 Solution

During quasi-static tether contraction, momentum conservation keeps ${\mathbf{V}}_{\mathrm{CM}}$ constant and conservation of angular momentum keeps $L$ constant. To leading order in the slow radial motion, the kinetic energy isIts derivative at fixed $L$, not fixed $\omega$, isTo compute the power supplied by the tension, use the instantaneous unit vector $\mathbf{n}$ from the lower to the upper ship. Their radial velocities relative to the centre of mass are $\pm (\dot{l}/2)\mathbf{n}$, and the tension forces are $\mp T\mathbf{n}$. ConsequentlyThe translational contributions cancel between the two ships and the tangential contributions are zero. The contraction mechanism supplies positive work, increasing the rotational kinetic energy.

The slow-contraction approximation matters: for a prescribed smooth $l(t)$ with a taut tether, the exact kinetic energy and tension include radial terms,These still satisfy $d{E}_{\mathrm{exact}}/dt=-T_{\mathrm{exact}}\dot{l}$. Thus the paper's displayed formula is the leading result for slow, smooth contraction, or is exact when the radial kinetic energy is constant; it is not an identity for every possible shrinking schedule.