= Solution
For the <uniform unloaded space-elevator cable>, let the constant <linear mass density> be $\lambda$ and the outer endpoint be $b=\eta'R$. In the co-rotating frame the outward effective <force> per unit length is $\lambda(\omega^2r-GM_E/r^2)$. The cable is just clear of the surface, with neither a ground load nor an endpoint counterweight, so both ends have zero applied <tension>. The integrated effective <force> must vanish:
$$
0=\lambda\int_R^b\left(\omega^2r-\frac{GM_E}{r^2}\right)dr=\lambda\left[\frac{\omega^2}{2}(b^2-R^2)+GM_E\left(\frac1b-\frac1R\right)\right].
$$
Using $GM_E=\omega^2\eta^3R^3$, this becomes
$$
0=(\eta'-1)\left[\frac{\eta'+1}{2}-\frac{\eta^3}{\eta'}\right].
$$
The root $\eta'=1$ is a zero-length cable. The physical nonzero cable has
$$
\boxed{\eta'=\frac{\sqrt{1+8\eta^3}-1}{2}\simeq23.6.}
$$
It extends well beyond the <geostationary orbit> so its outer <centrifugal force> balances the inner segment's weight. If a cable is anchored with a nonzero endpoint load or has a counterweight, that load changes the integrated balance; its length is not fixed by the zero-end-load formula.
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