Talk:Lagrange point

L1 vs L2 distance

On this diagram, shouldn't L1 and L2 be almost the same distance from the secondary body? Tom Ruen (talk) 07:02, 30 December 2021 (UTC)

Yes. But I can't fix that, since I'm no good at modifying images and animations. Does anyone else want to take a shot at it? Fcrary (talk) 07:29, 30 December 2021 (UTC)

Good point. I just replaced it. Tom Ruen (talk) 19:28, 30 December 2021 (UTC)

I'm a littler more concerned with the table in "Solar System Values." Just checking L1 and L2 for Sun-Earth I get 1.49155E9 m and 1.50153E9 m, respectively (for Earth-Sun distance equal to Earth's semi-major axis) but the table reports 148.11E9 m and 151.1E9 m, respectively for those same values. I am pretty sure my numbers are right. There seem to be some percentage errors on top of the 2-order-of-magnitude disagreement as well. Just as a gut-check I went to the James Webb Space Telescope (which orbits L2) page and that page gives an approximation for L2 as, "1,500,000 km." That agrees, to order-of-magnitude, with my calculations.

For completeness, I am using masses and distances off the respective Wikipedia pages: M_Sun = 1.9885E30kg - M_Earth = 5.97237E24kg - R_semi-major = 149598023km

71.120.2.107 (talk)mjd 71.120.2.107 (talk) 12:10, 2 June 2022 (UTC)

Sorry - figured this out - I was calculating distance of Lagrange point from smaller mass - the table indicates distance from larger mass. Thanks. 71.120.2.107 (talk)mjd 71.120.2.107 (talk) 13:42, 2 June 2022 (UTC)

redundant and conflicting data

In the Past and current missions table, the WIND satellite is listed twice. Once saying it is in L1 (which I believe it is) and once saying it is in L2. Since the page is actively being edited by people who probably know more than I do, I will leave it to you guys to figure it out and fix it. 2600:8801:8C0B:4800:E0F5:4086:FC3C:7E4C (talk) 20:39, 8 January 2022 (UTC)

WIND is on a L1 halo orbit. But before it entered that orbit, it spent some time around L2. That's why it's listed twice. Do you think we should add some text to make that more clear to readers? Fcrary (talk) 04:19, 9 January 2022 (UTC)

Is the L3 formula correct?

I was working on an assignment and I found this article looking for an analytical approximation for L3 location. The formula in the article was not making sense to me and I checked the reference [18], in which the formula appears in equation (20). The one in the reference does not look like the same and does make sense to me when plugging the numbers. 2001:1C00:B0C:F000:A0B3:5D53:E375:2471 (talk) 00:36, 4 December 2022 (UTC)

Indeed, it was wrong, I fixed it, thanks for noticing. Tercer (talk) 12:24, 20 July 2023 (UTC)

L1, L2, & L3 Quintic Equations

The correct polynomial for the distance between the secondary and L1 is

x^5+(µ-3)x^4+(3-2µ)x^3-(µ)x^2+(2µ)x-µ = 0

and

r = R x

The correct polynomial for the distance between the secondary and L2 is

x^5+(3-µ)x^4+(3-2µ)x^3-(µ)x^2-(2µ)x-µ = 0

— Preceding unsigned comment added by 131.176.243.11 (talk • contribs) 09:03, 19 July 2023 (UTC)

L1 Quintic Equation:

This equation is incorrect. To demonstrate, take the arbitrary test case where R = 6; M1 = 10, and M2 = 2, (µ = .8333) the quintic equation has a zero at x=0.8098. Since x = r/R, then r = 6*0.8098 = 4.8586. Plugging these values into the formula above the quintic equation describing the force balance, the left-hand-side becomes 10/(6-4.8586)^2-2/4.8586^2 = 7.5911. The right-hand-side equation becomes (10/(10+2)*6-4.8586)*(10+2)/6^3=0.0079. Clearly these aren't equal so the quintic equation is incorrect.

