# Intuitive explanation of the gravity of a sphere

1. Mar 26, 2004

### jacobfreeze

Does anyone have an intuitive explanation why the gravitational force of a uniform sphere on a distant point is the same as if all the sphere's mass were concentrated at the center? The usual integration over spherical shells goes through o.k. but the result is so simple it seems to me that maybe we could get there by insight instead of calculation...or at least part of the way there...
Do we have the same result for a disk or a ring and a point in the same plane? Is the gravitational force of a ring on a point in the same plane equivalent to the force from the same mass concentrated at the center of the ring?
I think this is true...so is there an intuitive way to get this result in the simpler case of a ring and a co-planar point, and does the spherical case then follow?

2. Mar 29, 2004

### tyroman

Here's a try...

Imagine a plane perpendicular to your line of sight and passing through the sphere's center cutting the sphere into two equal hemispheres. It is "intuitive" that the G force from the near hemisphere equals the pull from the far hemisphere (assuming the distance from you to the sphere is very great when compared to the sphere's diameter). With this intuited observation, you can conclude that the net G force of the two hemispheres lies somewhere in the disc which describes the surface where the two hemispheres meet.

Next consider this disc and draw an imaginary line from top to bottom through the center of the disc (also would be the center of the sphere). Apply the same logic as above in consideration of the G force from the right and left semidiscs and it becomes intuitive that their net G force lies somewhere on the vertical line dividing them (whose length is equal to the original sphere's diameter).

Finally consider this line, find its center (also would be the center of the sphere) and again applying the logic above, conclude that its net G force lies at its (and the sphere's) center... qed

Similar analysis could be applied to the other shapes you mention.

3. Mar 29, 2004

### jacobfreeze

Of course it's obvious by symmetry if you neglect the difference between the near and far sides of the sphere. But for a point near the sphere, where this difference isn't negligible, why does it work out so neatly that the net effect of gravity is the same as if all the mass were concentrated at the center? Is watching the result fall out of a tricky substitution in a multiple integral the best we can do?

4. Mar 29, 2004

### Mike2

Better yet, hollow out the exact center and then what is the gravitational field at the exact center of a sphere?

5. Mar 29, 2004

### jacobfreeze

The net force at the center, hollow or solid, is zero...equal vectors in all directions just cancel out. But what about my original question? I don't think anyone really knows why the gravitational force on a point outside a uniform sphere is the same as if all the sphere's mass were concentrated at the center. It's a very basic example of the kind of physics where we can calculate without understanding much of anything.

6. Mar 30, 2004

### tyroman

In the limit

to jacobfreeze;

My "try" at an "intuitive explanation why the gravitational force of a uniform sphere on a distant point is the same as if all the sphere's mass were concentrated at the center" hinges on the word "distant" in your original question.

I assumed that you realized that application of Newton's Law:

F=G(m1m2)/r2

to an element in the near hemisphere and its complementary element in the far hemisphere in a case where the difference between r for the near element and r for the far element is large when compared to ravg then F for the near element and F for the far element are not equal and their resultant would not be equivalent to a point midway between the elements.

My analysis relies on "(assuming the distance from you to the sphere is very great when compared to the sphere's diameter)" and is valid only in the limit.

7. Apr 1, 2004

### 2Pac

to jacobfreeze;

This occurs because the actual force of gravity always points towards the center (if the mass is of uniform density). If variations of density occur in the mass the center can not be defined as the center of gravity.

8. Apr 2, 2004

### jacobfreeze

No, no, no!!! Warning: This is a difficult question. It puzzled Newton for years. If you search Google with "gravitational force of a sphere" you'll find a link to mathworld where the relevant integral over spherical shells is worked out. If all you want from physics is the right answer falling out of a tricky calculation, there you have it. But if you're looking for something that feels more like understanding, I don't know where you should look, and that's what I'm asking.

9. Sep 4, 2009

### AxelBoldt

The fact that the external gravitational field of a sphere is the same as the gravitational field of a point source located at the sphere's center is a consequence of the Gauss Law.

That law states that the total flux of any gravitational (or electric) force field through any closed surface is proportional to the total matter (or charge) enclosed by the surface. This law is "intuitive" if you think of the matter (or charge) as the origin of field lines, and the total flux as counting the number of field lines passing through the closed surface. Any enclosing surface will capture "all" of the field lines, so will have the same total flux.

Now pick a large sphere S with the same center as our given massive sphere and apply the Gauss Law to it. The total flux through S only depends on the contained mass, and so it will be the same for our massive sphere and a point source of the same mass. The gravitational field of sphere and point source are both clearly radially symmetric. So if the two situations yield the same total flux through S, and both situations are radially symmetric (as is S), then the *local* flux at each spot of S must be the same for the two situations. This local flux is basically the gravitational field at that spot.

You can also use the Gauss Law to show the other half of Newton's sphere theorem: there's no gravitational field inside a massive shell. This time pick a small sphere S contained within our shell and centered at the center of the shell, and apply the Gauss Law to S. S does not contain any mass, so the total flux through S is zero. Radial symmetry then requires that the local flux at each spot of S is also zero.

Last edited: Sep 4, 2009