# Formula for perihelion speed as a function of circular speed and eccentricity

I'm trying to work out simple formulas for orbits that use a particular circular orbit as a standard for distance and speed, and then talk about other orbits in reference to the first.

So one way to vary an orbit is to keep the perihelion distance the same, but to change the eccentricity and so change the perihelion velocity. We know the final form of the equation should match certain easy cases

(e=eccentricity)

For circular orbits, $e=0$, our function should match $v=v_{circular}$

and

For escape velocity, $e=1$, our function should match $v=\sqrt 2 ~v_{circular}$

(using subscripts 0 for the circular reference orbit, and 1 is perihelion and 2 is aphelion.)

For the reference orbit, we have, $$\frac{mv_0^2}{r_0}=F=\frac{GMm}{r_0^2}$$ $$GM=r_0 v_0^2$$

Using conservation of energy, we get $$E_1=E_2$$

$$\frac{1}{2} m v_1^2-\frac{GMm}{r_1}=\frac{1}{2} m v_2^2-\frac{GMm}{r_2}$$

$$\frac{1}{2} v_1^2-\frac{GM}{r_1}=\frac{1}{2} v_2^2-\frac{GM}{r_2}$$

$$\frac{1}{2} v_1^2-\frac{r_0 v_0^2}{r_1}=\frac{1}{2} v_2^2-\frac{r_0 v_0^2}{r_2}$$

$$r_1=r_0$$

$$\frac{1}{2} v_1^2-v_0^2=\frac{1}{2} v_2^2-\frac{r_1 v_0^2}{r_2}$$

Using conservation of angular momentum, we get $$m r_1 v_1=m r_2 v_2$$ $$v_2=\frac{r_1}{r_2} v_1$$
$$\frac{1}{2} v_1^2-v_0^2=\frac{1}{2} (\frac{r_1}{r_2})^2 v_1^2-\frac{r_1 v_0^2}{r_2}$$

$$v_1^2-(\frac{r_1}{r_2})^2 v_1^2=2(v_0^2-\frac{r_1}{r_2} v_0^2)$$

$$v_1^2(1-(\frac{r_1}{r_2})^2)=2 v_0^2(1-\frac{r_1}{r_2})$$

So here, we can see the algebra is getting jammed up. If I divide both sides by $1-\frac{r_1}{r_2}$ then that excludes the formula from applying to when $r_1=r_2$, but the formula I want should apply in that case.

If we go ahead and do this division anyway, we get

$$v_1^2(1+\frac{r_1}{r_2})=2 v_0^2$$

$$1+\frac{r_1}{r_2}=1+\frac{1-e}{1+e}=\frac{2}{1+e}$$

$$v_1^2(\frac{2}{1+e})=2 v_0^2$$

$$v=v_0 \sqrt{1+e}$$

Which matches my easy cases at the top. And so should be a pretty good solution.

But the derivation here excluded $e=0$, so I'm troubled.

Any ideas?

1. Use conservation of angular momentum to say that $mv_1r_1 = mv_2r_2 \implies v_1 = v_2$ when $r_1 = r_2.$ Then, you can note that this fills in the discontinuity in $v=v_0\sqrt{1+e}$ when $e=0.$
2. Make a limit argument so that $r_1 \neq r_2$. Starting from the last non-jammed equation
$$v_1^2\left(1-\left(\frac{r_1}{r_2}\right)^2\right)=2 v_0^2\left(1-\frac{r_1}{r_2}\right)$$ $$v_1^2=2 v_0^2\left(\frac{1-\frac{r_1}{r_2}}{1-\left(\frac{r_1}{r_2}\right)^2}\right)$$ $$\lim_{r_1 \to r_2}v_1^2=\lim_{r_1 \to r_2}2 v_0^2\left(\frac{1-\frac{r_1}{r_2}}{1-\left(\frac{r_1}{r_2}\right)^2}\right)$$ $$\lim_{r_1 \to r_2}v_1^2=\lim_{r_1 \to r_2}2 v_0^2\left(\frac{1}{1+\frac{r_1}{r_2}}\right)$$ $$\lim_{r_1 \to r_2}v_1^2=v_0^2$$