Using MATLAB's symbolic toolbox, the correct equation was derived to be: x^5-(2+µ)x^4+(1+2µ)x^3+(µ-1)x^2-2(µ-1)x+µ-1. The only difference is the minus sign in front of the coefficient of the x-term. Taking again the previous test case, this equation has a zero at x=0.3414. Since x = r/R, then r = 6*0.3414 = 2.0487. Plugging these values into the formula above the quintic equation describing the force balance, the left-hand-side becomes 10/(6-2.0487)^2-2/2.0487^2 = 0.1640. The right-hand-side equation becomes (10/(10+2)*6-2.0487)*(10+2)/6^3=0.1640. These are equal, verifying the validity of the proposed equation.

L2 Quintic Equation:

Again, this equation is incorrect. First of all, instead of x, it has r as the variable. This results in mixed units throughout the equation as r is not dimensionless. Assuming the r's were supposed to be x's it is still incorrect, however. Taking the same test case as before, the quintic equation has a root at x=0.9025. Since x = r/R, then r = 6*0.9025 = 5.4150. Plugging these values into the formula above the quintic equation describing the force balance, the left-hand-side becomes 10/(6+5.4150)^2+2/5.4150^2 = 0.1450. The right-hand-side equation becomes (10/(10+2)*6+5.4150)*(10+2)/6^3=0.5786. Clearly these aren't equal so the quintic equation is incorrect.

Using MATLAB's symbolic toolbox, the correct equation was derived to be: x^5+(2+µ)x^4+(1+2µ)x^3+(µ-1)x^2+2(µ-1)x+µ-1. This is very similar to the quintic equitation for L1 except all the coefficients are positive. The only difference is the minus sign in front of the coefficient of the x-term. Taking again the previous example, this equation has a zero at x=0.4381. Since x = r/R, then r = 6*0.4381 = 2.6285. Plugging these values into the formula above the quintic equation describing the force balance, the left-hand-side becomes 10/(6+2.6285)^2+2/2.6285^2 = 0.4238. The right-hand-side equation becomes (10/(10+2)*6+2.6285)*(10+2)/6^3=0.4238. These are equal, verifying the validity of the proposed equation.

L3 Quintic Equation

The equation was not written. Once again using the MATLAB symbolic toolbox, the resulting quintic equation is x^5+(µ-8)x^4+(25-6µ)x^3+(13µ-37)x^2+2(13-7µ)x+7(µ-1). Taking again the previous example, this equation has a zero at x=0.0975. Since x = r/R, then r = 6*0.0975 = 0.5850. Plugging these values into the formula describing the force balance in the L3 section, the left-hand-side becomes 10/(6- 0.5850)^2+2/(2*6- 0.5850)^2=0.3564. The right-hand-side equation becomes (2/(10+2)*6+6-0.5850)*(10+2)/6^3=0.3564. These are equal, verifying the validity of the proposed equation. WaffleJet34 (talk) 03:57, 24 May 2023 (UTC)

effects on L3

I have a concern or two about this paragraph …

The L₃ point lies on the line defined by the two large masses, beyond the larger of the two. Within the Sun–Earth system, the L₃ point exists on the opposite side of the Sun, a little outside Earth's orbit and slightly farther from the center of the Sun than Earth is.

Hm, for what mass ratios can one of these two ("outside" and "farther") be true and not the other?

This placement occurs because the Sun is also affected by Earth's gravity and so orbits around the two bodies' barycenter, which is well inside the body of the Sun.

This seems to me much less important than what follows.

An object at Earth's distance from the Sun would have an orbital period of one year if only the Sun's gravity is considered. But an object on the opposite side of the Sun from Earth and directly in line with both "feels" Earth's gravity adding slightly to the Sun's and therefore must orbit a little farther from the barycenter of Earth and Sun in order to have the same 1-year period. It is at the L₃ point that the combined pull of Earth and Sun causes the object to orbit with the same period as Earth, in effect orbiting an Earth+Sun mass with the Earth-Sun barycenter at one focus of its orbit.

How accurate is that last ("in effect…")? Two distinct bodies are not generally equivalent to one body of their combined mass.

One of these days I'll do the necessary algebra, but not now. —Tamfang (talk) 23:42, 15 July 2023 (UTC)