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Following on from this question, I wish to know how to find the reverse - how to find eccentricity given a Semi-minor axis & altitude.

I want to use something based on

$$b=a\sqrt{1−e^2}$$

but I have no real way to solve for $e$, for I do not trust my answer,

$$e=a\sqrt{1}-b$$

could anyone please help?

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  • $\begingroup$ I assume that by altitude you mean $a$, the length of the semimajor axis? $\endgroup$ – Phiteros Oct 13 '16 at 17:35
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It looks like you did your math wrong. I'll write out the steps for you:

Square both sides: $b^2=a^2(1-e^2)$

Divide by $a^2$: $\frac{b^2}{a^2}=1-e^2$

Isolate $e^2$: $e^2=1-\frac{b^2}{a^2}$

Taking the square root results in a solution for eccentricity, based on the length of the semiminor and semimajor axes:

$e = \sqrt{1-\frac{b^2}{a^2}}$

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  • $\begingroup$ I'm currently writing a TI-BASIC program, and this will help tremendously. another question though: how would one do $\endgroup$ – Rory Yammomoto Oct 13 '16 at 18:48
  • $\begingroup$ 12rv=πab/P for b? $\endgroup$ – Rory Yammomoto Oct 13 '16 at 18:48
  • $\begingroup$ made a mistake, 1/2rv=πab/P. $\endgroup$ – Rory Yammomoto Oct 13 '16 at 18:51
  • $\begingroup$ @RoryYammomoto What do you want to solve that one for? Which variable? $\endgroup$ – Phiteros Oct 13 '16 at 19:02
  • $\begingroup$ Oh, for b. That one will be $b=\frac{2\pi a}{Prv}$ $\endgroup$ – Phiteros Oct 13 '16 at 23:18

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