Past exam of the physics 1 paper of the University of Oxford / 2013 / 8

A geostationary orbit has the same angular velocity as the Earth. Let $M_E=4\pi \rho R^3/3$ and $r_g=\eta R$. Balancing the required centripetal acceleration with the exterior gravitational field givesThe orbital plane must be equatorial and the motion prograde for the satellite to remain over one point. The uniform density assumption supplies $M_E$ in terms of the specified $\rho$ and $R$.

Taking the Earth's sidereal rotation period as $86164\mathrm{s}$, $\omega =2\pi /(86164\mathrm{s})$, and $G=6.6743\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$ givesThe orbital radius is therefore about $6.61R$, or an altitude of about $5.61R$. Replacing the sidereal day by $86400\mathrm{s}$ changes this estimate by less than $0.2\%$.

For the uniform unloaded space-elevator cable, let the constant linear mass density be $\lambda$ and the outer endpoint be $b={\eta }^{\prime }R$. In the co-rotating frame the outward effective force per unit length is $\lambda ({\omega }^{2}r-G{M}_E/r^2)$. 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:Using $G{M}_E={\omega }^2{\eta }^3{R}^3$, this becomesThe root ${\eta }^{\prime }=1$ is a zero-length cable. The physical nonzero cable hasIt 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.

An element of the rotating cable satisfiesWith $T(R)=0$, integration givesIt increases below $r_g$, decreases above $r_g$, and returns to zero at $b$. HenceThe 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.