Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 316 / 2 / i / Solution

For $0<e<1$, differentiate the logarithm of the proposed Poynting–Robertson drag invariant:Substitution of the orbit-averaged Poynting–Robertson drag rates gives, over the common denominator $a^2(1-e^2{)}^{3/2}$, the numerator $A[-(2+3e^2)+5e^2+2(1-e^2)]=0$. HenceEquivalently, eliminating time gives $d\log a/de=(2/5)(2+3e^2)/[e(1-e^2)]$, whose integral is $\log a=(4/5)\log e-\log (1-e^2)+\text{constant}$. An exactly circular orbit remains circular; the expression for $C$ is singular at $e=0$ and is replaced by that separate solution.