# Center of Mass/Gravity

1. Aug 16, 2004

### Cyrus

Hi, I was wondering if someone could explain the difference between the center of mass and the center of gravity. From my understanding, the center of gravity factors in g when doing the center of mass calculation to each point particle. So if g varies widely over a small distance in space, then the center of gravity will be significantly different than if we assumed g to be constant throughout the body. I know the center of mass will balance at that point. But if the cg and the cm do not coencide, where does the object now balance?

Thanks for your help on this issue,

-Cyrus Abdollahi

2. Aug 17, 2004

### Clausius2

What do you mean with balance?. I think the answer is always in the center of mass. I'm not sure of reading some day that center of gravity and center of mass could be distanced in some cm., in a body like a football stadium.

3. Aug 17, 2004

### pmb_phy

It is not meaningful to speak of center of gravity as being different as center of mass. When people use the term "center of gravity" they are refering to a point in a body which moves as a point like particle. But if the body is extended and tidal gradients come into play, then there is no such point. Therefore you can reasonably think of the center of mass as being the center of gravity.

Pete

4. Aug 17, 2004

### NateTG

There are certainly situations where the center of gravity and the center of mass are different, but they involve large scales. On scales where gravity can be effectively modled as constant acceleration, it's pretty clear that the two will be the same.

An example of a situation where the center of mass, and center of gravity are different, consider, for example, the center of gravity, and mass of the moon. Let's simplify by treating the moon as a sphere of constant density. The center of mass of the sphere is fairily clearly the center of mass of the sphere, but, because gravity is acting less on the far side of the moon than on the near side of the moon, the center of gravity of the moon is closer to the earth than the center of mass of the moon.

5. Aug 17, 2004

### pmb_phy

Why do you think that gravity is acting less on the far side of he moon than on the near side? For a spherical body with a spherically seymetric mass density, the gravitational field behaves as if all the mass is located at the center of the sphere.

In any case, that is not precisely what "center of gravity" refers to. The center of gravity is defined as that point in a extended body which moves in a gravitational field as if all the mass were concentrated at that one point.

Its obvious through the definition that this has no meaning if the gravitational field is not uniform. It really should be part of the definition but, unfortunately, it rarely, if ever, is.

Pete

6. Aug 17, 2004

### BobG

'g' never varies widely over a short distance. But it does vary. The Earth's gravitational field is spherical (approximately).

If you drew a line from the center of mass to the center of the Earth, the object would theoretically balance when both the center of mass and the center of gravity lie along that line (the radius). However, unlike the center of mass, the center of gravity can move.

On the surface of the Earth, in a dense atmosphere, you'll never be able to tell the difference.

Put an object into space with long appendages (such as solar arrays on satellites), the force of gravity on the appendages is not exactly equal to the force of gravity at the center of mass. In fact, unless the solar arrays were aligned perfectly perpendicular to the satellite's radius with the center of gravity laid perfectly on the radius vector, the force of gravity on each solar array will be different.

The result is that the object is extremely unlikely to balance at all. Over time, the satellite will orient itself so that it's long axis will coincide with the satellite's radius - an orientation that is much more likely to place the center of gravity, center of mass, and center of Earth along the same straight line.

There's other environmental perturbations that affect a satellite's orientation, which is why you're never going to keep a satellite balanced with its long axis perpendicular to the radius (Torque is equal to the force times distance from the center of mass times the sine of the angle between the long axis and the force). Keeping the long axis closely aligned with the radius (therefore reducing the sine value) reduces the torque associated with gravitational force and provides a much more stable orientation.

Since the motion which brought the long axis into alignment with the radius can never be completely damped out, the satellite winds up with a pendulum effect, with the long axis moving back and forth across the radius (you can restrict its range of motion to less than about 5 degrees in a real space environment).

7. Aug 17, 2004

### NateTG

I meant to discuss the gravitational interaction between the Moon and the Earth, but failed to be clear about it. I may still have outsmarted myself though, I certainly don't want to go though the math :(

The notion of center of gravity can be (mathematically) well defined in non-uniform gravitational fields. Although really, this all relies on a slew of simplifying assumptions.

