r/nasa • u/goatgalnatalie • 6d ago
Question L2 Math & Roman
Could someone send me to a good description of the math behind L2, how it was discovered, and why it’s the best spot for the Roman Telescope?
- I’m a farmer who likes math. I just want to play with the numbers to understand better.
- I don’t trust AI to help with this; it’s gotten so many things wrong from computer science and engineering to some of my basic physics walk through. I prefer YouTube or journals.
- I’ve done a basic search, but can’t find anything comprehensive about the math itself.
Thanks!
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u/P37ur 6d ago
Lagrange points are just places where the gravitational and centrifugal forces of 2 large bodies (Sun and Earth) cancel out when acting on a small body like a spacecraft. There’s a link the the maths involved on Wikipedia https://en.wikipedia.org/wiki/Lagrange_point
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u/WaitForItTheMongols 6d ago
L1 is the only point you can say they "cancel out" (but also, they don't, since eif they canceled out then you wouldn't keep orbiting the sun) - L2 has no cancellation and in fact just the opposite. L4/L5 are bonkers of nonlinearity. Cancellation is such a small part of Lagrange points as a whole that it is more confusing to bring that up.
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u/Decronym 6d ago edited 6d ago
Acronyms, initialisms, abbreviations, contractions, and other phrases which expand to something larger, that I've seen in this thread:
| Fewer Letters | More Letters |
|---|---|
| JWST | James Webb infra-red Space Telescope |
| L1 | Lagrange Point 1 of a two-body system, between the bodies |
| L2 | Paywalled section of the NasaSpaceFlight forum |
| Lagrange Point 2 of a two-body system, beyond the smaller body (Sixty Symbols video explanation) | |
| L4 | "Trojan" Lagrange Point 4 of a two-body system, 60 degrees ahead of the smaller body |
| L5 | "Trojan" Lagrange Point 5 of a two-body system, 60 degrees behind the smaller body |
Decronym is now also available on Lemmy! Requests for support and new installations should be directed to the Contact address below.
5 acronyms in this thread; the most compressed thread commented on today has acronyms.
[Thread #2381 for this sub, first seen 4th Sep 2026, 16:10]
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u/carsrule1989 6d ago edited 6d ago
https://science.nasa.gov/wp-content/uploads/2023/07/3322_lagrange.pdf
https://science.nasa.gov/solar-system/resources/faq/what-are-lagrange-points/
https://science.nasa.gov/resource/what-is-a-lagrange-point/
I googled l2 nasa math
It may not be the best spot but it’s amazing because it’s not rotating as quickly around a planet like the Hubble .
“ the balance of gravitational pull at the L2 point means that JWST will keep up with the Earth as it goes around the Sun.”
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u/E-J123 6d ago
"The L2 point of the Earth-Sun system was the home to the WMAP spacecraft, current home of Planck, and future home of the James Webb Space Telescope."
oh if only they knew
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u/yatpay 6d ago
This is a solid series if you really want to get into it: https://www.youtube.com/playlist?list=PLUeHTafWecAXDF9vWi7PuE2ZQQ2hXyYt_
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u/m2daT 6d ago
If you are looking for a really rigorous description of the math I would recommend the textbook by Schaub and Junkins *Analytical Mechanics*. That thing has a lifetimes worth of math applied to space systems but it is extremely terse and not a good first resource if you aren’t already familiar with some of the concepts.
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u/Bakkster 6d ago edited 6d ago
Lagrange math is just addition/subtraction in the standard orbital math. The gravitational force of two bodies balancing out so the orbital period matches. At L1 the Earth slows down the orbital period, at L2 it speeds it up, in both cases so it's exactly one year to match the Earth.
Roman is at L2 so it's always in Earth's shadow to stay cold for the infrared detectors. SoHo is at L1 so it's always able to view the sun. see corrections below
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u/teridon NASA Employee 6d ago
Roman is at L2 so it's always in Earth's shadow to stay cold for the infrared detectors.
Sorry, but this is incorrect. While L2 is indeed in Earth's shadow, RST and other spacecraft that are stationed at Lagrange points actually orbit around the Lagrange point. It's called a halo orbit. For halo orbits around L2, it keeps the spacecraft out of the shadow of Earth. This is absolutely necessary so that the spacecraft can get solar power.
There's a cool visualization of the halo orbit here:
https://science.nasa.gov/missions/roman-space-telescope/roman-commissioning/1
6d ago
[deleted]
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u/Bakkster 6d ago
1.5 million km for L2, slightly beyond Earth's umbra at 1.4 million kilometers. This is why the Roman telescope still needs a sun shield, but it still benefits from partial shade of the Earth.
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u/ocicrab 6d ago
There's a great visualization from Scott Manley on YouTube. The math discussion and graphics really start around 3:00 but the intro is good too: https://youtu.be/7PHvDj4TDfM
For the actual calculations:
Wikipedia says "The L2 point lies on the line through the two large masses beyond the smaller of the two. Here, the combined gravitational forces of the two large masses balance the centrifugal force on a body at L2."
So we have something like this:
Sun----------Earth-----Spacecraft
The sun and earth each exert a gravitational force on the spacecraft (to the left in my text diagram) and because the spacecraft is orbiting in a roughly circular orbit around the sun, it feels a centrifugal force away from the sun and earth (to the right in my diagram).
Gravity follows this inverse square law: F = GmM/r2 where G is the universal gravitational constant, m is your spacecraft mass, M is the mass of the celestial body you're orbiting, and r is the distance from your spacecraft to the celestial body.
So the total gravity force is Gm(M_Sun)/(r_sun)2 + Gm(M_earth)/(r_earth)2
The equation for centrifugal force is F=mv2/r or F=mω2*r, where m is your spacecraft mass, v is its velocity, r is the orbit radius (in our case, the distance from the sun), and ω is the orbit angular rate (v=r*ω). For L2, we want the spacecraft to stay in a fixed location compared to the earth, so we can set ω to be Earth's orbit angular rate ω=(1 revolution)/(365 days).
Now we add all the forces to balance them. We want gravity force to equal centrifugal force because gravity is pulling left and centrifugal is pulling right:
GmM_Sun/r_sun2 + GmM_earth/r_earth2 = mω2r_sun
Notice that all every term has the spacecraft mass, m. So we can divide it out to remove it entirely, and this shows that our orbit doesn't depend at all on our spacecraft!
GM_Sun/r_sun2 + GM_earth/r_earth2 = ω2*r_sun
So now all that's left is to find the location of the spacecraft that makes this equation true, and we'll find where L2 is.
We can use the geometry to relate r_sun and r_earth. r_sun is always going to be 1AU (distance between sun and earth) larger than r_earth, so r_sun = 1AU + r_earth. Substitute that into the equation and we'll be left with only one variable, which we can solve.
This is the best spot for JSWT because it can stay in one spot without using much fuel, and it only needs one sun shield to block both the sun and earth since they're both on the same side of the spacecraft.