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Center of gravity and Center of mass different?

by Macdman11
Tags: gravity, mass
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Macdman11
#1
Feb27-13, 01:10 PM
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I was developing a test for my students and always look for a difficult question as the only bonus question. I came across the question in the book, "When are center of gravity and center of mass different?". I thought about it before looking at the book's answer; and I thought most likely when gravity varies over the object and the book agreed. Then the students argued that it could happen if there is no gravity on the object. Since CoG is the average location of an object's weight, and the object isn't massless, wouldn't the CoM & CoG be different?
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rudolfstr
#2
Feb27-13, 01:40 PM
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Think about it, would something that doesn't have mass even have a CoM or a CoG?
Ryoko
#3
Feb27-13, 01:44 PM
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An object floating in space with no gravity still has a center of gravity. (I guess center of mass would be more appropriate.) The CoM would immediately become apparent if a force were applied such as thrust from an engine or some other external force. Also, if the object had a spin, it would rotate about the CoM.

Doc Al
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Feb27-13, 03:14 PM
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Center of gravity and Center of mass different?

Quote Quote by Macdman11 View Post
Then the students argued that it could happen if there is no gravity on the object. Since CoG is the average location of an object's weight, and the object isn't massless, wouldn't the CoM & CoG be different?
In the absence of gravity, would CoG have any meaning? (CoM would be perfectly well defined, of course.)
Doc Al
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Feb27-13, 03:17 PM
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Quote Quote by Ryoko View Post
An object floating in space with no gravity still has a center of gravity. (I guess center of mass would be more appropriate.)
In common usage, center of gravity and center of mass are used interchangeably, since a uniform gravitational field is assumed. But here we are distinguishing the two and not assuming uniform gravity.
sophiecentaur
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Feb27-13, 03:38 PM
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I really thought "CG" was only used colloquially these days but I'm sure people will shower me with learned references where it's used.
BruceW
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Feb27-13, 04:00 PM
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Quote Quote by Macdman11 View Post
Then the students argued that it could happen if there is no gravity on the object. Since CoG is the average location of an object's weight, and the object isn't massless, wouldn't the CoM & CoG be different?
Hehe, clever students. I guess it depends on how you define centre of gravity. The definition I like best is that the centre of gravity is the point around which there is zero torque due to gravity, which is the solution to this equation (where capital R is the possible solution):
[tex]\int \rho (\vec{r} - \vec{R})\wedge \vec{g} \ dV = 0 [/tex]
So if there is zero gravitational field, then any position in space is a solution, and so the solution is not unique, and in fact includes all space. I guess this is intuitive from the definition above, because if there is no gravity, it cannot create a torque around any point.

That makes me think of something - are we guaranteed that the centre of gravity is unique? When the gravitational field is zero everywhere, then it is not unique, but is there some mathematical law that says the solution to R in the integral above should be unique if the density and gravitational field are not zero everywhere?

Edit: there is also the complication of 'self-gravity' of the object, as rudolfstr brought up. But I think assuming zero 'self-gravity' is OK for this problem, even though it is not strictly possible in the real world.
CWatters
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Feb27-13, 05:34 PM
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If gravity isn't uniform across an object is it possible for it to have more than one center of gravity?
BruceW
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Feb27-13, 06:42 PM
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I thought about it a bit longer and if we use the definition "The centre of gravity is the point around which there is zero torque due to gravity", then simply due to what torque is, we will never get a unique 'centre of gravity'. Even if the gravitational field is uniform, the centre of gravity is a line of solutions, going through the centre of mass. From the equation I wrote above, if we rearrange, we get:
[tex]\vec{R} \wedge \int \rho \vec{g} \ dV = \int \rho \vec{r} \wedge \vec{g} \ dV [/tex]
This is now just a cross product of the form [itex]\vec{R} \wedge \vec{A} = \vec{B}[/itex] where R is the thing we are trying to find. So if A and B are zero vectors, then any R will be a solution. If B is zero, then R is just cA (where c is any real number, i.e. we have an infinite number of solutions, parallel to A). And if A and B are non-zero vectors, then as long as we can find one R, then the set of all possible solutions is R + cA
A.T.
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Feb28-13, 03:02 AM
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Quote Quote by BruceW View Post
The definition I like best is that the centre of gravity is the point around which there is zero torque due to gravity,
Quote Quote by BruceW View Post
if we use the definition "The centre of gravity is the point around which there is zero torque due to gravity", then simply due to what torque is, we will never get a unique 'centre of gravity'.
So your favorite definition of "centre of gravity" is one that doesn't actually define a centre? What is wrong with using the point, at which all the object's mass concentrated would give the same net gravitational force as the extended object experiences?
A.T.
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Feb28-13, 03:11 AM
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Quote Quote by CWatters View Post
If gravity isn't uniform across an object is it possible for it to have more than one center of gravity?
If you place a ring around a planet such that the net gravitational force on the ring is zero, then you can assume this zero force acting anywhere you want on the ring. But aside from the cases where net gravitational force on the object is zero, you get a unique center of gravity.
xAxis
#12
Feb28-13, 10:49 AM
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Newton's shell theorem proved that objects graviti is the same as if the whole mass was concentrated in its centre of mass. So in that respect there is no difference between COM and COG. Now as to the behavior of object in the gravitational field, the force on it depends only on distance. The question is now, is it always the distance of COM of the object? Will two differently shaped objects with the same mass always have absolutely same acceleration taking into account minute changes in the field? I believe they will.
A.T.
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Feb28-13, 10:54 AM
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Quote Quote by xAxis View Post
Newton's shell theorem proved that objects graviti is the same as if the whole mass was concentrated in its centre of mass.
Newton's shell theorem applies only to spherical objects.
Quote Quote by xAxis View Post
Now as to the behavior of object in the gravitational field, the force on it depends only on distance. The question is now, is it always the distance of COM of the object?
Not in a non-uniform field.
sophiecentaur
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Feb28-13, 11:00 AM
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All this assumes that the gravitational field is Uniform.
The Centre of Mass is independent of gravity so it is more consistent (the first moment of any object).
The point where the gravity from a nearby point mass 'acts' on a rod, will vary with the orientation of the rod so that is not something that could be relied upon. Where is the CM of the Earth-Moon system? Somewhere below the Earth's surface. If you take a 1kg mass and place it near the surface of the Moon, where does the mass of the Earth plus Moon 'act from' when interacting with the 1kg mass. Where is the 'Centre of Gravity? Certainly not just below the Earth's surface.
Sloppy terminology can lead to all sorts of errors.
D H
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Feb28-13, 06:15 PM
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Quote Quote by BruceW View Post
I thought about it a bit longer and if we use the definition "The centre of gravity is the point around which there is zero torque due to gravity", then simply due to what torque is, we will never get a unique 'centre of gravity'. Even if the gravitational field is uniform, the centre of gravity is a line of solutions, going through the centre of mass.
That's not quite the right definition. You're missing a key part. Take a point that is on your line but is not the center of mass. Now rotate the object a bit. Now you'll get a different line, and your point won't be on it. You'll instead get a torque about that chosen point.

