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i have a problem. we have a elliptical orbit. Velocities of perigee point and apogee point is given. I have to find the eccentricity and semi major axis.

I found eccentricty with conservation of momentum. but couldn't find the semi major axis. I tried

$V_p^2$ = ($\mu$ (1+e))/ $r_p$

i found $r_p$ and $r_a$ with that formula but when I do the eccentricity formula with that radius the eccentricity is not match with the I found before.

How can I obtain the semi major axis? I am pretty sure first eccentricity is the right answer.

$$h = momentum = r_a∗V_a= r_p∗V_p$$ i know the $$V_a/ V_p$$ and with ratio of that i used $$r_p/r_a=(1−e)/(1+e)$$ with that i found e.

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  • $\begingroup$ Welcome on Physics SE :) Please see physics.stackexchange.com/help/notation for the correct use of TeX to make your formulas a bit easier to read. Also, could you explain where the formula you use for $r_p$ and $r_a$ comes from? $\endgroup$ – Sanya Aug 10 '16 at 22:10
  • $\begingroup$ h = momentum = $r_a * V_a$ = $r_p * V_p$ i know the $V_p$ and $V_p$ with ratio of that i used $$r_p / r_a = (1-e) / (1+e) $$ with that i found e $\endgroup$ – letham Aug 10 '16 at 22:22
  • $\begingroup$ Hi letham, I edited your post, could you please change any mistakes I made? Thanks $\endgroup$ – user108787 Aug 10 '16 at 22:24
  • $\begingroup$ @letham , the following link tells you how to calculate semi-major axis, given eccentricity and radius at either apogee or perigee. en.wikipedia.org/wiki/Orbital_eccentricity $\endgroup$ – David White Aug 11 '16 at 1:04
  • $\begingroup$ @DavidWhite checked it already. the results are wrong with that. $\endgroup$ – letham Aug 11 '16 at 11:24
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Total energy is constant, so
$V_p^2-\frac{2GM}{r_p}=V_a^2-\frac{2GM}{r_a}$
$V_p^2-V_a^2=2GM(\frac{1}{r_p}-\frac{1}{r_a})$
$(V_p+V_a)(V_p-V_a)=2GM\frac{1}{r_p}(1-\frac{r_p}{r_a})=2GM\frac{1}{r_p}(1-\frac{V_a}{V_p})=2GM\frac{1}{r_p}(\frac{V_p-V_a}{V_p})$
$V_p+V_a=\frac{2GM}{r_pV_p}$
where I have used $r_pV_p=r_aV_a$.
Assuming you know $M$ then from this you can find $r_p$ then $r_a$.

Finally the semi-major axis is $a=\frac12(r_p+r_a)$.

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  • $\begingroup$ They don't have $M$, per se, they have $\mu$. $\endgroup$ – PM 2Ring Jun 24 '18 at 5:54

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