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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)

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  • $\begingroup$ Generally, it is impossible without knowing the mass of the star. If you know that, just substitute r for (1-e)a in the formula ... after you do that, you'll notice that you know all parameters of the equation and can calculate the speed. $\endgroup$
    – Tosic
    Nov 22, 2019 at 8:59
  • $\begingroup$ @Tosic i kinda feel a little dumb now tbh haha, I know the mass of the star, I had everything this whole time... If you put that as an answer I'll accept it ! $\endgroup$
    – A.D
    Nov 22, 2019 at 9:16
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    $\begingroup$ BTW, when doing calculations involving GM it's more accurate to use the standard gravitational parameter than to use separate values of G & M. $\endgroup$
    – PM 2Ring
    Nov 22, 2019 at 10:12
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    $\begingroup$ @PM2Ring alright, glad to gain some knowledge ! Just a question though: how can the sgp be known to a greater accuracy than GM ? (I'm guessing it has to do with how it's calculated but...) $\endgroup$
    – A.D
    Nov 22, 2019 at 10:29
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    $\begingroup$ @A.D That Wikipedia article briefly explains: "Conversely, measurements of the smaller body's orbit only provide information on the product, μ, not G and M separately." Also see space.stackexchange.com/a/39930 It's really hard to measure G to high precision. It's generally done using some variation of the Cavendish torsion balance experiment. Also see Bending Spacetime in the Basement. $\endgroup$
    – PM 2Ring
    Nov 22, 2019 at 11:35

2 Answers 2

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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.

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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 $$ v_p = \left( \frac{GM}{a}\frac{(1+e)}{(1-e)}\right)^{1/2}\ . $$

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    $\begingroup$ Beautiful! :-) $\endgroup$
    – uhoh
    May 1, 2021 at 23:47
  • $\begingroup$ This helped me a lot too! $\endgroup$
    – ALiCe P.
    Sep 17, 2021 at 6:56

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