# Why is mass considered to be concentrated only at a single point of the body

I can not understand the concept of "Center of mass"?

Why do we consider the mass of an object to be concentrated at a single point inside or outside of that body?

How can the mass of an object be concentrated outside the object?

• It's a definition. It is useful because the center-of-mass behaves as if it were subject only to the net external force (which is often zero or very simple). The behavior of an entire extended body can then be decomposed into a simple motion of the center-of-mass and potentially more complicated rotations of the body about its center of mass.
– hft
Sep 9, 2022 at 17:54
• Re, "how can the mass of an object be concentrated outside the object?" Where is the center of mass of a donut? As at least one answer below states, the mass of an object is not actually concentrated at its COM, but for many simple physics problems, it's provable that you can get the correct answer by pretending that the mass is concentrated there. Sep 9, 2022 at 18:34
• I wrote an answer to a similar question here: physics.stackexchange.com/a/726024/330899 Sep 9, 2022 at 18:36

Basically, the center of mass is used to simplify your calculations. If you shoot a rocket into the sky on New Year's Eve and it burns out, it will start to describe a parabola in the sky. After it explodes (and the outcome is nearly impossible to predict exactly), all the pieces for themselves will describe an individual parabola as well, but their center of mass will still further describe the original parabola (which is still easy to describe). Here you also see the center of mass being outside of the body, this also happens in a donut.

Another similar example is, when you want to describe the movement of $$n$$ bodies under their own gravitational fields ($$n$$-body problem). The problem can be simplified a lot changing into the coordinate system of the center of mass. The calculations can be seen here. The center of mass moves with a constant velocity as a result of the translation symmetry of the problem (just like angular momentum is preserved as a result of rotational symmetry): $$\mathbf{C}(t)=\mathbf{X}_0+\mathbf{V}_0t =\frac{\sum_im_i\mathbf{r}_i(t)}{\sum_im_i}.$$ The vectors $$\mathbf{X}_0=\mathbf{C}(0)$$ and $$\mathbf{V}_0=\mathbf{C}'(0)$$ are constants given by the initial conditions $$\mathbf{r}_i(0)$$ and $$\mathbf{r}_i'(0)$$ of the problem for all $$i$$.

You can see the center of mass manifesting in the Pluto-Charon system: Both bodies seem to orbit around a phantom point outside both of them, which is their center of mass. An animation based on photos of New Horizons can be seen here.

Newtonian Gravity is a good example, where the center of mass is used: According to the shell theorem by Newton, a spherically symmetric object causes the same gravitational field outside of it as if all of its mass were concentrated in its center. This simplifies calculations in orbital mechanics, but since celestial objects are not perfectly spherically symmetric due to being a bit flattened because of their rotation, there will be a little error.

The shell theorem also states that if you have a spherically symmetric shell, there won't be any gravitational pull on an object inside of it, regardless of its position, even if it is not the center. This can be used to calculate how long falling right through the Earth would take. If you're inside the Earth, then you can ignore all the mass with a larger distance to the center and pretend all the mass with a smaller distance to the center is concentrated there, so you can just use Newton's formula for gravity instead of using more complicated integration.

To expand a bit further: In General Relativity, this generalizes to the Birkhoff theorem, which states, that a spherically symmetric vacuum solution must be the Schwarzschild solotion.

• Re. your statement about movement in the gravitational field of multiple bodies. I believe that replacing multiple forces by one single force exerted by the total mass located at the c.m. only works for uniform forces. Otherwise it would imply that the gravitational force on a mass inside a shell could be calculated as if the mass of the shell were at the center, leading to a nonzero result contradicting the shell theorem. Sep 9, 2022 at 18:54
• I just did the calculations and indeed replacing all forces by one to the center of mass of all other bodies does not add up. I seem to have remembered that wrongly. Therefore thanks, I changed that part. Sep 9, 2022 at 19:18

Center of mass and rigid bodies. Mass is not concentrate at the center of mass. Mass has its own distribution.

