A rigid truss of length on an orbiting station is held transverse (perpendicular) to…

PHYSICS · Multi-part · Fermi University Physics · Relativity & Gravitation · Newtonian gravitation as a field theory · Newtonian gravitation as a field theory

The Geometry of Gravity, Ch The tidal field: what free fall cannot remove

Problem

A rigid truss of length ℓ=20.0 m\ell = 20.0\ \mathrm{m} on an orbiting station is held transverse (perpendicular) to the radius from the Earth's centre, at distance r=6.771×106 mr = 6.771\times10^6\ \mathrm{m} (an altitude of 400 km400\ \mathrm{km} above the surface). Treat the Earth as a point mass M⊕=5.972×1024 kgM_\oplus = 5.972\times10^{24}\ \mathrm{kg} and use G=6.674×10−11 N m2 kg−2G = 6.674\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}. Ignore the truss's own gravity and any non-gravitational forces. ((a)) Write Eij\mathcal{E}_{ij} in radially aligned axes as a diagonal matrix with numerical entries in s−2\mathrm{s^{-2}}, and confirm that its trace vanishes. ((b)) Find the tidal acceleration −EijSj-\mathcal{E}_{ij}S^j between the two ends of the truss.

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