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Cross-section A(r) / A(ground) along the tether (log scale); grey dashed = safety factor 1
Effective gravity along the tether: positive below GEO, zero at GEO, negative above (the tether is pulled up)
Energy per kilogram to climb from the ground to the climber altitude (red dot = climber)
Tether speed (inertial) against circular and escape speed at the release altitude
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The science
Rotating-frame force balance. A cable fixed to the equator co-rotates with Earth at angular speed Omega = 7.2921e-5 rad/s. At distance r from the centre a mass element feels the effective gravity
g_eff(r) = mu / r^2 - Omega^2 r, psi(r) = - mu / r - Omega^2 r^2 / 2, g_eff = dpsi/dr
which is inward below the geostationary radius r_g = (mu / Omega^2)^(1/3) = 42 164 km, zero at r_g and outward above it. Everything below GEO hangs from GEO; everything above pulls up on it.
Constant-stress taper. For a cable of density rho and design stress sigma = strength / SF, the tension is T = sigma A and the force balance of one element is d(sigma A)/dr = rho A g_eff, so
A(r) = A_0 exp[ (rho / sigma) (psi(r) - psi(R_E)) ], taper ratio A(r_g)/A_0 = exp[ (rho / sigma) (psi(r_g) - psi(R_E)) ]
The work per kilogram from the ground to GEO is psi(r_g) - psi(R_E) = 48.4 MJ/kg (13.4 kWh/kg). Because that number sits in an exponential, the specific strength sigma/rho decides everything: with 63 GPa and 1300 kg/m3 at safety factor 2 the cross-section only has to grow 7.4 times, for Kevlar it would have to grow 7e16 times. A cable of constant area survives only if strength / rho > 48.4 MN m/kg.
Above GEO. The same profile continues outward; the area shrinks again and the tether ends at a counterweight of mass m_c = sigma A(r_c) / (Omega^2 r_c - mu / r_c^2), so the shorter the tether above GEO, the heavier the counterweight. The tether mass is rho * integral of A dr (Simpson).
Release speeds. A payload released at radius r moves at the inertial speed Omega r. That equals the circular speed at GEO and exceeds the escape speed sqrt(2 mu / r) beyond r = (2 mu / Omega^2)^(1/3) = 53 124 km, so the cable could throw a payload away from Earth with no rocket.
Honest simplifications. Point-mass spherical Earth (no J2, no Moon or Sun tides), an equatorial anchor, a perfectly straight radial cable (Coriolis lean, climber-induced oscillations, atmospheric and thermal loads, orbital debris, radiation and the real materials problem of defects in nanotube ribbons are not modelled), no safety margin for the ground anchor. Material numbers are typical datasheet values; the carbon nanotube row is aspirational, not a product.
Checks run in the page. The geostationary radius against 42 164 km; Omega r against the circular speed at GEO; the work to GEO as a Simpson integral against the potential difference; the taper ratio from an RK4 integration of the force balance against the closed form; the element balance at five radii; the escape radius by bisection against the closed form; and the tether-mass integral at two resolutions.