# Gravitational field intensity inside a hollow sphere

It is quite easy to derive the gravitational field intensity at a point within a hollow sphere. However, the result is quite surprising. The field intensity at any point within a hollow sphere is zero.

What exactly is the reason behind this? Except for, of course, the mathematics behind it. Is there any logic why the field intensity should be zero within a sphere? For example, it is logical to say that the field intensity would be zero at the center, as all the intensities cancel out. However, this cannot be the case for any point within the sphere.

• See Newton's shell theorem. Related: physics.stackexchange.com/q/2481/2451 , physics.stackexchange.com/q/69812/2451 and links therein. Relativistic version: physics.stackexchange.com/q/43626/2451 Dec 3, 2014 at 16:21
• @Qmechanic a bit difficult to understand... Dec 3, 2014 at 16:23
• I like Qmechanic's answer, but my own intuition is this: suppose the shell consists of a lot of small "moons" covering the shell. You could be close to one and feeling its gravity, and the ones "across" the way have gravity that falls off at $d^2$, but the number of them increases with $d^2$, so they are just as powerful as the one you are next to. Dec 3, 2014 at 17:50

One intuitive way I've seen to think about the math is that if you are at any position inside the hollow spherical shell, you can imagine two cones whose tips are at your position, and which both lie along the same axis, widening in opposite direction. Imagine, too, that they both subtend the same solid angle, but the solid angle is chosen to be infinitesimal. Then you can consider the little chunks of matter where each cone intersects the shell, as in the diagram on this page: You still need to do a bit of geometric math, but you can show that the area of each red bit is proportional to the square of the distance from you (the blue point) to it--and hence the mass of each bit is also proportional to the square of the distance, since we assume the shell has uniform density. But gravity obeys an inverse-square law, so each of those two bits should exert the same gravitational pull on you, but in opposite directions, meaning the two bits exert zero net force on you. And you can vary the axis along which the two cones are drawn so that every point on the surface of the shell ends up being part of a pair like this, which leads to the conclusion that the entire spherical shell exerts zero net force on you.

• This "intuitive proof" is due to Newton.
– rob
Mar 19, 2017 at 5:34
• Note that this proof only works because the angles between the "line of sight" and the surface normal for each of the patches are the same; so both patches are "foreshortened" by the same factor of $\cos \theta$. Otherwise you could make this argument for the interior of any closed surface with a uniform charge density. Jul 9, 2018 at 18:48
• Gravity is stronger the closer you are to the mass, so the above example doesn't make sense. No matter where you are in the sphere (with the exception of the epicenter) you're closer to a wall point, and that mass would pull on you to a greater extent than the rest (assuming a perfectly symmetrical surface). Sep 2, 2022 at 16:29
• Never mind. I just saw another post indicating that although gravity has a stronger pull to the closer location, the combined mass and their calculated distances exactly cancels it out leaving a net 0 gravitational pull from the sphere itself. Trippy. Sep 2, 2022 at 16:38
• @mkinson - Yes, if you choose uniform solid angles on both sides, then the chunk that's further away is also larger in size, which cancels out its greater distance to produce the same net gravitational pull as the chunk that's closer but smaller in size. That's what I was talking about when I said "the area of each red bit is proportional to the square of the distance from you (the blue point) to it--and hence the mass of each bit is also proportional to the square of the distance, since we assume the shell has uniform density." Sep 2, 2022 at 16:42

A part of the problem is first asking the correct question. If you reference a point inside a spherical torus (the perimeter is comprised of mass), the effects of gravity upon a mass (point) located somewhere inside the sphere will vary. So long as you deal only with a point, that will be the case.

However, if you deal with a solid sphere, instead of a 'point' inside the sphere, you have to deal with a concentric region inside the sphere. No matter how you define that concentric region, so long as the center resides at the main sphere's center, the net pull of gravity towards the outer perimeter remains a net force of zero.

• That makes sense considering at any point if you cut the sphere laterally from where you're falling there will be a greater mass "below" you than "above", and you will therefore be pulled down until you hit the epicenter, where no matter how you slice the sphere all directions will contain equal mass. Sep 2, 2022 at 16:31