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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)
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)
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)
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)
I have a concern or two about this paragraph …
Hm, for what mass ratios can one of these two ("outside" and "farther") be true and not the other?
This seems to me much less important than what follows.
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)
This is the
talk page for discussing improvements to the
Lagrange point article. This is not a forum for general discussion of the article's subject. |
Article policies
|
Find sources: Google ( books · news · scholar · free images · WP refs) · FENS · JSTOR · TWL |
Archives:
1,
2Auto-archiving period: 90 days
![]() |
![]() | This article is written in American English, which has its own spelling conventions (color, defense, traveled) and some terms that are used in it may be different or absent from other varieties of English. According to the relevant style guide, this should not be changed without broad consensus. |
![]() | This ![]() It is of interest to the following WikiProjects: | |||||||||||||||||||||||
|
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)
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)
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)
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)
I have a concern or two about this paragraph …
Hm, for what mass ratios can one of these two ("outside" and "farther") be true and not the other?
This seems to me much less important than what follows.
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)