How to calculate gravitational force Suppose the centre of mass of 2 bodies coincide , then how will we calculate the gravitational force between the 2 bodies ??
 A: Suppose you have two densities of mass $\rho_1$ and $\rho_2$. The two centers of mass coincide if
$$
\vec R_{CM_1} = \frac{\int_{\mathbb{R}^3} \vec r \rho_1(\vec r) \, d^3r}{\int_{\mathbb{R}^3}\rho_1(\vec r) d^3r} \quad \vec R_{CM_2} = \frac{\int_{\mathbb{R}^3} \vec r \rho_2(\vec r) \, d^3r}{\int_{\mathbb{R}^3}\rho_2(\vec r) d^3r}
$$
are equal, but the force is not calculated as
$$
\vec F = -G \frac{M_1M_2}{|\vec R_{CM_1} - \vec R_{CM_2}|^3}\left(\vec R_{CM_1} - \vec R_{CM_2} \right)
$$
Instead you must sum the contribution of all the infinitesimal elements
$$
\vec F = -G \int_{\mathbb{R}^3}\int_{\mathbb{R}^3}\frac{\rho_1(\vec r_1)\rho_2(\vec r_2)}{|\vec r_1 - \vec r_2|^3}\left(\vec r_1 - \vec r_2 \right) d^3r_1 \, d^3r_2
$$
So you are not dividing by zero like in the first formula, the integral is well defined always. It happens that in a spherical distribution of mass, the integral formula reduce to the first one, but in that case the centers of mass never coincides.
A: You do this the same way that you always do it: you compute the integral of the forces of each element on each other element. When both objects are spheres (or have spherical symmetry) the math gets easier - but in general just knowing the location of the center of mass of an object is not sufficient to determine the force.
As an example - if you have a hole through the earth and you try to compute the force of gravity as you fall into the hole, you find that this force goes to zero as you reach the center - even though you are getting closer to the center of gravity. In fact, the force becomes:
$$F = \frac{GM_{earth}m}{r^2}\frac{r^3}{R^3}$$
For an object on the surface of the earth, $F=mg=\frac{GMm}{R^2}$, so
$$F= \frac{m g r}{R}$$
But in general, the way to do this is to compute the (vector) sum of all the forces between any two infinitesimal elements in the two objects. Which is hard when they are not spherical.
