<p>You can do this without having to know or derive the vis-viva equation, just by applying conservation of energy and angular momentum.</p>
<p>At perihelion and aphelion the velocities are purely tangential, so conservation of angular momentum yields
<span class="math-container">$$ r_p v_p = r_a v_a\ ,$$</span>
<span class="math-container">$$ a(1-e)v_p = a(1+e)v_a\ .$$</span>
Conservation of energy (potential plus kinetic) then gives
<span class="math-container">$$ -\frac{GM}{r_p} + \frac{v_p^{2}}{2} = -\frac{GM}{r_a} + \frac{v_a^2}{2}\ , $$</span>
<span class="math-container">$$ -\frac{GM}{a(1-e)} + \frac{v_p^{2}}{2} = -\frac{GM}{a(1+e)} + \frac{v_a^2}{2}\ . $$</span>
Eliminating <span class="math-container">$v_a$</span> gives
<span class="math-container">$$ v_p = \left( \frac{GM}{a}\frac{(1+e)}{(1-e)}\right)^{1/2}\ . $$</span></p>

CC BY-SA 4.0
2021-05-01T23:47:26.220Z
*Beautiful!* :-)
uhoh
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2021-09-17T06:56:37.383Z
This helped me a lot too!
ALiCe P.
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2021-04-30T19:22:29.850Z
2021-04-30T19:22:29.850Z
ProfRob
<p>The vis-viva equation is commonly written like this: </p>

<p><span class="math-container">$$v^2 = GM\left(\frac{2}{r} - \frac{1}{a}\right)$$</span>. </p>

<p>For <span class="math-container">$r=a(1-e)$</span>: </p>

<p><span class="math-container">$$v = \sqrt{GM\left(\frac{2}{a(1-e)} - \frac{1}{a}\right)} = \sqrt{GM\frac{1}{a}\left(\frac{2}{1-e}-1\right)} = \sqrt{GM\frac{1}{a}\left(\frac{1+e}{1-e}\right)}$$</span>.</p>

<p>The derivation of the vis-viva equation is not at all trivial and can be found <a href="https://en.wikipedia.org/wiki/Vis-viva_equation" rel="noreferrer">here</a>.</p>

<p>The product <span class="math-container">$GM$</span> is also called the <a href="https://en.wikipedia.org/wiki/Standard_gravitational_parameter" rel="noreferrer">standard gravitational parameter</a> and for solar system bodies is often known more accurately than <span class="math-container">$G$</span> and <span class="math-container">$M$</span> separately. For the Sun <span class="math-container">$GM_☉$</span> 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.</p>

CC BY-SA 4.0
2022-08-14T09:30:18.840Z
In units of years & AU, $GM_\odot$ is $4\pi^2$, so Earth's mean speed is $2\pi$ AU/year.
PM 2Ring
CC BY-SA 4.0
2019-11-22T10:02:33.693Z
2019-11-22T23:52:07.817Z
2019-11-22T23:52:07.817Z
Tosic
<p>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.</p>

<p>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. </p>

<p>(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)</p>

CC BY-SA 4.0
2019-11-22T08:59:56.407Z
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.
Tosic
CC BY-SA 4.0
2019-11-22T09:16:54.830Z
@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 !
A.D
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2019-11-22T10:03:25.450Z
No problem. There is no shame in asking, but in refusing to learn :-)
Tosic
CC BY-SA 4.0
2019-11-22T10:12:52.007Z
BTW, when doing calculations involving GM it's more accurate to use the [standard gravitational parameter](https://en.wikipedia.org/wiki/Standard_gravitational_parameter) than to use separate values of G & M.
PM 2Ring
CC BY-SA 4.0
2019-11-22T10:29:32.190Z
@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...)
A.D
CC BY-SA 4.0
2019-11-22T11:35:35.897Z
@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 https://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](https://en.wikipedia.org/wiki/Cavendish_experiment). Also see [Bending Spacetime in the Basement](https://www.fourmilab.ch/gravitation/foobar/).
PM 2Ring
<orbit><velocity>
Calculating object velocity at perihelion
A.D
# Calculating object velocity at perihelion

(Asked by: username_0 on 2019-11-22)

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)

**Tags:** orbit, velocity
> **Comments:**
> **username_1**: 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.  
> **username_0**: @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 !  
> **username_1**: No problem. There is no shame in asking, but in refusing to learn :-)  
> **username_2**: BTW, when doing calculations involving GM it's more accurate to use the [standard gravitational parameter](https://en.wikipedia.org/wiki/Standard_gravitational_parameter) than to use separate values of G & M.  
> **username_0**: @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...)  
> **username_2**: @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 https://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](https://en.wikipedia.org/wiki/Cavendish_experiment). Also see [Bending Spacetime in the Basement](https://www.fourmilab.ch/gravitation/foobar/).  

---

## 2 answers

### Answer by username_1 on 2019-11-22. Score: 7 | Accepted answer

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](https://en.wikipedia.org/wiki/Vis-viva_equation).

The product $GM$ is also called the [standard gravitational parameter](https://en.wikipedia.org/wiki/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.

> **Comments:**
> **username_2**: In units of years & AU, $GM_\odot$ is $4\pi^2$, so Earth's mean speed is $2\pi$ AU/year.  

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### Answer by username_3 on 2021-04-30. Score: 6

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

> **Comments:**
> **username_4**: *Beautiful!* :-)  
> **username_5**: This helped me a lot too!  

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