# Why does this volume integral vanish?

I am stuck on this problem concerning the gravitational potential of a body. The body has a mass density $$\rho(\mathbf x)$$ and I have to calculate a contribution to the total gravitational potential defined by $$f(\mathbf x) = \int_V d^3\!x' \, \rho(\mathbf x')\frac{ \mathbf x \cdot \mathbf x' }{|\mathbf x|^3}$$ where $$\mathbf x$$ is the position vector measured from the centre of mass of the body. Here $$V$$ refers to the volume of the body and it is assumed that $$\mathbf x$$ is very large, in the sense that $$|\mathbf x| \gg |\mathbf x'|$$, for $$\mathbf x' \in V$$.

I expect $$f(\mathbf x)$$ to vanish, but I can not explain why.

NOTE: the original question has been edited. For better clarity the mass density, originally denoted as $$d(x')$$, has been renamed as $$\rho$$.

• Is $x$ the fixed distance from the center of mass to the observation point, or is it a variable that depends on $x'$ somehow? Because in the former case, the integral is not zero in general, it's proportional to the position of the center of mass. – Javier Oct 14 '20 at 3:12
• @Javier $x$ is independent of $x'$ but it is the vector from the centre of mass to any point in P in the potential. Could you explain why would in the first case it's proportional to COM? – mathripper Oct 14 '20 at 8:34
• Hi! You write: I have to calculate a contribution to the total gravitational potential, defined by... Does the definition refer to a contribution or to the total potential? What do you mean by I have to calculate a contribution? Is the body placed in an external gravitational potential? And why do you expect f(x) to be zero? – Deschele Schilder Oct 17 '20 at 11:34

If $$\mathbf{x}$$ is independent of $$\mathbf{x}'$$, the integral is just

$$\int dV' d(\mathbf{x}') \frac{\mathbf{x}\cdot\mathbf{x}'}{x^3} = \frac{\mathbf{x}}{x^3}\cdot \int dV' d(\mathbf{x}') \mathbf{x}',$$

and the latter integral is just by definition the total mass times the position of the center of mass:

$$\frac{\mathbf{x}}{x^3}\cdot \int dV' d(\mathbf{x}') \mathbf{x}' = M \frac{\mathbf{x}\cdot \mathbf{X}_\text{CM}}{x^3}.$$

This is identically zero only if the coordinate system has the origin at the center of mass.

• Which can always be arranged (: – daydreamer Oct 14 '20 at 15:15
• Isn't x equal to zero (and thus $x^3$, so you get 0/0) if the COM lies on the origin, in which case the potential is undefined? Inside the Earth, for example, the force of gravity decreases linearly with decreasing r (the distance from the center), giving trouble at the center for the potential. Or is the potential at the center (of the Earth) a smooth function consisting of two parabolic functions, "glued" together? – Deschele Schilder Oct 17 '20 at 10:57
• @DescheleSchilder $\mathbf{x}$ is not the position of the center of mass, it's the vector connecting the point where you're calculating the potential to the center of mass. The potential at the center of the Earth is smooth, it goes like $r^2$. – Javier Oct 17 '20 at 15:05