Why center of mass? Why not center of charge?

[Isaac0427]

Hi,

In electrostatics, we have Coulomb's law for the electric field of a point charge. We can generalize to the electric field of a continuous charge distribution, using $$E=k\int r^{-2}dq.$$ It would seem, using the same argument that gets us to the previous equation, we could define the gravitational field as the sum of the fields of each point mass, as $$g=G\int r^{-2}dm.$$ Is this correct? If not, why can't we do this?

If the previous equation is correct, this brings me to some troubling questions. First, how do we go from that equation to saying the field is proportional to the inverse square of the distance from the distribution's center of mass (or, if that is only an approximation, why is that approximation good)? Second, why can we not do a similar thing with charge, i.e. say the electric field of a charge distribution consisting of point charges all of the same sign is proportional to the inverse square of the distance from the distribution's center of charge? Why do we do the integration with electrostatics when we do not with universal gravitation?

Thanks in advance!

 

[paulo84]

Uh...I maybe am being noobish here, but EM and gravity are very different. They're both among the 4 fundamental forces and they operate in very different ways.

 

[Dale]

First, how do we go from that equation to saying the field is proportional to the inverse square of the distance from the distribution's center of mass (or, if that is only an approximation, why is that approximation good)?

That is Newton’s shell theorem. The theorem is not an approximation but it relies on spherical symmetry. So its application to real planets and stars is an approximation. It is a good approximation because planets and stars are pretty well approximated by spheres.

why can we not do a similar thing with charge,

You can. It is a pretty common homework exercise.

Why do we do the integration with electrostatics when we do not with universal gravitation?

Charge distributions often are not spherically symmetric

 

[kuruman]

Is this correct? If not, why can't we do this?

Who says we can't (and don't)? Note that the the correct denominator is $|\vec{r}-\vec{r}'|$ for the potential not just $r$ and that integration is over primed coordinates.
https://en.wikipedia.org/wiki/Gravitational_potential#General_relativity

Second, why can we not do a similar thing with charge, i.e. say the electric field of a charge distribution consisting of point charges all of the same sign is proportional to the inverse square of the distance from the distribution's center of charge?

Because it will be incorrect to do so except as an approximation for large distances relative to some characteristic length of the distribution. Have you studied multipole expansions of potentials? What you are talking about is the leading monopole term. All this stuff is gone over in great detail in intermediate university courses in Classical Mechanics and Electricity & Magnetism.

 

[Stephen Tashi]

First, how do we go from that equation to saying the field is proportional to the inverse square of the distance from the distribution's center of mass

That's a good question. I hope you are thinking of the correct definition for "center of mass" instead of thinking of a purely geometric center - such as the center of a sphere. Begin by computing the gravitational field of two point masses $m_1, m_2$ and see if the result can be manipulated so it is equivalent to the gravitational field of a mass $m_1 + m_2$ located at their center of mass.

Second, why can we not do a similar thing with charge, i.e. say the electric field of a charge distribution consisting of point charges all of the same sign is proportional to the inverse square of the distance from the distribution's center of charge?

What definition do you have in mind for a "center of charge"?

 

[Isaac0427]

That is Newton’s shell theorem. The theorem is not an approximation but it relies on spherical symmetry. So its application to real planets and stars is an approximation. It is a good approximation because planets and stars are pretty well approximated by spheres.

Thank you very much.

There is still one thing confusing me—when we did classical mechanics, my teacher said that for all masses, the force of gravity can be thought to act entirely on the center of mass. We applied this to, say, nonuniform rigid rods. But, a no uniform rigid rod is not spherically symmetric, nor are many other masses we applied the center of mass concept to. Why could we do this?

 

[jbriggs444]

my teacher said that for all masses, the force of gravity can be thought to act entirely on the center of mass. We applied this to, say, nonuniform rigid rods. But, a no uniform rigid rod is not spherically symmetric, nor are many other masses we applied the center of mass concept to. Why could we do this?

