Gravitation and the Principle of Superposition — Question 5

Question 5

A finite rod cannot always be replaced by a point at its midpoint

A uniform thin rod of mass $M$ occupies $0\le x\le L$. A test particle is at $x=-a$, with $a>0$.

1. Integrate the gravitational field at the particle.

2. Compare with a point mass at the midpoint and recover the distant limit.

Question 5 - Solution

A finite rod cannot always be replaced by a point at its midpoint

(a) Derivation and calculation

Let $\lambda=M/L$. An element at $x$ has $dm=\lambda\,dx$ and is distance $a+x$ from the particle: $$dg_x=\frac{G\lambda\,dx}{(a+x)^2}.$$ $$g_x=\frac{GM}{L}\left[-\frac1{a+x}\right]_0^L =\frac{GM}{L}\left(\frac1a-\frac1{a+L}\right) =\boxed{\frac{GM}{a(a+L)}}.$$

(b) Result and interpretation

The midpoint approximation is $g_{\mathrm{mid}}=GM/(a+L/2)^2$. Thus $$\frac{g_x}{g_{\mathrm{mid}}}= \frac{(a+L/2)^2}{a(a+L)}=1+\frac{L^2}{4a(a+L)}>1.$$ For $L/a\ll1$, $$g_x=\frac{GM}{a^2}\frac1{1+L/a} \simeq\frac{GM}{a^2}\left(1-\frac La+\cdots\right).$$ Both fields approach $GM/a^2$, while the midpoint model underestimates the exact finite-distance field.

Physical check

Every element pulls toward the rod, so the result must have positive $x$ direction.