# The Death of the Centre of Mass Theorem.

1. Jun 17, 2005

### Rogue Physicist

Many people believe that the Centre of Mass Theorem is a powerful and useful tool in Newtonian Mechanics. In fact it is a farce.

(1) It is trivially true at distances in which the massive object is virtually a point-mass, such as between distant stars.

(2) It is completely incoherent and self-contradictory at distances in which the size of the object(s) approachs 1/20 the distance between them.

Try it yourself. I'll post the simple proof that it is nonsense after a few people try to guess what is wrong.

Last edited: Jun 17, 2005
2. Jun 17, 2005

### Tom Mattson

Staff Emeritus
Just get on with it already. This isn't sciforums, and we don't play those kinds of games here.

3. Jun 17, 2005

### panthera

prove it then...i use this many times and its one of the easiest method to solve complex problems

4. Jun 17, 2005

### mezarashi

First off, I would like to point out that there is nothing wrong with approximations. Newtonian Mechanics is an "estimate" now with the Relativistic framework.

Secondly, I don't see how it would compromise the theorem which states that the center mass of a system cannot change in the absence of an external force.

5. Jun 17, 2005

### Rogue Physicist

Consider a sphere. The center of mass is located at the Geometrical center. It is a fixed point. Let us place a test-particle a few diameters away from a solid sphere on the right along the x-axis. (make the radius of the sphere 1 unit, and put the sphere at the origin). According to the CMT, the mass acts as if it were concentrated at the centre and the distance will be measured centre to centre for purposes of using Newton's Gmm/d^2 formula.

(1) divide the sphere logically into two halves vertically. Each half will have its own centre of mass, located 3/8 radians down the axis of symmetry. The actual position isn't important, but because it is an average mean of the atoms in the solid, it is fixed relative to the geometric skin of the half-sphere. So the near half will be 3/8 rads toward the test-particle along the x-axis.

(2) Naturally, the CM for the other half will be the same distance away from the geometric centre of the sphere in the opposite direction away from the test-particle. The total force will be the vector addition of the two halves. But by inspection this is impossible. The increase in force for one half the mass now located closer cannot balance the decrease for the other half, because while the distances are equal, the forces have changed by an unequal amount. Gravity is an inverse exponential force.

(3) The actual force calculated by summing the halves separately will be larger than the one calculated treating the sphere as a whole. Conversely, the force calculated by dividing the sphere horizontally will be weaker than the 'whole' force, since vertical components from each half will cancel.

(4) The Centre of Mass Theorem contradicts itself, and also the Sphere Theorem as well, which is a special case of the CM.

6. Jun 17, 2005

### Rogue Physicist

This may be one side-effect resulting from the definition chosen for the Centre of Mass (i.e., the arithmetic mean average of position of particles weighted by the mass of each particle) but it is not really a formal part of the Theorem.

The main statement of the Theorem is that objects act under gravitational and intertial forces as if their mass was a point-mass located at the Centre of Mass so defined. Since the procedure for calculating the CM is not ambiguous but always produces a unique and fixed answer for any given configuration of particles (whose mass cannot change) in 3-space, Naturally a corollary of the definition or rather a direct effect of it is that the CM cannot change unless the mass distribution itself changes. The corollary to this is NOT true however. It is perfectly possible to rearrange the distribution of mass symmetrically *without* altering the position of the CM.

But this triviality is not what is being challenged here. The point is that the CM is not just 'an approximation'. To be reliable it has to have clear constraints on when it is inaccurate and an accompanying formula for the error. This of course is not in your physics textbook. When the proper constraints are applied to the CM Theorem (i.e., it's only really valid between masses whose distance dwarfs their size) it becomes a mere truism in any case, and is not needed.

