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I had a rather theoretical problem with the elliptical velocity equation while glancing the formulas and proofs. While doing the proof there was a statement of energy conservation : $$K.E.+P.E.=M.E.$$ $$\frac{1}{2}mv^2-\frac{GMm}{r}=-\frac{GM}{2a}=\frac{-GM}{r+r_p}$${where v is the velocity of the object on the ellipse, a is semi major axis ,r is the distance of the object from the sun(or here the distance of the apogee) and $r_p$ is the distance of the perigee}

Now according to this if the velocity is increased $r_p$ should decrease therefore making the orbit smaller(since r is constant also shown in picture).

enter image description here

Mathematically: $$\frac{\frac{1}{2}mv^2}{GM}=\frac{1}{r}-\frac{1}{r+r_p},$$if v increases then: $\frac{1}{r+r_p}$ should also increase since $\frac{1}{r}$ is constant. This means that $r_p$ should decrease. Howewer in Hohmann's transfer orbit increasing the velocity should increase the eccentricity therefore increasing $r_p$.

What am i doing wrong here?

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You have already gotten to the correct mathematical equation yet you are confused about the minus sign. $$\frac{\frac12mv^2}{GMm}=\frac1r-\frac1{r+r_P}$$ is correct. As $v$ increases, the RHS also increases. $\frac1r$ is a constant, so $-\frac1{r+r_P}$ has to increase, meaning $\frac1{r+r_P}$ decreases, $r_P$ increases. This agrees with the standard treatments saying that eccentricity increases.

Maybe you might like it better if you had rewritten it as $$r+r_P=\frac1{\frac1r-\frac{v^2}{2GM}}$$ so that, by dividing by a smaller number, you see that $r_P$ has to increase.

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  • $\begingroup$ Ohh right. I totally forgot about inverting the sign. Thanks a lot for the help. I actually just started with orbital mechanics and ran into a problem right after the proofs. Probably need to study more about it $\endgroup$
    – Star Gazer
    Commented Feb 1 at 8:22

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