
#1
Aug1604, 10:37 PM

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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
Aug1704, 10:18 AM

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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
Aug1704, 11:21 AM

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Pete 



#4
Aug1704, 12:04 PM

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Center of Mass/Gravity
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
Aug1704, 12:46 PM

P: 2,955

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
Aug1704, 03:52 PM

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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
Aug1704, 07:06 PM

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In a nonuniform (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 [tex]r_1[/tex] to radius [tex]r_2[/tex] 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: [tex]m=k \rho(r_2r_1)[/tex] and the force exerted by gravity on the rod [tex]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_1r_2}{r_1 r_2}[/tex] now we can solve for the center of gravity: [tex]\frac{m \times M_e \times G}{r_{cg}^2}= M_e Gk \rho \times \frac{r_1r_2}{r_1 r_2}[/tex] [tex]r_{cg}=\sqrt{r_1 r_2}[/tex] Obviously [tex]r_{cm}=\frac{r_2r_1}{2}[/tex] and we know that if [tex]r_2 \neq r_1[/tex] the two won't be equal. 



#8
Aug1804, 04:37 AM

P: 2,955

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. r_{1} and r_{2}. While the difference r_{2}  r_{1} = d is a constant, r_{1} r_{2} is not. Therefore the force on the rod is a function of two parameters, namely r_{1} and r_{2}. You've treated it as if it is a function of only one parameter by assuming that r_{1}r_{2} is not a function of position. There was no reason to set r_{cg} 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 [tex]\bolf F_g = \frac{GMm}{r^2}[/tex] 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 



#9
Aug1804, 11:38 AM

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The rod can be treated as a point mass at radius [tex]r_{cg}[/tex].
In fact the expression: [tex]F_{G}=\frac{m \times M \times G}{r_{cg}^2}[/tex] holds for any center of gravity. 



#10
Aug1804, 02:02 PM

P: 2,955

However please keep in mind that I'm using a definition given in Newtonian Mechanics, A.P. French. That definition states 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 



#11
Aug1804, 07:26 PM

P: 4,780

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.




#12
Aug1904, 03:43 AM

P: 2,955

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 nonuniform 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
Aug1904, 05:50 AM

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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
Aug1904, 11:56 AM

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The center of gravity of an object with mass [tex]m[/tex] is some point [tex]\vec{p}[/tex] so that the object may be treated as a point mass at [tex]\vec{p}[/tex] with mass [tex]m[/tex] rather than dealing with the entire space that the object occupies for purpses of gravity. The center of mass of an object of mass [tex]m[/tex] is some point [tex]\vec{p}[/tex] so that the object may be treated as a point mass at [tex]\vec{p}[/tex] for purposes of linear momentum. The CG and CM of a particular object are the same in constant Newtonian gravitational fields. 



#15
Aug1904, 12:04 PM

P: 2,955

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_wor...nservation.htm Pete 



#16
Aug1904, 12:06 PM

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Thanks Pete 



#17
Aug1904, 07:29 PM

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#18
Aug2304, 05:29 PM

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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. 


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