= Solution
An element of the rotating cable satisfies
$$
\frac{dT}{dr}=\lambda\left(\frac{GM_E}{r^2}-\omega^2r\right).
$$
With $T(R)=0$, integration gives
$$
T(r)=\lambda\left[GM_E\left(\frac1R-\frac1r\right)-\frac{\omega^2}{2}(r^2-R^2)\right].
$$
It increases below $r_g$, decreases above $r_g$, and returns to zero at $b$. Hence
$$
\boxed{r_{T_{\max}}=r_g=\eta R,\qquad T_{\max}=\lambda\omega^2R^2\left(\eta^3-\frac32\eta^2+\frac12\right).}
$$
The <tension> is greatest at the <geostationary orbit>, where local effective gravity is zero. A zero local effective <force> does not mean zero <tension>: that section transmits the loads of the entire inner and outer cable portions.
\Image[/past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2013-space-elevator-tension.png]
{title=Tension in the uniform unloaded space-elevator cable}
{description=The tension vanishes at both ends and peaks at the geostationary radius. The outer endpoint follows from the integrated rotating-frame force balance, using the specified uniform Earth density.}
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