Newton's Law of Gravitation — Question 6

Question 6

Distance bounds produce asymmetric force bounds

The product $GMm$ is known exactly. Center separation has the hard bound $.98r_0\le r\le1.02r_0$. Let $F_0=GMm/r_0^2$.

1. Give the exact allowed force interval relative to $F_0$.

2. Compare it with a first-order symmetric estimate and explain which is a guaranteed bound.

Question 6 - Solution

Distance bounds produce asymmetric force bounds

(a) Derivation and calculation

Normalize by $x=r/r_0$: $$\frac F{F_0}=x^{-2},\qquad \frac{d(x^{-2})}{dx}=-2x^{-3}<0.$$ The extreme forces occur at opposite distance endpoints: $$\boxed{\frac1{1.02^2}\le\frac F{F_0}\le\frac1{.98^2}} \quad\text{or approximately}\quad 0.961169\le F/F_0\le1.041233.$$

(b) Result and interpretation

Writing $x=1+\delta$, the expansion is $$(1+\delta)^{-2}=1-2\delta+3\delta^2+O(\delta^3).$$ For $|\delta|\le.02$, first order suggests $\pm4\%$, whereas the exact deviations are $$\boxed{-3.8831\%\le(F-F_0)/F_0\le+4.1233\%}.$$ The exact endpoints, rather than the linear estimate, are guaranteed hard bounds.

Physical check

The positive quadratic term explains the asymmetric force interval.