I am trying to understand the proof of why the force acting on a spherical shell and a particle is $$\frac{GMm}{r^2}$$ Where M is the mass of the sphere and m is the mass of the particle.

I am looking at Wikipedia in the section of "Outside a shell". The method there is to "cut" the sphere into small rings and then calculate the force of gravity applied from one ring to the particle. Then they calculate the sum of infinitely many of those rings when their width is infinitesimal using an integral: $$F_r=Gm\int\frac{\cos \phi \ dM}{s^2}$$

That part confused me. Isn't that integral should be a definite integral instead of an indefinite one? How can an indefinite integral represent a sum? and if it was a definite integral what will be its limits?

Please help me understand that part.
I am new to physics, I only know some simple mechanics so please keep that in mind.

  • $\begingroup$ It's not an indefinite integral, but often, physicists suppress the bounds of integration until the very end of the calculation, when the range of the integration variable becomes obvious. The bounds of $dM$ are usually quite difficult to see though, I'm not sure whether you'll get an answer for that. $\endgroup$ Nov 24, 2020 at 15:59
  • $\begingroup$ hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/… $\endgroup$
    – BowlOfRed
    Nov 24, 2020 at 16:40
  • $\begingroup$ @NiharKarve That's not a comment; it's an answer. $\endgroup$
    – Bill N
    Nov 24, 2020 at 18:37
  • $\begingroup$ as the limits were a bit obvious they didn't mention the limits. $\endgroup$
    – Anonymous
    Nov 25, 2020 at 2:50
  • $\begingroup$ @PranavAggarwal no, it is the exact opposite reason $\endgroup$ Nov 25, 2020 at 12:35

3 Answers 3


The theorem you should use is Gauss law for gravity. https://en.wikipedia.org/wiki/Gauss%27s_law_for_gravity

You need to know some calculus. Namelly the divergence theorem may be useful to convert Gauss Law from differential form to integral form:


To obtain the result you need to do the following:

  1. Calculate the gravitational field outside the spherical shell at distance $r$. For this you use the Gauss law in integral form. For simplicity, you may assume spherical symmetry. The magnitude of the result will be $GM/r^2$ and field is radial.

Gauss Law is

$$\int_{\partial V} \mathbf{g} \cdot d \mathbf{S} = - 4 \pi G M$$

where $\partial V$ is a closed surface in space enclosing mass $M$. If you assume radial symmetry, $\mathbf{g} = - g \mathbf{\hat r}$, where $\mathbf{\hat r}$ is radial unit vector, then the first term in the formula above is simply $- 4 \pi r^2 g$ and the equation reduces to

$$4 \pi r^2 g = 4 \pi G M$$

removing the $4 \pi$ in both sides we find

$$g = \frac{GM}{r^2}$$

  1. The force that mass $M$ makes on mass $m$ is then calculated simply multiplying the field in the position of the particle by the mass of the particle, that is,

$$F = m g = GMm/r^2$$

If there is no spherical symmetry, which happens if mass $M$ is not a sphere, the result still holds but the integral above cannot be performed so easily. It is a good exercise anyway.

  • 1
    $\begingroup$ The thing that confused me about the proof in Wikipedia was that they didn't write the limits of a definite integral and I thought it was an indefinite integral, other than that I understood the rest of the proof. So using newton's law of gravitation wasn't so bad it's just the notation that got me confused. Maby Gaus's law of gravity can make the proof more simple however the math is too advanced for me. $\endgroup$
    – Freud
    Nov 24, 2020 at 17:53
  • 1
    $\begingroup$ Well, it is perhaps the divergence theorem that you will learn in math that you need to understand the proof. But if you want, for now, you could think of the integral in the following simple manner. From symmetry, if you consider a spherical closed surface with the mass $M$ inside and in its center, then the field is constant in the surface and the integral is simply the product of the surface area $4 \pi r^2$ and the unknown field $g$. Then use the equation to get $g$. Hope it helps. $\endgroup$
    – Blue
    Nov 24, 2020 at 18:07

Often while calculating integrals, the bounds of integration are suppressed until the very end, when the integration variable is reparameterised in such a way that the range of new integration variable is apparent.

The integral over $dM$ is certainly a definite integral, since we integrate over a physical mass shell. But since $dM$ is like a "catch-all" term for mass elements of all shapes and sizes, it is usually very difficult to gauge what the bounds should be - until you reparameterise in terms of $s$, in this case, and it is physically apparent that $s$ ranges from $r-R$ to $r+R$.

The reason we can do this is that often, especially in classical mechanics, the integrals we perform are physically manifest, and not just some arbitrary formal manipulation of symbols (otherwise we'd probably have to keep track of the bounds at each stage)

  • $\begingroup$ sorry i misinterpreted the question $\endgroup$
    – Anonymous
    Nov 25, 2020 at 14:06
  • $\begingroup$ not exctly misinterpreted, actually i was thinking about the next step to the asked question $\endgroup$
    – Anonymous
    Nov 25, 2020 at 14:07

If you read further down in the derivation you find that cos(ф) and dM are defined in terms of, s. Then the limits for integration with, s, as the independent variable are given as: (r-R) to (r+R). (In physics, all integrals have limits.)

  • $\begingroup$ What do you mean by "In physics all integrals have limits"? velocity is the indefinite integral of acceleration and it does not have any limits. $\endgroup$
    – Freud
    Nov 24, 2020 at 20:55
  • $\begingroup$ @Freud What do you mean? You can't determine the velocity from the acceleration without supplying limits, i.e., the initial & final times. $\endgroup$
    – PM 2Ring
    Nov 25, 2020 at 7:48
  • 1
    $\begingroup$ I meant that if you have the acceleration as a known function of time than you can find the velocity as a function of time using integration. But now that I think about it I still left with a constant of integration. so I can see your point @PM 2Ring $\endgroup$
    – Freud
    Nov 25, 2020 at 8:18

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