29

L1, L2 and L3 are saddle points in the effective potential of the gravitational field in a rotating frame of reference. That is if you combine gravity (of Earth and Sun) with the centrifugal force on an object that is moving around a point at one orbit per year you find that there are three saddle points, and these are L1, L2 and L3 If an object is in orbit ...


22

According to my understanding Lagrange points L1, L4 and L5 can form because gravitation pull can cancel out here as these are between Sun and Earth (where gravitational pull is in two different directions). This is a misconception. The net gravitational attraction is directed toward the Sun at the Sun-Earth L1 point. The point at which gravitational ...


18

Hubble was in low earth orbit, and was always intended to be serviceable. In fact, the original plan for Hubble was to have the space shuttle carry it down from orbit and take it back up, but they decided that was too risky compared to servicing in orbit. JWST, on the other hand, will be at the Earth/Sun L2 Lagrange point, like WMAP and Planck before it. ...


13

In astronomy, a trojan is a small celestial body (mostly asteroids) that shares the orbit of a larger one, remaining in a stable orbit approximately 60° ahead of or behind the main body near one of its Lagrangian points L4 and L5. Trojans can share the orbits of planets or of large moons. https://en.wikipedia.org/wiki/Trojan_(celestial_body) Whether or not ...


12

The L1, L2, and L3 points are unstable in any orbital system. (source) The L4 and L5 points of a pair of bodies are only stable if the larger of the bodies is at least 25 times as massive than the smaller (source). The ratio of the Pluto/Charon system is only 8.7. Because of this, none of the Lagrange points are stable, and an object orbiting at any of ...


12

Stargazer's rough, quick method: Look at the sky at local solar midnight, halfway between sunset and sunrise. Locate the ecliptic. If you're a stargazer, you should know where that is: it's the line the zodiac constellations are strung along. Visualize the local meridian: it's just the line running north-south through the zenith. L2 is approximately where ...


11

To get the L2 position from Horizons, you need to ask it for "Apparent AZ & EL", which is quantity 4 in the Observer table custom settings. When setting the time span, you need to use UTC time, which is 3 hours ahead of Rio time. Here's some hourly data for 2021-Dec-30 at the Rio de Janeiro Observatory, from 21:00 UTC to 9:00 UTC, 6 PM to 6 AM ...


10

Gaia's original science and technology report (see page 221, see also the summary) gives an analysis of the Lissajous orbit. From what I understand Gaia will be placed in a small amplitude Lissajous orbit, giving it an orbital radius of $\sim400000$ km away from $\sim100000$ km along the Sun-Earth axis. In addition to the fact that this orbit is ...


7

My question: is there a way to determine the "stable" region in each Lagrange location? Particularly in the L4 and L5 regions. tl;dr: Yes but it's usually a fancy version of "trial and error". We are sometimes wrongly told that this abominable zero-velocity potential diagram show areas that are "bound" or "unbound" to Lagrange points, but it does not and ...


7

The Giant Impact Hypothesis, a theory of how the moon was created, says that once it exceeds 10% of the mass of your 'J', the orbit of an L4 or L5 (your 'T') destabilizes. Possible origin of Theia In 2004, Princeton University mathematician Edward Belbruno and astrophysicist J. Richard Gott III proposed that Theia coalesced at the L4 or L5 Lagrangian ...


7

No. Such an arrangement is at best "metastable". That is, although there are periodic solutions to the three body problem (stable orbits) an infintesimal perturbation (eg the proverbial butterfly flapping its wings) will push the system off the stable orbit and into chaos. Getting a planet to remain at the barycentre is like trying to balance a ...


7

Moons formed in a number of ways. Many moons are just like the debris that is found at L4 and L5, of the nearly 80 known moons of Jupiter, most are small, irregular and are probably captured asteroids. These small moons are much like the population of Trojans. There are a few larger moons, the “Galilean moons formed out of the dusty disk left over after ...


7

Running the JPL Horizons calculation as in the original question gives the RA and DEC of the L2 point: $$SOE 2021-Dec-30 00:00 06 36 55.95 +23 19 55.7 n.a. n.a. 0.00987312593538 -0.2824505 179.6621 /L 0.3347 0.8202458 352.89131 -70.88475 n.a. n.a. 2021-Dec-31 00:00 06 41 13.66 +23 15 44.7 n.a. n.a. 0....


6

The stability of this system depends the ratio of masses of the two stars. If the larger star is more than 25 times more massive than the smaller star, then L5 is potentially stable, and this remains the case even if the planet does not have negligible mass The calculation of the value is done in detail on physics.stackexchange and there you can establish ...


6

Why is the L3 Lagrangian point not perfectly stable? In the circular restricted three-body problem (CR3BP or CRTBP) an object at any of the first Lagrange points L1, L2, L3 is unstable mathematically. Yes, a ball at the exact top of a hill will sit there, but any tiny offset in position or tiny nonzero velocity will lead to it accelerating down the hill. As ...


