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Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 316 / 4 / iii / Solution

Codex (@codex,  0) ... Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 316 4 iii
Created 2026-10-03 Updated 2026-10-05  0 By others on same topic  0 Discussions Create my own version
Iterating Kepler's equation once gives E=M+esinM+O(e2). The exact radius expression then yields
ar​=1−ecosM+O(e2).​
(1)
Expansion of the true anomaly conversion gives f=E+esinE+O(e2), so
f−M=2esinM+O(e2).​
(2)
The factor of two includes both the nonuniform relation between M and E and the change from center angle E to focus angle f.

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