= Solution
For nonretarded attraction between planar half-spaces, the interaction <free energy> per area is $W(h)=-A_H/(12\pi h^2)$. The <Derjaguin approximation> gives $F=2\pi R_{\rm eff}W(h)$, with $R_{\rm eff}=R_1R_2/(R_1+R_2)=a/2$ for equal spheres. Thus
$$
\boxed{F=-\frac{A_Ha}{12h^2}=-\frac{(15.6\times10^{-20})(10^{-3})}{12(10^{-9})^2}=-1.30\times10^{-5}\,\mathrm N.}
$$
The <force> is \b[$13.0\,\mu\mathrm N$, attractive]. Scaling both radii and the gap by two gives $F'=F/2$, or \b[$6.50\,\mu\mathrm N$ attractive]. This agrees with the invariant energy $V\simeq-A_Ha/(12h)$: differentiating with respect to a gap that doubles makes the <force> halve. The approximation requires $h\ll a$ and neglects roughness and retardation.
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