For relatively close distances, such as at the surface of the earth, the Theorem and method for calculating the force using the CM is grossly inaccurate, and results in an actual miscalculation of the 'constant' portion of the Gravitational Constant (G). The current G in Newton's Gmm/d^2 formula for the force is functional as a coordinator of units, but is in fact not the real Gravitational Constant. That must be teased out of the apparent constant after the corrections have been made for the inaccuracy caused by using the CM to calculate the force.

Last edited: Jun 17, 2005
7. Jun 17, 2005

### Meir Achuz

Newton proved that large spheres with uniform mass distribuitions behave like point masses in gravity. Since you don't believe Newton's geometric proofs (They are tricky, but Newton was smarter than us.), you cause yourself, and maybe others, a lot of wasted time.

8. Jun 17, 2005

### panthera

this isnt clear!! distances of two CMs from test particle are not equal....u first draw the diagram...if distances are equal then forces MUST also be equal considering it as a homogenous solid sphere.u are contradicting yourself...forces here
differ because of diff in dist only...its the only changeable quantity and rest are const.....gravity is inverse square law NOT inverse exponential one...hey if u dont mind are u kidding...i think u r doing this purposefully as whole world knows that its a square law!! there is no q of increase or decrease...the resultant force on the test particle will remain same even if the sphere is split into two halves with two diff CMs...

the test particle is not within the sphere but outside...from where u are getting vertical and horizontal components of a three dimentional object????????? there are r,phi,theta as coordinates and not X and Y that u can treat it as a two dimentional object!!!

I am now bound to think that either u are kidding OR u need to clear the concepts...

Regards

Last edited: Jun 17, 2005
9. Jun 17, 2005

### Staff: Mentor

Nope. Your "center of mass theorem" only holds (in general) for objects in a uniform gravitational field. The net gravitational force on an arbitrarily-shaped object in a uniform gravitational field acts at the object's center of mass.

Newton showed that objects with spherical mass distributions attract each other as if their mass were concentrated at the center of the sphere. This doesn't apply to hemispheres.

10. Jun 17, 2005

### Rogue Physicist

Your responses are kind of funny:

No, this was just one of Newton's claims, in proposing his theory of gravity. Try to follow the discussion.

Read again carefully. We've divided the sphere into a near half and far half, without moving either half. Each has its own Centre of Mass, which by symmetry will be along the axis passing through both. (in this case the x-axis itself.) Since the halves are identical, each Centre of Mass will be an EQUAL distance from the original Centre of Mass of the whole sphere. The distance to the test-particle is independant, and irrelevant for this part of the argument.

Now consider the distance from each CM to the test-particle. In this case, the change in distance away from the original CM of the whole sphere for one half of the mass is equal and opposite to the change in distance for the other half of the mass. But the change in forces upon the two CMs is not equal, since (if you like it said this way) it varies as 1/d^2, the inverse square of the distance.

Your hairsplitting over whether the law of gravity is 'exponential' in the general sense versus calling it 'inverse square' is moot. Since the denominator has an exponent, it is an exponential law, in contrast to a linear law with an exponent of one. That was my point in referring to it as exponential in the general sense. Only a linear law would vary in force equally with the variation in distance, and only a linear law of gravity could be consistent with the Centre of Mass theorem.

11. Jun 17, 2005

### Rogue Physicist

What you are talking about is the Centre of Gravity concept. That is entirely different from the Centre of Mass theorem. The Sphere Theorem is different again. Three names, three theorems.

The Centre of Mass Theorem (which is what we are discussing) applies to objects of any shape. What you are talking about here is the Sphere Theorem. Two names: Two different theorems.

12. Jun 17, 2005

### Tom Mattson

Staff Emeritus
I have never heard of anything called "The Center of Mass Theorem". Why don't you type it out mathematically? The only real clue I have as to what you mean is here...

That is not a consequence of any theorem related to the center of mass of a system of particles. At least not a theorem of which I am aware. I know that a the center of mass of any system of particles of total mass M follows the same trajectory that a single particle of mass M would follow. And I know that a uniform spherical mass is gravitationally equivalent to a point mass M, located at the center, for all points outside the mass.

