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 Incoming links: Gauss planetary equation for the semi-major axis

Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 316 / 2 / i / Solution Created 2026-10-03 Updated 2026-10-06

Treat

a, e

as instantaneous osculating orbital elements of the fixed-central-parameter Kepler orbit. The perturbing work changes the specific orbital energy according to

\overset{ε}{˙} = v \cdot (\overset{ˉ}{R} r + \overset{ˉ}{T} θ) = \overset{ˉ}{R} \overset{r}{˙} + \overset{ˉ}{T} \frac{h}{r}, \overset{a}{˙} = \frac{2 a ^{2}}{μ} \overset{ε}{˙} .

The specific angular momentum and conic equation give

\overset{r}{˙} = \frac{μ}{h} e sin f, \frac{h}{r} = \frac{μ}{h} (1 + e cos f), h = μ a (1 - e^{2}) .

Substituting proves the Gauss planetary equation for the semi-major axis:

\overset{a}{˙} = 2 \frac{a ^{3}}{μ ( 1 - e ^{2} )} [\overset{ˉ}{R} e sin f + \overset{ˉ}{T} (1 + e cos f)] .

Here

f

denotes true anomaly, not the fractional luminosity used in Question 1. The equation follows from work alone; a normal perturbing acceleration has no instantaneous contribution because the velocity lies in the orbital plane. For slowly averaged evolution, the perturbation must be weak over one orbital period.

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