The teacher's statement is correct for determining the total force from gravity and its effective point of application in a uniform gravitational field. As long as we are working on the surface of the Earth, some 6000 kilometers from its center and dealing with objects that are small compared to that distance, the Earth's gravitational field is reasonably uniform and this is a good approximation. One can hold a barbell at arms length oriented at any various angles without detecting any variation in its weight or any net torque from gravity.

Deviations are present but they are difficult to detect. One of the more easily detected deviations is in torque. With sufficiently sensitive gear, I suspect that one could detect a tidal torque on a barbell suspended at a 45 degree angle. There is a similar torque from the moon's gravity on the not-quite-spherically-symmetric Earth causing the Earth's rotation to slow over time.

 

[Isaac0427]

The teacher's statement is correct for determining the total force from gravity and its effective point of application in a uniform gravitational field. As long as we are working on the surface of the Earth, some 6000 kilometers from its center and dealing with objects that are small compared to that distance, the Earth's gravitational field is reasonably uniform and this is a good approximation. One can hold a barbell at arms length oriented at any various angles without detecting any variation in its weight or any net torque from gravity.

Deviations are present but they are difficult to detect. One of the more easily detected deviations is in torque. With sufficiently sensitive gear, I suspect that one could detect a tidal torque on a barbell suspended at a 45 degree angle. There is a similar torque from the moon's gravity on the not-quite-spherically-symmetric Earth causing the Earth's rotation to slow over time.

Ok, so just to confirm I am getting this right, if we had two weird blobs of mass both with a known CM, the force of gravity would not be (inverse-square) proportional to the CM to CM distance. Is this correct?

 

[jbriggs444]

Ok, so just to confirm I am getting this right, if we had two weird blobs of mass both with a known CM, the force of gravity would not be (inverse-square) proportional to the CM to CM distance. Is this correct?

Correct. You would need to compute a messy integral. At a large distance, the force would be approximately inverse square. But at close range, that approximation would break down.

 

[Dale]

my teacher said that for all masses, the force of gravity can be thought to act entirely on the center of mass.

This is true only in a uniform external field.

 

[Isaac0427]

Correct. You would need to compute a messy integral. At a large distance, the force would be approximately inverse square. But at close range, that approximation would break down.

And it would be a double integral for the force, right? One for the field of the first mass, one for the force on the second, right? But if the masses were both spherical and uniform, the force would be the inverse square of the CM-to-CM distance (times GMm). Is this all correct? If so I understand this perfectly now, thank you!

 

[jbriggs444]

And it would be a double integral for the force, right? One for the field of the first mass, one for the force on the second, right? But if the masses were both spherical and uniform, the force would be the inverse square of the CM-to-CM distance (times GMm). Is this all correct? If so I understand this perfectly now, thank you!

Yes!

 

[Stephen Tashi]

There is still one thing confusing me—when we did classical mechanics, my teacher said that for all masses, the force of gravity can be thought to act entirely on the center of mass. We applied this to, say, nonuniform rigid rods. But, a no uniform rigid rod is not spherically symmetric, nor are many other masses we applied the center of mass concept to. Why could we do this?

Others have explained this. I'll only point out that you have asked two different questions in this thread.

1) Can the gravitational field at a point in space that's due to an object be computed as if the entire mass of the object is at its center of mass?

2) Can the forces of gravity on an object be computed as if the gravitational field is acting only on the total mass of the object located at the object's center of mass ?

 

[Isaac0427]

From this, this has stemmed another question; even in a nonuniform field, for gravity or for electricity, is there a point on a charge/mass distribution in any case in which the net force can be thought to act all on that spot? I had an idea about this; some kind of "center of force" defined as $$x_{CF}=\frac{\int xdF}{\int dF}.$$ Am I getting somewhere or does that not work? I can see that for a uniform gravitational field, that would boil down to the center of mass formula.

