How far is it from the Venus?

Does Mercury gives too big perturbations for a stable Lissajous orbit?

  • $\begingroup$ As I know, L1, L2 and L3 are so highly unstable, that even this Lissajous orbital needs periodic fuel burns. But maybe somebody knows more. $\endgroup$
    – peterh
    Commented Sep 12, 2014 at 8:04
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    $\begingroup$ Some interesting comments are in dx.doi.org/10.1134/S0010952510050084 (a Venus L2 halo orbit for a Near-Earth Asteroid watch satellite). And no, Mercury is a non-entity here. $\endgroup$ Commented Sep 12, 2014 at 9:20

5 Answers 5


I get SVL1 as 1,007,927 kilometers from Venus' center. But that's too much precision. Since Venus' orbit isn't perfectly circular, you don't have that many significant digits. I usually say about a million kilometers.

I've made a spreadsheet calculating L1 and L2 distances from various bodies orbiting the sun: Sun plus Mercury, Venus, Earth, Mars, Ceres, Jupiter, Saturn, Uranus, Neptune, and Pluto.

The spreadsheet also gives various planet moon L1 and L2 distances.

The spreadsheet is based on equations found on pages 133 to 138 of Szebehely's Theory of Orbits - The Restricted 3 Body Problem.

I don't think Mercury is that much of an influence. I would guess pressure from sunlight would have a greater effect.

But even if the sun and venus and probe were a perfect circular 3-body system, L1 would not be stable. You would need station keeping propellent in any event.


Lagrange points move with the distance between primary and secondary body's barycenter, so the exact distance between Venus and $\text{SVL}_1$ (Sun-Venus Lagrange point 1) would depend on Cytherean distance to the Sun in its orbit. For mean distance though, we can simplify this to Cytherean semi-major axis ($108,208,000\ \text{km}$ or $0.723327\ \text{AU}$). Using following formula (refer to The Lagrangian Points, An Application of Linear Algebra, Hannah Rae Kerner, 2013 for how it is derived):

$$L_1 = R \left(1 -\left(\sqrt[3\ \ \ ]{\frac{\alpha}{3}}\right)\right)$$

Where $\alpha$ is the ratio of $M_2$ (secondary body's mass) to the combined mass of the system ($M_1 + M_2$), or $\frac{M_2}{M_1 + M_2}$.

So in our case, since Cytherean mass is $0.815\ M_E$ and the Sun's mass $333,000\ M_E$ ($M_E$ is Earth's mass), this mass ratio $\alpha$ is $2.4474 \cdot 10^{-6}$ and the mean $\text{SVL}_1$ distance to Venus is $-0.009344\ R$ or, to make it fit our formula, $0.99057\ R$ from the Sun (primary body).

So, in kilometers and astronomical units (AU), mean distance of $\text{SVL}_1$ is then:

  • $-1,011,090\ \text{km}$ or $-0.00676\ \text{AU}$ from Venus, or
  • $107,196,910\ \text{km}$ or $0.71657\ \text{AU}$ from the Sun

I calculated this using more precise values without cutting decimals off, but to confirm it, let's cheat a bit and use one of the online Lagrange point calculators. It assumes circular orbit (so only mean distance will be somewhat precise, like the one I calculate), and the figures for $a$ of $108,208,000\ \text{km}$, $M_1 = 1\cdot M_\odot$ (one solar mass) and $M_2 = 1\cdot M♀$ (mass of Venus) come out as:

  • $-1,007,985\ \text{km}$ or $-0.00674\ \text{AU}$ from Venus, or
  • $107,200,015\ \text{km}$ or $0.71659\ \text{AU}$ from the Sun

So fairly close, but I'm unsure what values that calculator uses as input, or at which point it cuts off decimal places so some small-ish discrepancy was expected.

If we also apply Cytherean orbital eccentricity to our calculations ($e = 0.0067$), we get a movement of the $\text{SVL}_1$ distance to Venus from $-1,004,259\ \text{km}$ or $-0.006713\ \text{AU}$ at perihelion to $-1,017,920\ \text{km}$ or $-0.006804\ \text{AU}$ at aphelion.

