that gravity cancels out for all points that are inside a sphere. How is this proven mathematically?

  2. jcsd
  3. May 17, 2006 #2
    yes, all you need is an intergral
  4. May 17, 2006 #3


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    whomever you heard that from is not to be trusted for accurate knowledge of physics.
  5. May 17, 2006 #4
    Perhaps your source was correct and you misheard:

    Gravity does cancel out for all points inside a hollow spherical shell. That is, assume there is no air inside a soccer ball, the gravity from the soccer ball on an ant floating anywhere inside the ball will sum to zero. This can be shown by adapting the integral form of Gauss's Law replacing the magnetic field with the gravitational field. If you are not yet familiar with vector calculus in order to use Gauss's Law directly, then you can simply think about it this way. In the exact middle of the hollow shell, there will be a gravitational pull from every point on the shell, but each point will have a different direction. In fact, for every point with a gravitational pull in one direction, there is a point on the exact opposite side of the ball with the same amount of pull but in the opposite direction.

    As you move from the middle to some other point inside the ball, you move closer to one edge of the ball than the other, and thus the forces from one side become stronger than the other. However, there is a competing change, as you move towards one side of the ball the amount of material behind you gets large, as you get closer to the material in front of you, these two competing changes will tend to cancel each other. What Gauss's Law beautifully proves, and what you may not guess intuitively, is that these two competing changes actually cancel each other exactly. Thus there is no net gravitational force anywhere inside the hollow spherical shell

    But for a solid sphere, like, for instance, the earth. Gravity certainly does not cancel out. What would this mean for tunnels? And given that the earth is not a perfect sphere what does this mean for valleys and other places of lower radius?

    Last edited: May 17, 2006
  6. May 17, 2006 #5


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    see for example

  7. May 17, 2006 #6

    Doc Al

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    Just to expand on this a bit. If you pretend the earth has a spherically symmetric distribution of mass, then at any point within the earth (say a distance r from the center) the gravitation field depends only upon the mass contained within the sphere of radius r. The mass at points > r can be considered a collection of spherical shells and, as you explained, they contribute nothing to the field at points inside them.
  8. May 17, 2006 #7


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    And, if you want to prove it mathematically using Newton's general law of gravitation and calculus techniques, here is most of it:
    1. In spherical polar coordinates, let the position of a mass particle inside the ball be given by [itex](\hat{r},\theta,\phi)[/itex] (measured from the C.M)
    where [itex]\phi[/itex] is the angle between the vertical and the particle's position vector.
    Let the density be constant for all sphere particles and the radius of the sphere R.

    2. Consider a test particle P having mass m and position vector [itex]r\vec{k}[/itex], i.e, a distance r along the "vertical"

    3. We need to sum up all forces acting on P from sphere particles, i.e, compute the integral:
    [tex]\vec{F}= -G \rho{m} \int_{0}^{R}\int_{0}^{\pi}\int_{0}^{2\pi} \frac{\hat{r}^{2} \sin \phi((r-\hat{r}\cos\phi) \vec{k} - \hat{r} (\sin \phi ( \cos \theta\vec{i} + \sin \theta \vec{j} ))} {( \hat{r}^{2} + r^{2} - 2r \hat{r} \cos \phi )^{ \frac{3}{2} } } d \theta {d \phi} d \hat{r} [/tex]
    where G is the universal gravitation constant and [itex]\rho[/itex] is the density of sphere particles.

    4. It is easy to see that the horizontal plane components vanishes; the [itex]\phi[/itex]-integration is then best handled by integration by parts.
    In the [itex]\hat{r}[/itex] integration, take care of whether you have [itex] r<\hat{r}[/itex] or [itex]r>\hat{r}[/itex]
    Last edited by a moderator: Jun 9, 2011
  9. May 17, 2006 #8
    I love this stuff.. So suppose the earth was a hollow sphere, there would still be gravity on the outside surface, but what if you dug a hole and jumped inside? you would have some velocity from gravitational acceleration while you were above the surface, so would you continue at a constant velocity inside the sphere? also, could air be pulled inside the sphere?
  10. May 17, 2006 #9


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    It would be VERY small though.

    To make a shell ~100miles thick, you'd need to remove 90% of the Earth's mass. g would drop proportionally - it would be ~ 0.98m/s^2.

    Now that we see gravity would be very small, we can see that our fall through the hole would be quite slow, as would the air.

    I'd be interesteed in edge effects though. If your acceleration were 0.98m/s^2, how fast would you be going after a 100 mile fall? And once you exited, yes, I guess you'd continue at that speed.
    Last edited: May 17, 2006
  11. May 17, 2006 #10

    Tom Mattson

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    Just to avoid confusion: It's not "BIG G" (Newton's universal gravitation constant) but rather "little g" (the acceleration due to the Earth's gravity) that changes.
  12. May 17, 2006 #11


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    you're absolutely correct. i was thinking of gravitational sheilding like there is for E&M. since there is no negative gravitational charges, there is no gravitational sheilding.

    but, being an inverse-square field, Gauss's Law applies and inside any hollow sphere, there is no net graviational field (unless there is some other big object nearby, the hollow shell does not sheild the field from that.
  13. May 18, 2006 #12
    How does the calculation of force within a charged sphere differ from that within a massive sphere? Does not the gravitational case only cancel for geometrical symmetries?
  14. May 19, 2006 #13

    Doc Al

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    Since both are governed by inverse square forces, the calculations are essentially identical. The electric field within a spherically symmetric shell of charge and the gravitational field within a spherically symmetric shell of mass are both zero.
    Yes, the shell theorem applies to spherically symmetric distributions of mass (or charge).
  15. May 19, 2006 #14
    How does the calculation differ between the two cases for a point outside the sphere?
  16. May 19, 2006 #15

    Doc Al

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    Again, the calculations are essentially identical. The gravitational field outside a spherically symmetric mass distribution is given by GM/r^2 (where M is the total mass of the sphere); similarly, the electric field outside a spherically symmetric charge distribution is given by kQ/r^2 (where Q is the total charge of the sphere).
  17. May 20, 2006 #16
    What if gravity obeyed an inverse-cubed law, would the force inside the sphere still net to zero?
  18. May 21, 2006 #17
    I thik you mean inverse-square law. I don't think there's no such thing as an inverse cubed law.
  19. May 21, 2006 #18


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    He was asking a hypothetical.

    The point is that the gravitational force is always radially outward from the center of the sphere. No matter how rapidly it dies off, it still has the same value at any given point at the surface of a spherical shell of some arbitrary radius r.
  20. May 21, 2006 #19
    I doubt it. I don't know enough 3-d math to prove it but just doing a mental expiriment... If gravity obeyed, say, an inverse r^10 law, it seems to me that being very near the edge of the sphere would create a (relatively) huge gravitational force toward that part of the sphere, while the points on the sphere surface that are really far away would have that much less of an effect.
  21. May 21, 2006 #20
    Why were 2 of my comments deleted?

    I made 3 posts around 10:45.
    ONLY one is showing.
    None were offensive or redundant.

    Since when do moderators sensor legitimate comments on PF?

    What the hell is going on here?

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