The former theorem applies to all mass distributions, but says nothing of gravitation. The latter theorem speaks of gravity, but only applies to uniform spherical distributions. But you seem to be citing a theorem that is supposed to be about gravitation, and apply to nonspherical objects. As far as I know, that is not a theorem at all, but a false statement.

13. Jun 17, 2005

### Q_Goest

Gravitational attraction between two point masses is commonly given as:

1) F=G*m1*m2/r^2

If we looked at any two symetrical points within a mass (m1), you might think their combined gravitational atraction to some other mass nearby (m2) would be as if the two symetrical points in mass m1 were both at the center of mass m1. Assume for a moment the distance from the center of mass m1 for these two points is x, then the force on mass m2 is:

2) F=G*.5m1*m2/(r+x)^2 + G*.5m1*m2/(r-x)^2

In other words, imagine a dumbell with two spherical masses connected by a massless bar who's length is 2x. (Note also the axis of the dumbell is aligned with the axis between m1 and m2.) Imagine the total mass of this dumbell is m1, so each of the two spherical masses is .5*m1 The center of mass of this object is at the geometrical center, half way between the two masses. One would then have to say that if the mass of this dumbell was m1, then the force it produces on another mass, m2, should be equal to a single mass of m1 at the center of mass of the dumbell. You could equally suggest each mass is just a differential mass, dm. In this case, each differential mass should counteract the equal amount of mass that it is symetrically opposite to in m1. So:

Let:
a = r-x
b = r+x
C = G*.5m1*m2

Then putting these into equation 2 above gives:

F = C/b^2 + C/a^2

F = (a^2*C) / (a^2*b^2) + (b^2*C) / (a^2*b^2)

F = (C*a^2 + C*b^2) / (a^2*b^2)

multiply top and bottom by 2 and replacing C:

F = 2*(G*.5*m1*m2)*(a^2+b^2)/(2*a^2*b^2)

F = G*m1*m2* ((a^2+b^2)/(2*a^2*b^2))

So 1/r^2 should equal ((a^2+b^2)/(2*a^2*b^2)) if the two point masses in the dumbell have the same gravitational attraction as a single mass at the center of the dumbell. But these two are not mathematically the same, so the two point masses do NOT have the same gravitational attraction to m2 as a single mass at the geometrical center. Note that the geometrical center is generally called the "center of mass". I think this is what RP is saying is the center of mass theorem as opposed to the center of gravity, but not sure.

It would seem there's a discrepency here that I'm sure someone can explain, but I believe you'll need to do a lot more math than I did. Obviously, the gravitational field around a dumbell is different than the gravitational field around a spherical object, but this doesn't answer the question RP raises.

RP, I think you need to do the math on this including the calculus to put your thoughts into better perspective. I believe you're right about the gravitational field inside a sphere, but I also have to believe this has all been done before and I'd bet someone here understands this enough to explain better than I.

14. Jun 18, 2005

### Staff: Mentor

I believe this is what RP is stating as his "center of mass theorem". It's just not true.

RP is confusing "center of gravity" with "center of mass". They are only coincident in a uniform gravitational field.

15. Jun 18, 2005

### Q_Goest

Problem is, the "dumbell" I'm refering to could equally be two halves of a sphere. There's something wrong with equation 1. Is equation 1 only an aproximation? I seem to remember seeing someone mention it is, and that the exact equation is an infinite series. If so, that would help clear up the confusion.

16. Jun 18, 2005

### Staff: Mentor

I'm not getting your point. Equation 1 is just Newton's law of gravity for point masses. No problem there. To find the net force on any extended body, like the dumbell or sphere, you must add up the forces on each element. In general it is not true that the net force equals what you'd get by treating all the mass as concentrated at the center of mass. (It happens to be true for spherical mass distributions.)

Maybe you can restate what you think is the confusion.

PS: You cannot model a sphere as a two-piece dumbell and expect to get the correct answer! If you replace the halves of a sphere by two point masses--it's no longer a sphere.