Or simply $-1,011,090\ \text{km} \pm 0.67\%$.

I'll let others describe $\text{SVL}_1$ stability and how much it would be perturbed by Mercury, but I suspect gravitational perturbations by Earth and Jupiter to be more severe. Of course, just that the distance of Lagrange point 1 from the primary and secondary body varies with their distance between each other during one orbit of the secondary body (Venus) around the primary one (Sun) due to being non-circular makes this $\text{SVL}_1$ saddle point rather unstable. Lissajous or indeed Halo orbits around these gravitationally flat points in space only make stationkeeping somewhat easier and these instabilities (and satellite's own due to precision burn requirements) easier to control. They don't cancel them out in any way, regardless how feeble they might be.

  • $\begingroup$ Kerner's equation is a first order approximation. But since $\alpha$ is typically tiny, higher powers of the term are even tinier. Adding more terms doesn't make it that much more precise. Kerner's is a good approximation. The online calculator you link to gives a slightly better approximation. $\endgroup$
    – HopDavid
    Commented Sep 12, 2014 at 18:45
  • $\begingroup$ @HopDavid All our numbers are within ~ 0.3% w/ error at less than half the Cytherean orbital eccentricity, so as you say it really doesn't matter for them to be that precise. A typical Lissajous / Halo orbital box around L-point of as large body as Venus will be significantly larger in size than those ~ ± 3,000 km. I even used mass ratios as input to calculate α, so no one should expect decimal precision here. I just wanted to confirm that my maths aren't off by too much by also plugging the numbers in that online calculator, with main purpose of my answer to show one way of calculating it. ;) $\endgroup$
    – TildalWave
    Commented Sep 12, 2014 at 19:06
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    $\begingroup$ Yes, your answer is good and I gave it a +1. I put in an answer as I saw this as an opportunity to share a spreadsheet I'm proud of. $\endgroup$
    – HopDavid
    Commented Sep 12, 2014 at 23:56

Lissajous orbits and Halo orbits around L1 or L2 are very closely related. For Lissajous orbits, the in-plane period (left-right) and out-of-plane period (up-down) are uncoupled. It will still oscillate around the Lagrange point, but in a kind of "oscilloscope squiggle".

But for some range of distances the two periods will become equal and you end up with an (essentially) closed halo orbit which is roughly a 3D-bent ellipse shape.

Halo orbits are then just a subclass of Lissajous orbits where the two periods become equal.

In a pure circular, restricted 3 body scenario (CR3BP), there are some halo orbits that are truly stable. See answers to Are some Halo Orbits actually Stable?, one of which links to this answer to Is it possible to have stable orbits around Lagrange point L1? in Physics SE. These would then be a sub-class of a sub-class of Lissajous orbits.

However, since gravity is a long-range force everything attracts everything (eventually) so even those "stable halo orbits" of the CR3BP are not going to be stable in the real solar system. For three-body orbits associated with the Sun and Venus, I think the major perturbers will be Earth and Jupiter, since Mercury's mass is so small. But then again periods of L1/L2 halo orbits are often roughly half of the planet's orbit, and that puts the synodic period of Venus + Mercury (145 days) on the same scale as the halo orbit's period (of order 112 days) offering the possibility of some resonant effects in some cases, so this will be hard to answer with certainty without a real, full-blown calculation.


I had tried to solve this problem in a different way.
- Sun-Mercury 240000km (L1&L2 distance from orbiting body center) - Sun-Venus no SVL1&SVL2.
- Sun-Earth 1640000km.
- Sun-Mars 1180000km.
- Sun-Jupiter 58250000km.
- Sun-Saturn 71380000km.
- Sun-Uranus 76830000km.
- Sun-Neptune 126900000km.
- Earth-Moon 65000km.

Sun-Venus never has L1&L2 because of "gravity field with different steering".

