So we have a uniform rod of mass M and length L and we have placed a point mass of mass m, which is distance a away from the midpoint of the rod.
Now setting the coordiates axes to work with, I put the rod along the Y axis with its mid point coinciding with the Origin while the point mass m is put along the X axis a distance units away from the rod.
The point mass is P, and a differential mass element dm of the rod situated at y=length(OQ) where Q is the position of the mass element, is taken, and its corresponding force was found to be
$ \displaystyle{ d \vec{F} = \frac{G m dm} {{y}^{2}+{a}^{2}} (- cos \theta \hat{i} + sin \theta \hat{j}) } $
where $\theta$ is the angle OPQ,
Now using the linear mass density,
$ \displaystyle{ \frac{dm}{dx}=\frac{M}{L} } $
and writing $cos \theta$ and $ sin \theta $ in terms of y,
$\displaystyle{ cos \theta = \frac{a}{\sqrt{{y}^{2}+{a}^{2}}} } $
and
$\displaystyle{ sin \theta = \frac{y}{\sqrt{{y}^{2}+{a}^{2}}} }$
the Forces comes out to be,
$\displaystyle{
{F}_{x} = – \frac{GmMa}{L} \int _{-\frac{L}{2}}^{\frac{L}{2}} { \frac{dy} {{({y}^{2}+{a}^{2})}^{\frac{3}{2}}} } }
$
and,
$\displaystyle{
{F}_{y} = \frac{GmM}{L} \int _{-\frac{L}{2}}^{\frac{L}{2}} { \frac{y dy} {{({y}^{2}+{a}^{2})}^{\frac{3}{2}}} } }
$
Which gave the following results:
$\displaystyle{
{F}_{y} = 0 } $
as expected, due to the symmetry.
and,
$\displaystyle{
{F}_{x} = – \frac{GmM}{a \sqrt{{a}^{2} + {(\frac{L}{2})}^{2}}} }
$
which I didn't expect to come out since I believe that if the it was assumed that the whole mass of the rod is concentrated at its mid point then the force should have been
$\displaystyle{
{F}_{x} = – \frac{GmM}{{a}^{2}} }
$
So I am curious as to why or why not this result should follow.
Best Answer
Newton invented integral calculus to prove the point mass premise for spherical objects. Gravitational force as a point mass is in general not valid for other shapes, although it is approximately true at large distances. At $a>>L$, your result approaches what you would expect.