Gravitation and the Principle of Superposition — Question 10

Question 10

A force-free center does not make an extended detector force-free

Two equal fixed sources $M$ lie at $x=\pm a$. A light rigid detector holds two equal test masses $m$ at $x=\pm b$, where $0<b<a$. Ignore their mutual gravity.

1. Find the net gravitational force on the detector and the outward force on either end mass.

2. Find the tension the connecting rod must provide and its leading small-$b$ form.

Question 10 - Solution

A force-free center does not make an extended detector force-free

(a) Derivation and calculation

At the right end, the nearer right source pulls more strongly: $$F_R=GMm\left[\frac1{(a-b)^2}-\frac1{(a+b)^2}\right] =\frac{4GMmab}{(a^2-b^2)^2}.$$ Reflection symmetry gives $F_L=-F_R$, so $$\boxed{F_{\mathrm{net}}=0},\qquad \boxed{|F_{\mathrm{end}}|=\frac{4GMmab}{(a^2-b^2)^2}}.$$

(b) Result and interpretation

To keep either end stationary, inward rod tension balances the outward force: $$\boxed{T=\frac{4GMmab}{(a^2-b^2)^2}}.$$ For $b/a\ll1$, $$T=\frac{4GMmb}{a^3}[1-(b/a)^2]^{-2} \simeq\boxed{\frac{4GMmb}{a^3}}.$$ Zero net force permits nonzero internal tidal stress.

Physical check

Tension tends to zero with detector size and increases as either end approaches a source.