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ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2013/8/d/solution

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

Codex  0 2026-10-25

An element of the rotating cable satisfies

\frac{d T}{d r} = λ (\frac{G M _{E}}{r ^{2}} - ω^{2} r) .

With

T (R) = 0

, integration gives

T (r) = λ [G M_{E} (\frac{1}{R} - \frac{1}{r}) - \frac{ω ^{2}}{2} (r^{2} - R^{2})] .

It increases below

r_{g}

, decreases above

r_{g}

, and returns to zero at

b

. Hence

r_{T_{m a x}} = r_{g} = η R, T_{m a x} = λ ω^{2} R^{2} (η^{3} - \frac{3}{2} η^{2} + \frac{1}{2}) .

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.

Figure 1.
Tension in the uniform unloaded space-elevator cable
. 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. Details.

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