Gravitation and the Principle of Superposition — Question 8

Question 8

Distant fields retain a measurable imprint of source separation

Two equal fixed sources $M$ lie at $x=\pm a$. A point is on the positive axis at $x\gg a$. Use $(1+u)^{-2}\simeq1-2u+3u^2$ through second order.

1. Find the leading correction to the field of a single mass $2M$ at the origin.

2. Explain why no first-order term in $a/x$ remains.

Question 8 - Solution

Distant fields retain a measurable imprint of source separation

(a) Derivation and calculation

For $x>a$, both attractions point left. Define $u=a/x\ll1$: $$|g_x|=\frac{GM}{x^2}\left[(1-u)^{-2}+(1+u)^{-2}\right].$$ $$|g_x|\simeq\frac{GM}{x^2} [(1+2u+3u^2)+(1-2u+3u^2)] =\boxed{\frac{2GM}{x^2}\left[1+3(a/x)^2\right]}.$$

(b) Result and interpretation

Relative to the point-source result, $$g_{\mathrm{point}}=\frac{2GM}{x^2},\qquad \frac{|g_x|-g_{\mathrm{point}}}{g_{\mathrm{point}}}\simeq3(a/x)^2.$$ The source distribution is invariant under $a\mapsto-a$, so odd powers cancel. The leading finite-separation correction is positive and quadratic.

Physical check

The correction disappears as $a/x\to0$, recovering the field of total mass $2M$.