You forgot to add the qualifier "for any orientation of the object". In a uniform gravity field, that qualifier does make the center of gravity unique, and it is the center of mass.

This definition doesn't work in a non-uniform gravitational field.


Quote Quote by A.T. View Post
What is wrong with using the point, at which all the object's mass concentrated would give the same net gravitational force as the extended object experiences?
That definition doesn't work in a uniform gravity field. The center of gravity is indeterminate per this definition in a uniform gravity field because every point qualifies as the center of gravity. This definition is not unique in a non-uniform gravity field; the location of the center of gravity changes as an object changes orientation.

This definition is used occasionally for space-based applications. For example, a space elevator would need its center of gravity rather than its center of mass at geosynchronous altitude.


Quote Quote by xAxis View Post
Newton's shell theorem proved that objects graviti is the same as if the whole mass was concentrated in its centre of mass. So in that respect there is no difference between COM and COG.
That's only true for objects with a spherical mass distribution. It's not true in general. A couple of examples: The Earth and the Moon.

The Earth has a non-spherical gravitational field thanks largely to its equatorial bulge. That non-spherical gravitational field is essential for how our sun synchronous satellites work. Place a satellite in such an orbit and the orbital plane will rotate by just the right amount over the course of a year so as to maintain near-ideal lighting conditions underneath the satellite. http://trs-new.jpl.nasa.gov/dspace/b.../1/04-0327.pdf.

My other example is the Moon. The Moon's gravity field is rather lumpy thanks to a number of mass concentrations (mascons) on the near side of the Moon. This lumpy gravity field can make for some rather bizarre orbits. http://science.nasa.gov/science-news...6nov_loworbit/
A.T.
#16
Mar1-13, 01:35 AM
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What is wrong with using the point, at which all the object's mass concentrated would give the same net gravitational force as the extended object experiences?
Quote Quote by D H View Post
That definition doesn't work in a uniform gravity field. The center of gravity is indeterminate per this definition in a uniform gravity field because every point qualifies as the center of gravity.
That is true, we can fall back to the center of mass in that case. We could generalize the definition, such that self gravity is considered too. Wouldn't that always yield a unique center of gravity, even without external fields and in uniform fields?
BruceW
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Mar1-13, 03:12 PM
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Quote Quote by A.T. View Post
What is wrong with using the point, at which all the object's mass concentrated would give the same net gravitational force as the extended object experiences?
Why would you define it that way? Using that definition: It is not unique (only in exceptional circumstances). And it is not a point around which the torque due to gravity is zero (only in exceptional circumstances). Unless I am missing something...
BruceW
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Mar1-13, 03:42 PM
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Quote Quote by D H View Post
That's not quite the right definition. You're missing a key part. Take a point that is on your line but is not the center of mass. Now rotate the object a bit. Now you'll get a different line, and your point won't be on it. You'll instead get a torque about that chosen point.

You forgot to add the qualifier "for any orientation of the object". In a uniform gravity field, that qualifier does make the center of gravity unique, and it is the center of mass.

This definition doesn't work in a non-uniform gravitational field.
My definition works in a non-uniform gravitational field. My definition is that the torque around the centre of gravity is zero. So if I actively rotate the object, its centre of gravity will change such that it remains at a point around which there is zero torque. I say my definition. But it isn't mine, of course. We are all standing on the shoulders of giants.


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