For this question I use Newtonian gravity only. Relativistic gravitational/kinematic effects are ignored.

It is known that the gravity surrounding a point particle of mass $M$ can be described by gravitational potential $\Phi$ where $$\Phi=-\frac{GM}{r}$$ $r$ is simply the distance from the particle. The gravitational acceleration $\mathbf{a}$ of a test particle (negligible mass) is defined as $$\mathbf{a}=-\nabla\Phi$$ Because Newtonian gravity is linear in potential, the potential of any arbitrary distribution of multiple point masses is the sum of the potential of all the masses.

I came across a question that I can't seem to answer it myself when trying to analyze two-body problem using the concept of gravitational potential. Assume that there are two point particles $A$ and $B$ with mass $M$ at points $(-a,0,0)$ and $(a,0,0)$ in Cartesian coordinates, initially at rest relative to each other. $\Phi$ in this case is time dependent because both particles will move towards each other due to mutual gravitational interaction. However let consider initial situation where $\Phi$ is $$\Phi=-\frac{GM}{\sqrt{(x+a)^2+y^2+z^2}}-\frac{GM}{\sqrt{(x-a)^2+y^2+z^2}}$$ and the acceleration at a point is $-\nabla\Phi$, as expected. Acceleration becomes ill-defined at points $(-a,0,0)$ and $(a,0,0)$ because it contains the term $0/0$. I think this is not a problem if you interpret this as the acceleration of a test particle, because it is infinitely close to either $A$ or $B$ and therefore experiences infinite acceleration. But, if I need to calculate the acceleration of $A$ itself, obviously the wrong answer is obtained because you can calculate the acceleration using inverse square law and it is finite. One way around this problem is to consider only the potential due to $B$ $$\Phi=-\frac{GM}{\sqrt{(x-a)^2+y^2+z^2}}$$ This gives answer in agreement with inverse-square law. But it raises the conceptual question: Why can we ignore the self-potential of the particle being accelerated? (This is not an issue for massless test particle.)

  • $\begingroup$ Hi exc3. Welcome to Phys.SE. I removed your last question. Please only ask 1 question per post. $\endgroup$
    – Qmechanic
    Commented Apr 19, 2022 at 10:22
  • 1
    $\begingroup$ Why do we need to argue that it can be "ignored"`? I don't think it obvious that a point mass should "attract itself" and that we need to argue why that is not so. $\endgroup$
    – ACuriousMind
    Commented Apr 19, 2022 at 10:31
  • $\begingroup$ @ACuriousMind In electromagnetism we do need to consider self attraction. Why not in Newton gravitation? The thought makes sense. $\endgroup$
    – my2cts
    Commented Apr 19, 2022 at 12:17
  • $\begingroup$ This might be relevant: physics.stackexchange.com/q/439786/122952 $\endgroup$
    – NDewolf
    Commented Apr 19, 2022 at 12:54

1 Answer 1


Just suppose that you just have a particle A with mass M, and A will just stay where it is or move in a straight line at constant speed. A cannot attract itself. And for $\Phi=-\frac{GM}{r}$, it should be noted that the domain of function is $r>0$ with $0$ excluded.


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