Gravitationally Equivalent Point Mass

We want to shrink the rod to a point such that the gravitational field at $x=2$ is the same as before.

Our first instinct might be to shrink it at the point $x=0.5$ . However, this is wrong because the parts of the rod closer to $x=2$ contribute (quadratically) more to the field than the parts of the rod away from it. This is why the point mass should be closer to $x=2$ than $x=0.5$ .

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To see this rigorously, we want to pick a point $x$ such that $\frac{M}{(2-x)^2} = \int^{1}_{0}{\frac{M \, dl}{(2-l)^2}}$

The integral can be easily solved: $\int^{1}_{0}{\frac{M \, dl}{(2-l)^2}} = \int^{1}_{2}{\frac{-M \,dy}{y^2}} = \frac{M}{2}$

So, now that we have $\frac{M}{(2-x)^2} = \frac{M}{2} \implies (2-x)^2 = 2$ , it can be clearly seen that $x > \frac{1}{2}$

The portion from x=0.5 to x=1 will experience a greater force than the portion from x=0 to x=0.5 because it is closer to the point mass. Therefore, if the rod is to be replaced by a point mass, then that point mass should be placed somewhere in between x=0.5 to x=1 m.