What we can say is that, for rigid bodies (or bodies in a static configuration, that do not undergo deformation during your analysis) the effects of volume force due to gravity are equivalent to a single force (the weight of the object) acting in the center of mass.

The same is true for inertia due to translation (or the time derivative of the momentum, depending on your description of the problem) is equivalent to a force equal the product of the acceleration of the center of mass and the total mass of the system (assuming that the total mass of the system is constant in time).

On the other hand, if you need to study the evolution of a deformable body under dynamic conditions, knowing the total mass and the position of the center of mass is not enough, but you need to know the mass distribution.

Definition of center of mass, $$G$$ The center of mass of a system, of total mass $$m$$, is defined as the weighted-average position of its points, i.e.

• for a system of discrete point masses: $$\mathbf{r}_G = \dfrac{\sum_i m_i \mathbf{r}_i}{\sum_i m_i} = \dfrac{\sum_i m_i \mathbf{r}_i}{m}$$

• for a system of with continuous distribution of masses: $$\mathbf{r}_G = \dfrac{\int_V \rho \mathbf{r}}{\int_V \rho}$$

Center of mass outside the system. As an example, a square frame (with a hole in the middle) with uniform mass distribution has the center of mass in the center of the hole. The same happens for a hula-hoop (the circle someone you to do some gym), or for a carton box, a bottle, and many other objects.

Center of mass and equation of motion of a rigid body. Having said that, a rigid body in in 3d-space has 6 degrees of freedom, 3 degrees of freedom for describing the position in space of one of its points $$P$$, and 3 degrees of freedom to describe its orientation in space.

It turns out that these equations have a very simple form we refer them to the center of mass, given the external force $$\mathbf{R}^e$$ and moment $$\mathbf{M}_G^e$$

$$\dfrac{d\mathbf{Q}}{dt} = \mathbf{R}^e$$
$$\dfrac{d\mathbf{\Gamma}_G}{dt} = \mathbf{M}_G^e$$

where $$\mathbf{Q} = m \mathbf{v}_G$$ is the momentum of the system, being $$\mathbf{v}_G$$ the velocity of the center of mass, and $$\mathbf{\Gamma}_G = \mathbb{I}_G \cdot \mathbf{\omega}$$ is the angular momentum of the system w.r.t. $$G$$, being $$\mathbb{I}_G$$ the inertia tensor of the system w.r.t. $$G$$, $$\omega$$ the angular velocity or the rigid system.

I can not understand the concept of " Center of mass " Why do we consider the mass of an object to be concentrated at a single point inside or outside of that body?

The "center-of-mass" is a useful definition, because the center-of-mass moves as if it were only subject to the net external force--all the internal forces cancel. Often the net external force is very simple or zero, which means the motion of the center-of-mass is very simple or uniform.

The concept of center-of-mass is helpful because we can break the problem into parts. One which is simple (the motion of the center-of-mass) and one which is more complex (e.g., the rotation of the rest of the body about the center-of-mass).

In equations, we define: $$\vec r_{center-of-mass} = \frac{1}{M}\sum_i m_i \vec r_i\;,$$ where $$M$$ is the total mass.

Assuming none of the $$m_i$$ change (fixed masses), the velocity of the center-of-mass is: $$\vec v_{center-of-mass} = \frac{1}{M}\sum_i m_i \vec v_i\;,$$ and the acceleration of the center of mass is: $$\vec a_{center-of-mass} = \frac{1}{M}\sum_i m_i \vec a_i = \frac{1}{M}\sum_i F_{on\;i} = \frac{1}{M}\sum_i F_{on\;i,\;external} = \frac{\vec F_{net,\;external}}{M}\;,$$ where the second-to-last equality comes from the fact that for each internal force there is a Newton's third law force pair that cancels and so we only need the external forces in the sum.