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Calculating semi-minor axis of an ellipse

Writer Andrew Henderson
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I'm coding a solar system animation and so far it's done, but the the orbits of the planets are circular. To make the simulation more realistic, I want to use elliptic orbits. So I visited Mercury (planet) and Solar System Exploration: Planets: Mercury: Facts & Figures pages, but the semi-minor axis of the Mercury (or any other planet) isn't given.

Then I looked for a way to calculate it and found a formula

$$\epsilon=\sqrt{\frac{a^2-b^2}{a}}=\sqrt{1-\left(\frac{b}{a}\right)^2}$$

in Ellipse - Wikipedia, the free encyclopedia page, but my math knowledge falls short in applying it to my situation.

Mercury
Semi-major axis 57,909,050 km
Eccentricity 0.205630
Aphelion 69,816,900 km
Perihelion 46,001,200 km

Is there a way to calculate the semi-minor axis of Mercury's orbit with the given values?

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1 Answer

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The formula you have got can be rearranged to yield the semi-minor axis,

$b = a \sqrt{1-\epsilon^2}$, where a is the semi-major axis. Substituting each value in the formula gives

$b = (57,909,050) \sqrt{1-0.205630^2}$
$b = 56,671,523 \space km$

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