"The centre of mass of a system of particles is the point that moves as though (1) all of the system's mass were concentrated there and (2) all external forces were applied there." (Resnick et al, p.215)

"The gravitational force on a body effectively acts at a single point, called the centre of gravity of the body." (Resnick et al, p.330)

According to these concepts:

If we have a body large enough that we could not assume that the centre of mass and the centre of gravity doesn't coincide. Then, by definition of the centre of gravity, the gravitational force does not act on the centre of mass. This means that we have an external force which we can't consider to be acting on the centre of mass.

Q (1)-Does the centre of mass still move as if the gravitational force were applied there?

Q (2)-Aren't this concepts the best definitions of the center of mass and centre of gravity?

Q (3)-What would be the axis rotation if I apply a force to one corner of this body?

Source: Fundamentals os physics / Robert Resnick, David Halliday, Robert Resnick - 10th edition.


2 Answers 2


Center of Mass is a mathematical concept used for ease of analysis because we are familiar with Equations of Motion for a point particle. Center of Gravity is a quite similar concept but very special to gravitational forces acting on a body.

Now, Center of Mass is the averaging of the position of masses while Center of Gravity is the averaging the position of Gravitational Forces acting on masses.

Since we observe events in the classical regime, Gravitational Field or $g$ value(which is just a practical representation of field - not to be confused with the Field itself) is uniform around us on Earth. So, it is likely to coincide with the center of mass.

But, the gravitational field varies as we move away from earth or as we go high. But even at the top of Mount Everest (8848 metres), the gravitational field strength is still 99.6% of its standard value. So, there will be very slight difference always w.r.t to Center of Mass and Center of Gravity which is negligible for most of the purposes.

However, if you have a varying gravitational field w.r.t position, say, $\Phi(\vec{r})$, then the centre of Gravity is going to be different from the centre of mass of the body or the system which we wish to study.

Now, answering your questions:

Q (1)-Does the centre of mass still move as if the gravitational force were applied there?

Yes, the motion along a path is the same as if net gravitational force acts on the centre of mass. But since both the centres are not the same, it will add up to a rotational motion of the body/system.

Q (2)-Aren't this concepts the best definitions of the center of mass and centre of gravity?

This question, I tried to establish the concepts above.

Q (3)-What would be the axis rotation if I apply a force to one corner of this body?

Depends on the problem, force and the body itself!


When you apply a force to an extended object (with density profile $\rho(\vec r)$), you can treat is as accelerating the center of mass:

$$ r_{i,\, cm} = \frac{\int_Vr_i \rho(\vec r)d^3r}{\int_V\rho(\vec r)d^3r}$$

The numerator is the 1st moment of the mass distribution. The denominator, which is the zeroth moment, is just the total mass:

$$ m = \int_V\rho(\vec r)d^3r$$

If that force acts from a point, $\vec r_0$, then there is also a torque:

$$ \tau_i = \epsilon_{ijk}(r_{j, 0} - r_{j, cm}) F_k $$

(Sorry for the index notation, but it has a purpose). The torque leads to a changing angular momentum:

$$ \dot L_i = \tau_i $$


$$ L_i = I_{ij}\omega_j $$

where the inertia tensor is related to the 2nd moment of the mass distribution:

$$ I_{ij} = \int_V (r^2\delta_{ij}-r_ir_j)\rho(\vec r)d^3 \vec r$$

So you get acceleration and torque of this mass distribution simplified down to a motion of and rotation around, a point.

What if that force is caused by gravity, in a non-uniform field?

The temptation is just find the g-weighted 1st moment:

$$ r_{i,\,cg} =\frac 1 m \int_V{ \rho(\vec r)g(r)r_id^3 r }$$

In a uniform field, it reduces to the center-of-mass. In a uniform gradient, it works well, too.

But there is a problem: $g(r)$ is supposed to be a vector, and if I put in index on it, as in $g_j(r_i)$, in the integral: there is no way to get rid of the $j$ and come out with a vector on the RHS to match the $i$ indexing the vector on the LHS.

You could define a CG for each component of the field:

$$ r_{i,\,j-cg} =\frac 1 m \int_V{ \rho(\vec r)g_j(\vec r)r_id^3 r }$$

and hope they are all equal. The difficulties related to this are addressed in https://en.wikipedia.org/wiki/Centers_of_gravity_in_non-uniform_fields, which I will not repeat, but in summary, it says the CG is the point about which a force needs to be applied to give the correct torque....which may not be unique, nor even defined, in all cases.

In practice, field configurations do admit a center-of-gravity, in which the motion can be described from a potential energy:

$$ U(\vec r_{cm}) = m\phi(\vec r_{cm}) $$

and the force is the gradient:

$$ F_i = -\nabla_i U=m\nabla_i\phi $$

Rotation depends on the quadruple moment:

$$ Q_{ij} = \int_V{(3r_ir_j-r^2\delta_{ij})\rho(\vec r)d^3r}$$

coupling to the gradient of gravitational field:

$$ U = Q_{ij}\nabla_i\nabla_j\phi $$

which leads to a torque versus angle. (Note: being a tensor moment, it has 4 extrema with respect to orientation, while a dipole as 2: aligned and anti-aligned.)

A classic example of this is the Shuttle Radar Topography Mission (SRTM):

enter image description here

There are 2 masses (the shuttle and an antenna) on each end of a 60 meter boom. In the operational configuration, the boom is roughly 45 degrees off vertical, which is the point of maximum instability. The heavier shuttle wants to be lower where the field is stronger.

In order to keep the system aligned, an attitude correction was applied every 20 minutes. The original design was a compressed nitrogen jet at the antenna (see question 3) applying a small correcting torque.


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