Talk:Astronomy on Mercury

When is there retrograde motion

Let T be the orbital period of a planet, e be the eccentricity and k be the spin-orbit resonance. When the planet reaches aphelion at a certain moment, define the longitude facing the sun to be 0. At moment t, the longitude facing the sun is the difference between the orbital revolution angle Θ and the rotation angle θ. We have θ = ⁠2πt/T/k⁠, and Kepler's second law implies ${\dfrac {d\Theta }{dt}}={\dfrac {2\pi }{(1-e^{2})^{\frac {3}{2}}}}{\dfrac {1}{T}}(1-e\cos \Theta )^{2}$ (see below), so the longitude facing the sun Δθ has ${\dfrac {d\Delta \theta }{dt}}={\dfrac {2\pi }{(1-e^{2})^{\frac {3}{2}}}}{\dfrac {1}{T}}(1-e\cos \Theta )^{2}-{\dfrac {2\pi }{T/k}}$. Retrograde motion occurs when ${\dfrac {d\Delta \theta }{dt}}$ changes its sign, namely when $(1-e\cos \Theta )-{\sqrt {k}}(1-e^{2})^{\frac {3}{4}}$ changes its sign. Note that the minimum value (obtained when the planet reaches aphelion) LHS is always negative. Since the maximum of 1−ecos Θ is 1+e (when the planet reaches perihelion), there is retrograde motion if and only if $1+e>{\sqrt {k}}(1-e^{2})^{\frac {3}{4}}$, or e is greater than the unique real root of $x^{3}-3x^{2}+\left(3+{\dfrac {1}{k^{2}}}\right)x-\left(1-{\dfrac {1}{k^{2}}}\right)=0$.

k	Minimum eccentricity to see retrograde motion
1:1	0 (always exists)
3:2	0.1910588914... (the eccentricity of Mercury is greater than this value)
2:1	0.3106016499...
5:2	0.3936079089...
3:1	0.4552115689...

Calculating ⁠dΘ/dt⁠: Suppose that the equation of the planet when orbiting sun is r = ⁠1/1-ecos Θ⁠, then Kepler's second law implies ${\dfrac {1}{2}}r^{2}{\dfrac {d\Theta }{dt}}={\dfrac {dS}{dt}}={\dfrac {S_{ellipse}}{T}}={\dfrac {\pi }{(1-e^{2})^{\frac {3}{2}}}}{\dfrac {1}{T}}$. 129.104.241.193 (talk) 23:08, 17 May 2024 (UTC)