Easy way to get excentricity

Hi,
it has been few days since I looked on some problems from astronomy contest. One of the problem was about excentricity of Pluto.


This has to do with ellipses since according to Kepler’s laws, planets are orbiting on ellipses, shapes just a little different from circles, at least when you consider their equation.

For circle equation is x2+y2=1

For ellipse equation is x2/a+y2/b=1

Sun is always the focus of the elipse, above those two points are -c,0 and c,0. For planets those ellipses are much less excentric which means that in the equation above, “a” and “b” are fairly similar.

In the problem I knew only perihelium and afelium of Pluto.

e = \frac{\varepsilon}{a}=\frac{\sqrt{a^2-b^2}}{a}

“e” is excentricity. “a” is semi-major axis. “b” is semi-minor axis. “ε” is linear excentricity (not really important).

Since the equation goes as the one above you need both semi-major and semi-minor axis to get the result. From knowing afelium and perihelium I easily got semi-major axis. To get semi-minor you must know that the distance from focal point to the top or bottom of ellipse is equal to semi-major axis, from this you can use pythagorean theorem and then all this information insert into the equation. All went right and with perihelium of 29.66 AU and afelium of 49.32 AU the excentricity is 0.246 which is just right, if excentricity is equal to 1 than it is parabola and if greater it is hyperbola.

Dragallur

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