 

[Stephen Tashi]

is there a point on a charge/mass distribution in any case in which the net force can be thought to act all on that spot?

Are you thinking of a mass distribution that is a "rigid body"? Or could the mass distribution be a cloud whose density varies in time in response to internal and external forces?

 

[Isaac0427]

Are you thinking of a mass distribution that is a "rigid body"? Or could the mass distribution be a cloud whose density varies in time in response to internal and external forces?

Yes, a rigid body.

 

[jbriggs444]

Yes, a rigid body.

Sometimes a "center of gravity" as distinct from a "center of mass" will be used to refer to the point at which the total gravitational force from a non-uniform field could understood to act on a rigid object.

Sometimes people are sloppy and use the two terms as if they are equivalent (which they are if the gravitational field is uniform).

 

[Isaac0427]

Sometimes a "center of gravity" as distinct from a "center of mass" will be used to refer to the point at which the total gravitational force from a non-uniform field could understood to act on a rigid object.

Sometimes people are sloppy and use the two terms as if they are equivalent (which they are if the gravitational field is uniform).

Would that center of gravity be defined by the equation in post 14?

 

[Stephen Tashi]

some kind of "center of force" defined as $$x_{CF}=\frac{\int xdF}{\int dF}.$$ .

It isn't clear what the notation "$dF$" would mean. Presumably the bottom line of the integration is integration of a function involving force as a function of position over a volume of space.

is there a point on a charge/mass distribution in any case in which the net force can be thought to act all on that spot?

We have to make a specific interpretation of "can be thought to act". Let's formulate it as "Does there exist a (vector valued) function giving a force $F(t)$ and a position valued function $X(t)$ that the motion of a rigid body in a possibly nonuniform gravitational field is identical to the motion of the rigid body acted on only by the force $F(t)$ applied at the position $X(t)$?"The motion of a rigid body between two instants of time can be written as a translation of the body followed by the rotation of the body about an axis. This is a result from "kinematics" -i.e. motion without any consideration of what forces produce the motion. If I am allowed to apply forces to the body at two different positions then, inuitively, I could apply one force $F_1(t)$ to the body's center of mass $X_1(t)$ to produce the translation and apply a second force force $F_2(t)$ to some other point on the body to produce a torque that rotates the body. From that intuition, would it follow that there is some way to apply a single force at a single location to produce the equivalent result?

The integration you suggested seems to compute a position. The intuitive analysis suggests that you also need to define a force acting at that position.

 

[Isaac0427]

The integration you suggested seems to compute a position. The intuitive analysis suggests that you also need to define a force acting at that position.

Right, but the force is in the equation. That’s what F is.

 

[Stephen Tashi]

Right, but the force is in the equation. That’s what F is.

Whatever final answer "$X_{CF}$" your integration produces, it has physical units. If the units of $X_{CF}$ are Newtons then it doesn't produce an answer for position in meters. We need both $X(t)$ and $F(t)$. You need to define a (vector) force whose components are in Newtons acting on a (vector) position who units are meters.

 

[jbriggs444]

Would that center of gravity be defined by the equation in post 14?

I believe that you can get the answer with a similar equation. You would be calculating total torque and then doing a division (in the sense of a vector cross product) of total torque by total force yielding a moment arm which is at right angles to both.

Edit: On further consideration, this general approach only controls for two degrees of freedom in the usual case (and none if the net force is zero). One would want something else to get the third.

 

[kuruman]

some kind of "center of force" defined as
$$x_{CF}=\frac{\int xdF}{\int dF}.$$

Remember that force is a vector. For this expression to have meaning the force anywhere must be in the same direction. Well, in a one-dimensional but non-uniform gravitational field, this expression defines the center of gravity (not the center of mass) of a uniform object. For example, a 1 m rod sticking radially out on the surface of a neutron star has its center of gravity at a different point from the center of mass which is at the midpoint.