Gravitation and the Principle of Superposition — Question 7

Question 7

A semicircular arc pulls toward its occupied half

A uniform thin semicircular arc of radius $R$ and mass $M$ occupies angles $0\le\theta\le\pi$ above the $x$ axis. Find the field at the circle center.

1. Integrate the Cartesian field components.

2. Compare its magnitude with placing all the mass at the top of the arc.

Question 7 - Solution

A semicircular arc pulls toward its occupied half

(a) Derivation and calculation

The arc length is $\pi R$, so $dm=(M/\pi)d\theta$. Resolving sourceward contributions, $$\vec g=\frac{GM}{\pi R^2}\int_0^\pi (\cos\theta\,\hat i+\sin\theta\,\hat j)\,d\theta.$$ $$g_x=\frac{GM}{\pi R^2}[\sin\theta]_0^\pi=0,\quad \boxed{g_y=\frac{GM}{\pi R^2}[-\cos\theta]_0^\pi=\frac{2GM}{\pi R^2}}.$$

(b) Result and interpretation

Placing the entire mass at the top would produce $$g_{\mathrm{point}}=GM/R^2,\qquad \boxed{g_{\mathrm{arc}}/g_{\mathrm{point}}=2/\pi\simeq0.63662}.$$ The reduction is the mass-weighted vertical projection: $$\langle\sin\theta\rangle=\frac1\pi\int_0^\pi\sin\theta\,d\theta=\frac2\pi.$$

Physical check

Left-right symmetry removes horizontal components; the surviving field points toward the occupied upper half.