6

Most discussions of Lagrange points use the simplification of the circular restricted three-body problem, where two objects (e.g., the Sun and the Earth) are massive and in fixed circular orbits about the center of mass, and we are interested in the possible motions of a third particle with effectively infinitesimal mass (e.g., a spacecraft or asteroid). ...


5

Note: Answering from a comment posted on Space Exploration The classical stability analysis of these libration points assumes that we are examining the motion of a particle whose dynamics are perturbed by the gravitational impacts of a primary and secondary mass, so as a bottom-line-up-front type of answer, the mass of T is negligible - so any large ...


5

It does indeed seem counterintuitive that $L_4$ and $L_5$ would be at the same time both high points of potential as well as stable points in the system. In fact, a quick look at an example contour plot with all five Lagrangian points demonstrated would also suggest that $L_4$ and $L_5$ would be unstable: In the picture you can see that $L_1 - L_3$ are ...


5

Objects are not place at the Sun-Earth L1 or L2 Lagrange points. They are instead placed in pseudo orbits about these points. These pseudo orbits intentionally avoid being directly in line with the Earth and the Sun for two key reasons. One reason is that these points are directly in the line between the Sun and the Earth. An object at the Sun-Earth L1 would ...


5

One of the answers to the researchgate question What are the equation of motion for elliptical restricted three body problem? says: https://www.sciencedirect.com/search?qs=equation%20of%20motion%20for%20elliptical%20restricted%20three%20body%20problem&show=25&sortBy=relevance which returns a large number of papers, most paywalled. The answer also ...


4

The Lagrangian point $L_2$ is very close to the most distant point from Earth with an umbra. $L_2$ is like the radius of the Hill sphere at $r=a\sqrt[3]{\frac{m}{3M}}$ for circular orbits, with $m$ the mass of Earth, $M$ the mass of the Sun, and $a$ the distance Earth-Sun. The ratio $\frac{m}{3M}$ of the Earth and the triple mass of the Sun is almost exactly ...


4

Bad news, this type of SPK file has a different sort of interpolation that is not supported by the jplephem package (Hermite interpolation vs Chebyshev polynomials). You can find this out by doing: In [1]: print(len(kernel.segments)) 1 In [2]: print(kernel.segments[0]....


4

tldr; L2 is a very stable thermal environment as well as good instantaneous sky visibility and high observing efficiency. The main reason space telescopes are placed in an L2 orbit is because L2 is a stable thermal environment. Telescopes in Earth orbit can receive sunlight and earthlight in different directions, meaning that the telescope would have to ...


4

For this answer, I'll consider space telescopes to be telescopes that operate in space and that are intended to look at objects at the extremes of the solar system and beyond. This excludes Earth observation satellites, satellites that monitor the Sun, and satellites sent to another planet to observe that other planet as "space telescopes". Every space ...


4

To add to the other answers, there is a theory (unproven, but fairly widely accepted as plausible) saying that the Earth in fact once, early during the solar system's formation, had a roughly Mars-sized co-orbital companion planet (called Theia) at its L4 or L5 point, and/or possibly in something like a horseshoe orbit. The theory suggests that eventually, ...


3

The Wikipedia article cites "Zdeněk (1962)" for the statement that the dust responsible for the Gegenschein has a possible concentration at L2. I haven't been able to obtain that paper, but I can't really see why that would be the case, since L2 is not dynamically stable. However, the dust consists of millimeter-sized grain (see e.g. this APOD image), which ...


3

From this source I get: The size of these islands varies. Each planet in the solar system has its own Lagrangian points. The islands of stability get bigger farther from the Sun and also for more massive planets. The ones associated with Earth are roughly 500,000 miles (800,000 kilometers) wide. The biggest zones (at least in the solar system) ...


3

The angle between the inner and outer edges of the penumbra is the same as the Sun's apparent angular diameter. At 1.0 au from the Sun, that's 32 arcmin or 0.0093 radian. At the distance from the Moon to EM-L2, this angle spans 0.0093 * 64700 km = 602 km. The Moon's diameter is 3474 km, so the umbra diameter would be 2872 km and the penumbra diameter would ...


3

The L1, L2, L3 points are not stable. (full stop) Small deviations grow exponentially, even in the perfectly circular restricted three-body problem. In reality, we have a non-circular many-body problem, when (1) these points are not strictly well-defined and (2) deviations from the simple case have to be accounted for to obtain the exact trajectories (and to ...


3

No Near-Earth objects (NEOs) may be temporarily captured to oribt Earth as discussed in the reference quoted in the question. Such a capture must invariably involve the NEO to pass by the L1 or L2 points, which are the saddle points in the co-rotating binary potential of the Earth-Sun system. Thus, these objects have just high enough co-rotating energy (...


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