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 $$x^3-3x^2+\left(3+\dfrac{1}{k^2}\right)x-\left(1-\dfrac{1}{k^2}\right)=0$$. 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)