Is the inverse square law exact near a spherical body?

In summary: Doing so will lead to the inverse square law being exactly correct for Newtonian gravity outside of a spherically symmetric object.
  • #1
fizzy
193
17
I'm forking this off another thread where I brought it up but it was getting OT.

"Inverse square law is exact everywhere outside of spherically symmetric objects"
It is good enough for a first approximation but it is certainly not exact.

Consider a test mass one radius from a spherical body. Work out the contributions form two points diametrically opposed on the surface, once is distant by r , the other by 3r. Compare that to the same masses placed a the centre of mass.

1/r^2+1/(3r)^2 = 2/(2r)^2 ??

A similar inequality and of the same sign will apply for all pairs of points symmetrically placed either side of the plane through the c.o.m. and perpendicular to the line joining the test mass and the centre. Summing all such pairs to represent the whole shows the inequality holds for the whole body.

The inaccuracy reduces as the separation becomes much larger than the diameter of the sphere.

mfb replied:
It is exact. You can't limit the calculation to two points, you have to integrate over the whole shell. Do it and see what you get.
Alternatively: It is a direct consequence of Gauss' law.

This is a well-known result from classical mechanics and not the topic of this thread.

Well I was not considering just two points. I extended the two point conceptually to integrate over the whole body. There may be a flaw in my logic, so maybe someone can point it out.

Because of the non linear nature , the nearer point will have attraction greater than the centre of mass proxy point and that difference will greater than deficit of the more distant symmetrically placed sibling point.

For all such pairs of points there is an inequality of the same sign. There will be exactly the same number of such points in either hemisphere so whichever way we integrate we end up with an inequality w.r.t the point mass at the centre .
 
Last edited:
Physics news on Phys.org
  • #2
Google "shell theorem". Newton himself proved you wrong. The maths is quite messy if you aren't familiar with Gauss' Theorem; it's a one-liner if you are.
 
  • #3
Thanks Ibix, can't see what was wrong with my logic but doing the integral seems clear.
 
  • #4
You haven't laid it out completely, but you appear to be summing ##F=GMm/r^2## for each elementary point on the sphere. This is wrong - force is a vector, so you need to sum ##\vec F=(GMm/r^2)\vec{\hat r}##. Assuming your test mass is on the z axis, the x and y components will clearly cancel by symmetry, so you should only be summing the z components, which will, in general, have a trigonometric factor you appear not to have accounted for. Since that's always less than one your inequality won't hold all around the sphere.
 
  • Like
Likes Dale
  • #5
Original discussion

Let's put our observer above the north pole to establish a coordinate system. The points at the north pole and south pole lead to a larger attraction than an equal mass at the center, but two opposite points at the equator will lead to a smaller attraction: They have a larger separation and the smaller absolute forces don't point in the same direction, so we lose something there as well.
If you integrate over the whole sphere you get a force exactly equal to a point mass in the center.
fizzy said:
It is good enough for a first approximation but it is certainly not exact.
Don't make statements like that if you are not really certain.
 
  • Like
Likes Ibix
  • #6
yes, effectively there will be a difference in angle between the back of front elements of each pair that I had not accounted for. That could be it.
 
  • #7
fizzy said:
Summing all such pairs to represent the whole shows the inequality holds for the whole body.
Instead of summing you need to integrate. For a body of finite density the mass at a point is 0, and thus the error that you describe sums to 0. So instead you need to consider differential volumes ##dV## with mass ##dm=\rho dV##. When you do so you will find that the inverse square law is indeed exact outside of a spherically symmetric object for Newtonian gravity.

fizzy said:
I extended the two point conceptually to integrate over the whole body. There may be a flaw in my logic, so maybe someone can point it out.
You don’t integrate points, you integrate differential volumes.
 

1. Is the inverse square law always accurate near a spherical body?

Yes, the inverse square law is always accurate near a spherical body. This law states that the intensity of a force, such as gravity, is inversely proportional to the square of the distance between two objects. As long as the distance between the objects is small compared to the radius of the spherical body, the inverse square law holds true.

2. How does the inverse square law apply to objects of different sizes?

The inverse square law applies to objects of different sizes because it is based on the distance between the objects, not their size. As long as the distance between the objects is small compared to the radius of the spherical body, the law will hold true regardless of the size of the objects.

3. Does the inverse square law only apply to gravity?

No, the inverse square law applies to any force that follows an inverse square relationship, such as the force of electric or magnetic fields. This law is a fundamental principle in physics and is applicable to a wide range of phenomena.

4. How does the inverse square law affect the strength of a force?

The inverse square law states that as the distance between two objects increases, the strength of the force between them decreases by a factor of four. This means that the force becomes weaker as the distance between the objects increases.

5. Are there any exceptions to the inverse square law near a spherical body?

In most cases, the inverse square law holds true near a spherical body. However, there are some exceptions, such as when the spherical body is rotating or when there are other external forces acting on the objects. In these cases, the inverse square law may not accurately describe the relationship between the objects.

Similar threads

  • Special and General Relativity
Replies
24
Views
1K
  • Special and General Relativity
Replies
32
Views
1K
  • Mechanics
Replies
5
Views
1K
Replies
13
Views
897
  • Classical Physics
Replies
1
Views
987
  • Introductory Physics Homework Help
Replies
8
Views
877
  • Astronomy and Astrophysics
Replies
3
Views
2K
  • Classical Physics
2
Replies
48
Views
2K
Replies
5
Views
6K
  • Introductory Physics Homework Help
Replies
10
Views
2K
Back
Top