Suppose we have a thin rod of density $\rho$ and length $L$, and we want to calculate the acceleration due to gravity at one end of the rod. Let this end of the rod be at $x=0$ and the other end be at $x=+L$. The force exerted on a unit mass test particle at $x=0$ by an infinitesimal segment of the rod of length $dx$ at position $x$ from the end is $$\frac{G \rho dx}{x^2}$$ This is the same as the gravitational acceleration due to this infinitesimal segment of the rod. The forces are all in the same direction, along the axis of the rod, so we can add them up. Hence, integrate over the whole rod, i.e. find the integral $$\int_0^L \frac{G \rho}{x^2} dx$$ This does not converge because of the singularity at $x=0$. It suggests that the acceleration due to gravity of the rod at its end is infinite. Another way to do the calculation would be to consider a point outside the rod at a distance $a$ from the end of the rod then let $a\rightarrow 0$. Using its position as the origin, the gravitational acceleration at this point is given by $$\int_a^{a+L} \frac{G \rho}{x^2} dx = \frac {G \rho L}{a(a+L)}$$ Nevertheless, this still gives a singularity when we let $a\rightarrow0$. The fundamental issue is that, with an inverse square law, we seem to get infinities when we consider the surface of an object or a point inside it, at least when using Cartesian coordinates as appropriate for something like a rod. How do we deal with such a situation? Do we simply ignore it and rule the calculation as illegitimate? Or is there some formal reason why this reasoning is wrong? And what technique should we use to get a sensible answer for the acceleration due to gravity at a point in contact with the end of a rod?


1 Answer 1


The fundamental problem is that the gravitational force is not well-defined at points in space where the mass density (expressed as mass per volume) is infinite. In your case, since you are defining the rod to have zero volume, its mass per volume is infinite. The infinity that you are encountering is closely related to the infinity you get from the gravitational field of a point mass $m$ when you take $r \to 0$.

If you want to get a sensible result for the gravitational field at the end of a rod, you can instead look at the case of a cylinder with finite radius $R$. Since we now take the rod to have a volume of $\pi R^2 L$ (not zero, as you had in your calculations), it is possible to find the gravitational field at the center of one of the end-caps. You can use techniques very similar to those in your question to do this, using multiple integrals and cylindrical coordinates to execute the calculation. Since it is a standard exercise (try it!), I will not provide a full derivation, but instead simply cite the result: $$ F = \frac{2 G M \left(\sqrt{R^2+a^2} -\sqrt{(L+a)^2+R^2}+L\right)}{L R^2} $$ where $M$ is the total mass of the cylinder. So long as $R > 0$, this is perfectly well-behaved at $a = 0$. Conversely, if you take the limit $R \to 0$, you will recover your own result — which is divergent at $a = 0$.

  • $\begingroup$ +1 for the first sensible attempt to solve this issue that I have seen on the internet. However, if I replace M with $p L\pi R^2 $ where p is the density and plot the force for finite small positive values of p, R and L with respect to a, the force appears to increase with increasing distance a from the end of the rod. Does that seem possible? $\endgroup$
    – KDP
    Mar 20 at 15:27
  • $\begingroup$ @KDP: I think you've done something wrong with your plot. Making your substitution we have$$F = 2 \pi \rho \left(\sqrt{R^2+a^2} -\sqrt{(L+a)^2+R^2}+L\right)$$and so$$\frac{dF}{da} = 2 \pi \rho \left( \frac{a}{\sqrt{R^2+a^2}} - \frac{L+a}{\sqrt{(L+a)^2+R^2}}\right).$$It's not too hard to show that this expression is negative so long as $\rho$, $a$, $r$, and $L$ are all positive. $\endgroup$ Mar 20 at 15:52
  • $\begingroup$ Yes, it appears I made an error in my plot. The nice thing about your result is if we set a=0 reinstate G and the mass of the test particle (m) so that $F = 2 G m \pi \rho ( R - \sqrt{L^2 +R^2} + L)$ and take the limit as ${L \to \infty}$ the result is $2Gm \rho \pi R$ which is finite. Seriously, this answer deserves more upvotes :-) $\endgroup$
    – KDP
    Mar 20 at 16:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.