In a non-uniform (newtonian) gravity field it's cerntainly possible to show that the center of gravity and center of mass (if they exist) must be in different locations.
If we imagine a long rod of uniform density in space that goes from radus $$r_1$$ to radius $$r_2$$ radially outward from the earth where the radii are measured from the center of the earth. Then we can calculate the total mass of the rod:
$$m=k \rho(r_2-r_1)$$
and the force exerted by gravity on the rod
$$F=\int_{r_1}^{r_2} M_e G k\rho \frac{1}{x2} dx = M_e G k \rho \times -(\frac{1}{r_2} - \frac{1}{r_1})=M_e Gk \rho \times \frac{r_1-r_2}{r_1 r_2}$$
now we can solve for the center of gravity:
$$\frac{m \times M_e \times G}{r_{cg}^2}= M_e Gk \rho \times \frac{r_1-r_2}{r_1 r_2}$$
$$r_{cg}=\sqrt{r_1 r_2}$$
Obviously
$$r_{cm}=\frac{r_2-r_1}{2}$$
and we know that if $$r_2 \neq r_1$$ the two won't be equal.

8. Aug 18, 2004

### pmb_phy

NateTG - I believe that you've made two errors above.

The first error has to do with the expression for the force on the rod. The force on the rod is a function of position which is reflected in two numbers, i.e. r1 and r2. While the difference r2 - r1 = d is a constant, r1 r2 is not. Therefore the force on the rod is a function of two parameters, namely r1 and r2. You've treated it as if it is a function of only one parameter by assuming that r1r2 is not a function of position. There was no reason to set rcg as you did since it must also appear on each side of the force expression. It would have proven more useful for you to choose a point on the rod as a reference point and then represent the force as a function of r = distance from reference point to Earth, and d = length of rod. You'd then have the force as a function of position, r. It would then be your claim that the position, r, would behave as a point particle moving in the same gravitational field. That assumption would be incorrect because the second mistake you've made is in an invalid assumption. The expression

$$\bolf F_g = \frac{GMm}{r^2}$$

is only valid for a point particle. You cannot equate the expression for the force on the rod with this expression for the gravitational force on a point particle, since a rod is not a point particle. Therefore since a rod is not a point particle it won't move like a point particle in a gravitational field and it therefore fails the defining criteria for "center of gravity".

Pete

Last edited: Aug 18, 2004
9. Aug 18, 2004

### NateTG

The rod can be treated as a point mass at radius $$r_{cg}$$.

According to the same argument the center of gravity of any solid cannot exist. Clearly there is a problem with that argument.

No, I have not. $$r_1$$ and $$r_2$$ are not inside the integral so I don't really care whether they are functions of position or not.

Um... No. Assuming (as I explicitly did in the post) that there is a center of gravity for the rod, the center of gravity must satisfy the equation that I posted.
In fact the expression:
$$F_{G}=\frac{m \times M \times G}{r_{cg}^2}$$
holds for any center of gravity.

10. Aug 18, 2004

### pmb_phy

As I've explained to you, that is incorrect. Why do you think that a particle will fall at the same rate as a rod? Rotate the rod and let it be perpendicular to a radial line from the center of the Earth. Then the force will be different and it will accelerate differently. The acceleration of an extended object in a gravitational field is a complex problem to solve. One thing is for certain. Extended objects do not fall at the same rate as a particle. Yet you keep using the expression for a point particle, i.e. F = GMm/r2 for the extended body.
As I've explained abovel, the center of gravity is only a meaningful quantity when the gravitational field is uniform. In that case there are no tidal forces which make the body move off of a geodesic (in GR lingo).
There was nothing in my comment about $$r_1$$ and $$r_2$$ being inside the integral. You gave the force on the rod as a function of $$r_1$$ and $$r_2$$. The force depends on two variables, not one. Give it a try. Calculate the force on a rod 2000 km long. Let the closer end of the rod be at 5,000 km. Then try it with the closer end at 10,000 km. If you closely analyze this problem then you'll see that you can find the force as a function of one variable which describes the position of the rod and one which describes the length. I.e. F = F(r,d). But what you cannot validly do is set this force to F = GMm/r2 or whatever parameter you choose to describe the distace from the center of the Earth.
And you've based this on an invalid assumption, i.e. you've based this on the invalid assumption that a body falls in the same way as a point particle. It does not. Hence the fact that center of gravity only has meaning in a uniform g-field.

However please keep in mind that I'm using a definition given in Newtonian Mechanics, A.P. French. That definition states
French is obvioulsy refering the the center of mass of seperated (point?) objects. Please let's put this off until I can think about this more since I can't understand what French means by this. And today I had an injection in my back for pain so I'm an idiot for even posting today since I should be lying down.

I'll get back to you. Okay?

In the meantime you should actually try your result with a few examples. What you'll find is that the location of the center of gravity will vary on the body with the distance the rod is the the center of the Earth. So, according to you, the center of gravity will not be a unique point on the rod. Is that what you hold to be true??

