What would be the gravitational field due to a hollow sphere (mass: $M$, radius: $r$) on its surface?

Consider an elemental mass of the surface. Let field due to $dm$ be $E_1$ and due to $M - dm$ be $E_2$.

At a point just inside the sphere, $E_1 = E_2$, as field inside has to be zero(the directions will clearly be opposite by symmetry).

At a point just outside the sphere, $E_1 + E_2 = GM/r^2$ which implies $E_1 = E_2 = GM/2*r^2$.

Now net field on dm will be equal to $E_2$ as field on self is zero.

Therefore net field on the surface would be $GM/2*r^2$. But the standard integration method yields $GM/r^2$, which I believe is wrong as we add the field of $dm$ on itself. So what will be the field?

  • 1
    $\begingroup$ You're in luck because there is literally a theorem called the shell theorem dealing with that very topic $\endgroup$
    – Slereah
    Dec 5, 2017 at 12:34
  • 2
    $\begingroup$ en.wikipedia.org/wiki/Shell_theorem I already looked at it, it talks about field inside and outside the sphere. I'm concerned with the field on the sphere. And what's wrong with my logic. $\endgroup$
    – ymuf
    Dec 5, 2017 at 12:38
  • $\begingroup$ The standard method is correct and you are wrong. Field due to the elementary mass is $Gdm/r_{1}^{2}$. You go wrong where you assume that the large mass and elementary mass would produce an equally large field. In fact you are wrong that the sum of the fields would be $GM/r²$ it is just the field of the mass M. Unless you are on the exact location of dm then its field is zero and the total field is right as you said. $\endgroup$
    – Communisty
    Dec 5, 2017 at 12:50
  • $\begingroup$ Your approach is correct! What is the "standard integration method" you are referring to? $\endgroup$
    – lesnik
    Dec 5, 2017 at 12:52
  • $\begingroup$ @lesnik check link for shell theorem $\endgroup$
    – ymuf
    Dec 5, 2017 at 13:13

2 Answers 2


The above-mentioned shell theorem explains how to calculate the gravitational field inside and outside the shell.

It's not really correct to ask what is the field exactly on the surface of the shell. Because the function $g(r)$ is not continuous at $r = R$.

But it is correct to ask what gravitational force is acting on each part of the shell from the rest of it. If the force acting on a small part of the shell is $g * dm$, then it's natural to say that the gravitational field on the surface is $g$.

So, you take a small part of the shell $dm$ and want to find the gravitational force acting on this part of the shell. Your approach is correct, the force is $ G M dm / 2 R^2$.

I can suggest another approach that gives same result.

Let's calculate the potential energy of the shell. Let's take a very small part of it and pull it out to infinity. We have spend some energy to do it: $G M dm/R$. Then we take another part of the shell, and so on. But we will take the small parts from different sides of the shell, so that the remaining mass still forms a spherical shell. It would be just thinner and thinner until is dissolves to nothing.

The energy we have to spend depends on the remaining mass. $dA(m) = G m dm /R$. This is a linear function and is easy to integrate. Total energy we spent would be $A = G M^2/2R$. And the total gravitation energy of the shell is $W=-GM^2/2R$.

Now let's inflate the sphere by $x$. It's energy would increase by: $$x * dW/dR = x * G M^2/2R^2$$

We can also calculate the energy we spent pulling each piece of the shell up by $x$. The force acting on each small piece $dm$ is proportional to $dm$: $f = g*dm$. So, the total work is $$\sum_{dm} x * g * dm = x*g*M$$

Now compare the work done and increase of energy: $$x * GM^2 / 2R^2 = x * g *M$$ $$g = GM/2R^2$$


Following on the above, if we let M be a constant, it seems we can then integrate GMdm/2R^2 over dm (from 0 to M) to obtain a sum or total of gravitational forces on the surface of the sphere, obtaining GM^2/2R^2.


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