Calculating object velocity at perihelion

Question

I would like to calculate the velocity of an asteroid orbiting around a star (Sun) at the perihelion of its orbit. I know the excentricity of the ellipse and its semimajor axis.

I have found that the vis-viva equation is used to calculate the velocity of an object on an elliptical orbit and that the perihelion is at distance r = a(1-e). However I (simply enough) cannot see how to mathematically combine these two pieces of information in order to get the velocity at the perihelion.

(I am not so much looking for just a formula but rather a proof/intuition regarding how to get from the vis-visa equation for velocity to a perihelion velocity equation)

Answer 1

The vis-viva equation is commonly written like this:

$$v^2 = GM\left(\frac{2}{r} - \frac{1}{a}\right)$$.

For $r=a(1-e)$:

$$v = \sqrt{GM\left(\frac{2}{a(1-e)} - \frac{1}{a}\right)} = \sqrt{GM\frac{1}{a}\left(\frac{2}{1-e}-1\right)} = \sqrt{GM\frac{1}{a}\left(\frac{1+e}{1-e}\right)}$$.

The derivation of the vis-viva equation is not at all trivial and can be found here.

The product $GM$ is also called the standard gravitational parameter and for solar system bodies is often known more accurately than $G$ and $M$ separately. For the Sun $GM_☉$ is about 1.327E+20 m³ s⁻² which in different units is 1.327E+11 km³ s⁻² or about 1.0 AU³ year⁻².

Answer 2

You can do this without having to know or derive the vis-viva equation, just by applying conservation of energy and angular momentum.

At perihelion and aphelion the velocities are purely tangential, so conservation of angular momentum yields $$ r_p v_p = r_a v_a\ ,$$ $$ a(1-e)v_p = a(1+e)v_a\ .$$ Conservation of energy (potential plus kinetic) then gives $$ -\frac{GM}{r_p} + \frac{v_p^{2}}{2} = -\frac{GM}{r_a} + \frac{v_a^2}{2}\ , $$ $$ -\frac{GM}{a(1-e)} + \frac{v_p^{2}}{2} = -\frac{GM}{a(1+e)} + \frac{v_a^2}{2}\ . $$ Eliminating $v_a$ gives $$ v_p = \left( \frac{GM}{a}\frac{(1+e)}{(1-e)}\right)^{1/2}\ . $$