If a person were to stand on a flat disc of thickness 1 meter but infinite diameter, would they experience finite or infinite downward force?

There is an infinite amount of mass, all of which attracts the person but most of which is also infinitely far away. Do the two "cancel out" or does the infinite mass win?

For a precise definition of "infinite diameter", take the limit of a circular disc with diameter $x$ as $x$ approaches infinity, with the person standing on top of the disc's center of gravity.

  • $\begingroup$ The answers all look at the vertical component of the gravitational force. It might be worth looking at the horizontal part too, to see if the solution is stable or if the vanishing horizontal component is an artifact of taking the limit of the force at a point over the exact center. $\endgroup$
    – user5174
    Jun 24, 2016 at 7:00
  • $\begingroup$ @Hurkyl: See here: physics.stackexchange.com/questions/167753/… . It looks like it's not actually convergent, but the divergence is only logarithmic, which I guess means the principal value is physically meaningful in a wide range of situations anyway... $\endgroup$
    – Micah
    Jun 24, 2016 at 15:05
  • $\begingroup$ @user5174 There is no horizontal part of the force for an infinite disk, because wherever you are, you are always effectively at the centre. $\endgroup$
    – KDP
    Mar 11 at 10:22

3 Answers 3


If the disk has infinite diameter it is nothing but an infinite plane. For any finite thickness we can consider a layer of mass whose superficial density is $\sigma$. Moreover, if the plane is infinite it does not matter if you are one meter or one kilometer away from the plane. Wherever you look at the plane you will see the same structure. So the gravitational field cannot depend on the distance from the plane. It must be uniform and their lines must be perpendicular to the plane.

Applying the Gauss Law for a cylindrical Gaussian surface whose symmetry axis is perpendicular to the plane you get $$\oint \vec g\cdot \mathrm d\vec A=2Ag=-4\pi G m,$$ where $A$ is the basis/top area of the cylinder and $m$ is the mass inside it. Therefore the gravitational field reads $$g=-\frac{2\pi Gm}{A}=-2\pi G\sigma,$$ and this is a finite quantity.

Edit: Just plugging in some numbers to see what we get. The gravitational constant is $G=6.67\times 10^{-11}\, ~\mathrm{Nm^2/kg}.$ If the layer of mass is $d$ meters thick and made of a material with the same mean mass density of the Earth ($\rho=5.5\cdot 10^3\, ~\mathrm{kg/m^3}$) it will give $$\sigma=\frac{5.5\cdot 10^3\cdot d\cdot A}{A}=5.5\cdot 10^3\cdot d\,~\mathrm{kg/m^2}.$$ Therefore, the magnitude of the gravity calculated above is $$g=2.3\cdot 10^{-6}\cdot d\, ~\mathrm{m/s^2}.$$ In order to give $9.8\, ~\mathrm{m/s^2}$ the disk would have to be $4.3\cdot 10^3$ kilometers thick. Have in mind that the diameter of the Earth is about $12\times 10^3$ kilometers.

  • 1
    $\begingroup$ @Winther You are right. I meant the thickness is negligible regarding the finite or infinite value of the force but I shall make this point clearer. Thanks. $\endgroup$
    – Diracology
    Jun 23, 2016 at 20:11
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    $\begingroup$ Maybe because I'm not a physicist... what units is this in, what is $G$ in this case, and what value of $\sigma$ would be Earth-like (i.e. a landscape made out of Earth-like materials would cause what resulting gravity)? $\endgroup$
    – corsiKa
    Jun 23, 2016 at 21:01
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    $\begingroup$ @philip_0008 - you can't do that because that result is only valid when the mass is spherically distributed (and you are outside the sphere). For this case, you really need to do the integral - and since most of the mass is far away to the side, it doesn't provide much downward force... $\endgroup$
    – Floris
    Jun 24, 2016 at 3:53
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    $\begingroup$ The argument that 'regardless of the distance you will see the same structure, so therefore the field must be uniform' is (as posed) flawed, because it would also apply to an infinitely long, thin cylinder, which does not have a uniform field. It's roughly right, but it needs rather more finesse. $\endgroup$ Jun 24, 2016 at 6:03
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    $\begingroup$ The argument is flawed because you don't see the same structure: the greater your distance from the plane, the greater your distance from the plane. $\endgroup$
    – user5174
    Jun 24, 2016 at 6:35

You can do the integral, and will discover that the answer is "finite" - because not only does the distance to mass increase, but so does the angle.

Consider an annulus at radial distance $r$: if you have mass per area $\sigma$, the total mass at that distance is $2\pi r \sigma$; if the vertical distance to the center of the disk is $h$, the vertical component of force goes as $\frac{F\cdot h}{r}$.

I will leave you with this hint. See if you can write the integral from here. You will find that it depends on $h$ and $\sigma$ only. This is the same result (and the same analysis) as you would get to prove the electric field in front of a uniformly charged plane is finite (with very similar looking equations).


As the question is finite versus infinite, I guess we don't need the exact result for a finite disc (though it is not difficult to calculate).

The simple intuitive answer is that even though the disk mass is infinite, most of the forces from the bits of disk going out to infinity will cancel out due to symmetry, so the answer is finite.

Let us assume we are a height $h$ above the disk. Let us assume that $R$ is a number much larger than $h$. Let us now assume that we break the force into two pieces: piece 1 is the force from a very large disk with radius $R$, and piece 2 is the force from the rest of the disk that goes from R to infinity. The force from the very large disk is clearly finite as there is finite mass. The force from the rest of the disk is now simpler to calculate since $R$ is much larger than $h$. So the question is whether the force from the remaining mass results in a finite or an infinite force.

If we imagine an annulus (thin ring) with depth of 1 meter (like the disk), radius $r$ (larger than $R$) and thickness $dr$ (an infinitesimal) then we can calculate the gravitational force from this ring. As mentioned above, a very important effect is that there will be a lot of cancellation: far north bits of the annulus will mainly cancel out far south bits for example. The part that survives is the vertical component of the force only. So the vertical component of the force will introduce a factor of $h/r$ (when $r$ is much larger than $h$).

The vertical component of the gravitational force from the annulus then looks like (using the inverse square law and noting mass is density times volume):

$$dF = \mathrm{constant} \cdot r \cdot \frac{dr}{r^2}\cdot\frac{h}{r},$$


$$dF = \mathrm{constant}\cdot\frac{dr}{r^2}.$$

Integrating over all such rings out to infinity leads to a finite result as the integral from a finite number to infinity of $1/r^2$ is finite.