Last edited: Jun 18, 2005
17. Jun 18, 2005

### Q_Goest

We treat any 3 dimensional geometry as a point mass when calculating the force of gravity at a large distance, so this should apply to a hemispherical shape as well as a spherical shape.

If we consider a spherical shape is the same as two hemispherical shapes, then at any distance from the sphere, regardless of whether we consider the sphere a single element or two hemispherical elements, the gravitational force calculated should be the same.

I created a spreadsheet that calculated the force and difference between these two given the math I did above. Regardless of how far one goes out, even if we consider the force of gravity of Mars on Earth for example, there is a very slight difference in the two calculations. The difference will be larger, the closer you get, but it always remains and is calculable.

I might suggest two possible reasons.
1) The equation given (equation 1) is not exact. I seem to remember someone mentioning this here, but haven't see it anywhere else.
2) One might be able to mathematically show that even at a significant distance such as from Earth to Mars, the gravitational field of a hemisphere is not equal to that of a point mass, so the slight deviation of both hemispheres from the assumption of being point masses rectifies the error that's calculated.

EDIT:
Yes, I think that's it exactly. That's what I mean by #2 above and I think that's where RP is thinking there's a paradox. The gravitational field of a hemisphere is not perfectly symetrical, so although equation 1 will be a close aproximation at a large distance, it won't be exact. There's some deviation in the n'th decimal place depending on how far you go out, and when calculating gravitational force at a distant point far from these two hemispheres, you still need to take into account that the hemisphere's gravitational field is not symetrical.

Last edited: Jun 18, 2005
18. Jun 18, 2005

### Staff: Mentor

If the object is sufficiently far enough from the gravitating body to be considered a point mass, then its geometry is irrelevant. (Note that if you far enough away, the gravitational field can be treated as uniform.) In estimating the gravitational force that the moon exerts on an object I hold in my hand, the shape doesn't matter.

If, on the other hand, you are talking about the object's gravitational field, then, once again, if you are far enough away then the shape doesn't matter. For smaller distances, it does. The field of a two-piece dumbell is a dipole field; as distance goes to infinity, it becomes the field of a point mass. The sphere is a special case: Its field is the same as that of a point mass at any distance (beyond its surface). At close distances, a dumbell is a poor model for a sphere.

Right. Two hemispheres equals a sphere. But two hemispheres does not equal a dumbell.

I will look over your math when I get a chance. If your expression for the field from a dipole does not approach that of a point mass as distance goes to infinity, you did something wrong. (Either way, a dipole is not a sphere!)

19. Jun 18, 2005

### Q_Goest

Yes, as the distance goes to infinity, the equation for a point mass equals the equation I came up with for a dipole. Similarly, as distance goes to infinity for a hemisphere, the gravitational field becomes uniform.

If you model two hemispheres as a dipole, (which seems on the surface to be a reasonable assumption) you find it is only an estimate of the actual field created by a hemisphere. The slight deviations then, if taken into account, should cancel out the error of the initial assumption.

I'm afraid Newton's law of gravity will have to remain standing...

20. Jun 18, 2005

### Staff: Mentor

OK, let's expand this to look at the on-axis field at some distance from this dumbell:
$$F = \frac{0.5 G m_1 m_2}{r^2} [\frac{1}{(1 +x/r)^2} + \frac{1}{(1 -x/r)^2}]$$
Taking the first few terms of a binomial expansion (for r >> x):
$$F = \frac{0.5 G m_1 m_2}{r^2} [\frac{1}{(1 + 2x/r + (x/r)^2)} + \frac{1}{(1 - 2x/r + (x/r)^2)}]$$
Which becomes:
$$F = \frac{G m_1 m_2}{r^2} (1 + (x/r)^2)$$
Which approaches your equation 1 as r goes to infinity.

PS: Looks like you already figured this out while I was typing my post.

Last edited: Jun 18, 2005