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I need to calculate the eccentricity of a satellite orbit with only the apside altitudes (height above the surface of the Earth), 400 and 1500 km. However, I don't know the semi-major axis ($a$) or semi-minor axis ($b$).

$$\text{eccentricity} = (1-(b/a)^2)^{1/2}$$

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closed as off-topic by peterh, StephenG, James K, J. Chomel, Sir Cumference Apr 23 '18 at 19:26

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  • $\begingroup$ Why would you need to do this? $\endgroup$ – James K Apr 20 '18 at 23:15
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    $\begingroup$ I'm voting to close this question as off-topic because this was cross posted on Physics SE. $\endgroup$ – StephenG Apr 20 '18 at 23:50
  • $\begingroup$ @JamesK I need the eccentricity and semi-major axis to input into an orbit generation/propagation code. $\endgroup$ – user12120 Apr 21 '18 at 0:56
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The major axis is just the sum of the distance at perigee plus the distance at apogee. This is 400 km + 1500 km + the diameter (not radius) of the Earth, and the semi-major axis is half of that.

However, the semi-minor axis is harder to determine. Therefore, why not use an alternative method to calculate the eccentricity?

$$e = \frac{r_a-r_p}{r_a+r_p}$$

Here, the $r$'s are the distance to the center of the Earth, not the apside altitudes. Can you take it from here?

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