For an orbit around the earth with period equal to a sidereal day with eccentricity $e=0$ and inclination $i=0$, the path on earth would be a single point on the equator. I have done some simulations that suggests that if $e>0$ and $i>0$ the path would take the form of the figure '8'. If $e\gg 0$ one loop in the figure would be less than the other and greater $i$ gives a taller '8'.

My question is: what would the figure be if $e=0$ and $i=90^{°}$?

  • 1
    $\begingroup$ Intuitively, a large circle, but hopefully someone can do the math. See the Analemma Tower project: dezeen.com/2017/03/23/… $\endgroup$
    – safesphere
    Nov 13, 2018 at 22:13


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