Pete

Last edited: Aug 18, 2004
11. Aug 18, 2004

### Cyrus

keep it simple! :-), I asked about the difference between the center of gravity and the center of mass, if you want to have detailed discussions that based on this subject post it on another forum, this is confusing me!! :-( and is not my question. Sigh..... your getting into things like plannets and such, but for now i just want a practical straight forward anwser I can use to grasp the difference between the two concepts.

Last edited: Aug 18, 2004
12. Aug 19, 2004

### pmb_phy

That's the goal, as always.
To answer a person's question we have to discuss it among ourselves sometimes. In this particular case I want to get a better understanding on the exact meaning of "center of gravity".
I disagree. This is exactly your question. You asked I was wondering if someone could explain the difference between the center of mass and the center of gravity and that is the exact question I/we are attempting to answer.
The very meaning of "center of gravity" has to do with a 1 or more bodies in a gravitational field. So when you ask about the center of gravity your asking about bodies in a gravitational field. NateTG and I have different opinions on this. NateTG was demonstrating his point to me by using a concrete example using a rod in the gravitational field of the Earth. We certainly can't answer you if we don't completely understand it, can we?
Me too.

I gave you the precise definition of "center of gravity" that is found in a widely known mechanics text by a well known author, i.e. A.P. French. He holds that the center of mass and the center of gravity can be different when a body is in a non-uniform gravitational field because the different objects are subject to different forces, or for a single body, different parts of the body are subject to different forces.

But I don't exactly understand French's definition or how to apply it. I read NateTG's example and it doesn't seem meaningful to me. I would have expeceted the center of gravity of an extended body to be a particular point within the body which is independant of the gravitational field it is in or where it is in the gravitatioal field. NateTG's example gives a center of gravity which is not a point fixed in a body. Therefore NateTG and I are discussing it to better understand it and to better answer your question. One of us is wrong. If it is me then I want to know.

Pete

13. Aug 19, 2004

### Staff: Mentor

The center of gravity of an extended body (of mass M), at least as I learned it, is not (in general) a fixed point in the body but is relative to an external point (P). If you imagine a point mass (m) placed at a given point (P) outside the body, it can be shown that the resultant force between the mass m and the extended mass acts along a line between P and the extended mass. The "center of gravity" (relative to point P) is that point G along that line for which you get the same force as if all the extended mass (M) where concentrated at a single point. I believe this is similar to the definition that NateTG is using.

14. Aug 19, 2004

### NateTG

I believe phb_pmy, Doc Al, and I agree on this:

The center of gravity of an object with mass $$m$$ is some point $$\vec{p}$$ so that the object may be treated as a point mass at $$\vec{p}$$ with mass $$m$$ rather than dealing with the entire space that the object occupies for purpses of gravity.

The center of mass of an object of mass $$m$$ is some point $$\vec{p}$$ so that the object may be treated as a point mass at $$\vec{p}$$ for purposes of linear momentum.

The CG and CM of a particular object are the same in constant Newtonian gravitational fields.

15. Aug 19, 2004

### pmb_phy

To be honest, when I'm not 100% sure on something I prefer not to have an opinion. I've been trying to understand French's definition of center of mass and would rather not have an opinion for now.

Regarding center of mass, I think it has a usefulness beyond considerations of linear momentum. Einstein's 1906 photon in a box experiment comes to mind. And, surprisingly, center of mass is also defined for radiation! :surprise:

See -- http://www.geocities.com/physics_world/sr/momentum_conservation.htm

Pete

16. Aug 19, 2004

### pmb_phy

ARe you speaking about the passive effects of gravity or the active effects? I.e. are you refering to how an object behaves in a g-field or are you refering to the gravitational field of an object?

Thanks

Pete

17. Aug 19, 2004

### Staff: Mentor

The center of gravity is a way to summarize the effect of an external field on an extended object. So, in your terms, it describes the "passive" effects of gravity.

18. Aug 23, 2004

### Gonzolo

Hi, here is my answer, irrespective of what was said by the others :

When a gravitationnal field is constant throughout a body constant (g = 9.8 m/s2 everywhere), center of mass and center of gravity are the exact same point of the body.

When g is not constant throughout the body, such as in a mountain (g is greater at a lower altitude). The 2 points are not the same. The center of gravity is lower.

The center of mass is the point around which the object would spin if it was in free space (the point common to any axis of rotation).

The center of gravity is the point which would have the same acceleration as the body if it was falling and had the same total mass concentrated into it.

Last edited by a moderator: Aug 23, 2004
19. Aug 23, 2004

### Gonzolo

Imagine a perfect pipe going from the Earth to the Moon. Its center of mass is clearly half way, but the center of gravity is much nearer to Earth. A ball at the center of gravity of the pipe would feel the same acceleration as the pipe, towards the Earth..