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    $\begingroup$ I have edited so that it is no longer an unreadable blob of text, but I'm still having a hard time understanding what you are writing here. Are you answering the question?. $\endgroup$
    – user10509
    Commented Dec 4, 2018 at 16:42
  • $\begingroup$ Hi 李昱廷 and welcome to space! I think you are using a translation from Chinese to English, and "gravity field with different steering" isn't very helpful. See if you can do some more reading to find a better English term, because that translation doesn't really say anything useful in English. People are generally helpful with language problems here, so if you can find a link that explains and supports your thinking, and add the link to your answer that would be great! $\endgroup$
    – uhoh
    Commented Dec 5, 2018 at 2:49
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    $\begingroup$ Then L1&L2 could be easily calculated. (L1)Find a position between Sun and planet where orbital speed of Sun less orbital speed of planet equal to the synchronization speed of the planet. $\endgroup$
    – 李昱廷
    Commented Dec 5, 2018 at 14:24
  • $\begingroup$ Hello @李昱廷 I just saw all of your comments today! I realize that you don't know how replies works. If I type a comment, you automatically receive a notification flag because this is your post. But if you want to reply to me, you have to type @uhoh in your message. If you don't I'll never know that you are answering. You can read more at How do comment @ replies work? $\endgroup$
    – uhoh
    Commented Apr 18, 2019 at 14:07
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    $\begingroup$ @uhoh I got it,thanks. $\endgroup$
    – 李昱廷
    Commented Apr 18, 2019 at 19:05

I had post my answer last year but seems no one interested about that.Anyway,I'll give an other relational eqution about L1 of all planets in solar system. I'll use the Sun-Earth L1 of Senbehely's answer(149.75 borrow from Hopdavid's spreadsheet) to caculate all L1 of planets.

(Units:ten thousand Kilometers)

MercuryL1=149.75 x 0.387AU x 0.381=22.08

VenusL1=149.75 x 0.73AU x 0.935=102.21

EarthL1=149.75 x 1AU x 1=149.75

MarsL1=149.75 x 1.523AU x 0.475=108.36

JupiterL1=149.75 x 5.205AU x 6.825=5319.7

SaturnL1=149.75 x 9.537AU x 4.564=6518.15

UranusL1=149.75 x 19.2AU x 2.44=7015.5

NeptuneL1=149.75 x 30.05AU x 2.575=11587.5

The following is Victor Szebehely's answer V.S. solution of relational eqution

Mercury:22.04 vs 22.08

Venus:100.79 vs 102.21

Earth:149.75 vs 149.75

Mars:108.25 vs 108.36

Jupiter:5191.1 vs 5319.7

Saturn:6426.1 vs 6518.15

Uranus:6951.9 vs 7015.5

Neptune:11520 vs 11587.5

(Victor Szebehely's answer borrow from Hopdavid's spreadsheet and thanks)

the relational eqution is:
Distance of planet's L1=Earth's L1 x Orbital radius ratio x Cubic root of mass ratio

This equation is simple and more accurate(If the L1 of Earth is accurate).Because it doesn't have errors due to excessive mass of the planet.

I have to apologize because I am swearing to answer in this forum.So that you should hardly understand what I said.

  • $\begingroup$ Can you explain how this answers the question(s)? I don't see it. $\endgroup$
    – user10509
    Commented Oct 6, 2019 at 20:52
  • $\begingroup$ Distance of planet's L1=Earth's L1 x Orbital radius ratio x Cubic root of mass ratio $\endgroup$
    – 李昱廷
    Commented Oct 6, 2019 at 23:58
  • $\begingroup$ For example: If you wanna caculate Jupiter's L1 distance by using Earth's L1 data: Orbital radius ratio=Jupiter's orbit radius divide by Earth's orbit radius. Cubic root of mass ratio=Jupiter's mass divide by Earth's mass. Then we could get: Jupi's L1=149.75 x (5.2AU/1AU) x (Cubic root of 317.87)=149.75 x 5.2 x 6.825=5319.7 $\endgroup$
    – 李昱廷
    Commented Oct 7, 2019 at 0:24

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