<p>It looks like you did your math wrong.  I'll write out the steps for you:</p>

<p>Square both sides:
$b^2=a^2(1-e^2)$</p>

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

<p>Isolate $e^2$:
$e^2=1-\frac{b^2}{a^2}$</p>

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

<p>$e = \sqrt{1-\frac{b^2}{a^2}}$</p>

CC BY-SA 3.0
2016-10-13T18:48:06.260Z
I'm currently writing a TI-BASIC program, and this will help tremendously. another question though: how would one do
Rory Yammomoto
CC BY-SA 3.0
2016-10-13T18:48:46.023Z
12rv=πab/P for b?
Rory Yammomoto
CC BY-SA 3.0
2016-10-13T18:51:38.593Z
made a mistake, 1/2rv=πab/P.
Rory Yammomoto
CC BY-SA 3.0
2016-10-13T19:02:43.037Z
@RoryYammomoto What do you want to solve that one for?  Which variable?
Phiteros
CC BY-SA 3.0
2016-10-13T23:18:21.013Z
Oh, for b.  That one will be $b=\frac{2\pi a}{Prv}$
Phiteros
CC BY-SA 3.0
2016-10-13T23:20:01.617Z
you can substitute that into the equation for $e$ to get $e=\sqrt{1-(\frac{2\pi}{Prv})^2}$
Phiteros
CC BY-SA 3.0
2016-10-14T15:57:24.940Z
Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/46805/discussion-between-rory-yammomoto-and-phiteros).
Rory Yammomoto
CC BY-SA 3.0
2016-10-17T16:31:23.530Z
this does not seem to work. It turns out my original answer for (1/2)RV=πAB/P for B, (((1/2)RVP)/(πA)) was correct enough.
Rory Yammomoto
CC BY-SA 3.0
2016-10-13T17:33:31.700Z
2016-10-13T17:33:31.700Z
Phiteros
<p>Following on from <a href="https://astronomy.stackexchange.com/questions/1137/calculate-apsides-without-knowing-eccentricity">this question</a>, I wish to know how to find the <em>reverse</em> - how to find eccentricity given a Semi-minor axis &amp; altitude. </p>

<p>I want to use something based on</p>

<p>$$b=a\sqrt{1−e^2}$$</p>

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

<p>$$e=a\sqrt{1}-b$$</p>

<p>could anyone please help?</p>

CC BY-SA 3.0
2016-10-13T17:35:55.827Z
I assume that by altitude you mean $a$, the length of the semimajor axis?
Phiteros
<orbital-elements><orbital-mechanics>
Calculate eccentricity with altitude and semiminor axes
Rory Yammomoto
# Calculate eccentricity with altitude and semiminor axes

(Asked by: username_0 on 2016-10-13)

Following on from [this question](https://astronomy.stackexchange.com/questions/1137/calculate-apsides-without-knowing-eccentricity), 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?

**Tags:** orbital-elements, orbital-mechanics

> **Comments:**
> **username_1**: I assume that by altitude you mean $a$, the length of the semimajor axis?  

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## 1 answers

### Answer by username_1 on 2016-10-13. Score: 1

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:

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

> **Comments:**
> **username_0**: I'm currently writing a TI-BASIC program, and this will help tremendously. another question though: how would one do  
> **username_0**: 12rv=πab/P for b?  
> **username_0**: made a mistake, 1/2rv=πab/P.  
> **username_1**: @RoryYammomoto What do you want to solve that one for?  Which variable?  
> **username_1**: Oh, for b.  That one will be $b=\frac{2\pi a}{Prv}$  
> **username_1**: you can substitute that into the equation for $e$ to get $e=\sqrt{1-(\frac{2\pi}{Prv})^2}$  
> **username_0**: Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/46805/discussion-between-rory-yammomoto-and-phiteros).  
> **username_0**: this does not seem to work. It turns out my original answer for (1/2)RV=πAB/P for B, (((1/2)RVP)/(πA)) was correct enough.  

---

