Lagrange Points Calculation Work

Click For Summary
SUMMARY

This discussion focuses on the calculation of Lagrange points, specifically addressing the mathematical expressions and forces involved in the orbital mechanics of a small mass at L1 and L2 relative to two larger masses. Key equations derived include the expression for orbital period T, the net force on a small mass at L1, and the necessary substitutions to simplify these equations. The conversation highlights a common confusion between L1 and L2, emphasizing the importance of correctly applying Newton's Law of gravity and centripetal force in these calculations.

PREREQUISITES
  • Newton's Law of Gravity
  • Centripetal Force Calculations
  • Orbital Mechanics
  • Mathematical Manipulation of Equations
NEXT STEPS
  • Study the derivation of Lagrange point equations in detail.
  • Learn about the stability of Lagrange points L1 and L2.
  • Explore the application of orbital mechanics in satellite positioning.
  • Investigate the differences between L1 and L2 in practical scenarios.
USEFUL FOR

Students and professionals in astrophysics, aerospace engineering, and orbital mechanics, as well as anyone involved in satellite design and positioning strategies.

Samuelriesterer
Messages
110
Reaction score
0
(Sorry text is hard to read, please see attached document for an easier read)

I am having trouble with #6, I'm not sure if what I have going on is entirely correct. Also #7 is a little confusing.

Problem Statement & work done:

For an object in orbit around a second, there are five LaGrange points where a small mass will orbit the second body with the same period as the first due to the combined gravity of the two objects. The masses at these LaGrange points are effectively remain in fixed positions relative to the first body and they all orbit the second body as if they were attached to a frame.

Three of these points lie on the line passing through the two bodies. For these points, we can use Newton's Law of gravity to get the force on a small mass at L1 from each of the two bodies. The sum of these two forces is the net force which can be represented as the centripetal force.

1 - Use Newton's law of gravity and centripetal force to get an expression for the period of mass m in orbit about mass M distance R away. (geometry!)

GMm/R^2 = mV^2/R → T = sqrt[(4pi^2R^3)/(GM)]

2 - If a small mass at L1 is to have the same period as mass m, write an expression for its orbital speed in terms of that period and the orbital radius (R - r1).

v = (2pi(R-r1))/sqrt[(4pi^2R^3)/(GM)]

3 - Write the expression for the total force on a small mass u at L1 exerted by the large mass M distance R-r1 away and the smaller mass m distance r1 away in the opposite direction.

F_net = GMu/(R-r1)^2 – Gmu/r1^2

4 - Set the net force equal to u times the centripetal acceleration using the speed from [2].

GMu/(R-r1)^2 – Gmu/r1^2 = uv^2/(R-r1)

5 - Substitute for T2 from [1] and cancel all common factors.

GMu/(R-r1)^2 – Gmu/r1^2 = (uGM(R-r1))/R^3

6 - Clear all fractions by multiplying all terms by the denominators and then divide by M.

(GMu(r1)^2)/(R-r1)^2 – (Gmu(R-r1)^2)/((r1)^2(R-r1)^2) = (uGM(R-r1))/R^3 →
(GMu(r1)^2-Gmu(R-r1)^2)/((r1^2)(R-r1)^2) = (uGM(R-r1))/R^3 →
R^3(GMu(r1)^2 – Gmu(R-r1)^2) = GMu(r1)^2(R-r1)^3

7 - Next, determine the power to which radii or combinations of them are raised
in each term. (For example, (R+r)2r3 would be counted as the power 5.) Then divide
every term by R to that power and rewrite the expressions using p = r/R as shown
below:


(r(R-r)^2)/R^3 – p(1+p)^2

You should get something equivalent to:

(m/M)(1+p)^2 = (1+p)^3 p^2 - p^2
 

Attachments

Physics news on Phys.org
This treats M as fixed in space. The usual treatment allows that m and M orbit each other around a common mass centre.
Leaving that aside, I see no way to get 1+p terms. Looks like there's been a confusion between L1 and L2. The example given
Samuelriesterer said:
(r(R-r)^2)/R^3 – p(1+p)^2
should read ##(r(R-r)^2)/R^3 = p(1-p)^2##.
For L2, the sign of the left hand side of eqn 4 is also switched. I believe this accounts for all the differences between the target answer and your own answer.
 
Thanks for the reply. You are right, that is the equation, I had a typo on the minus sign. And L2 is the same with reversed signs. Please take a look at my upload file, I have worked out the problem but am really unsure about the results. Thanks so much!
 

Attachments

Similar threads

  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 16 ·
Replies
16
Views
1K
  • · Replies 15 ·
Replies
15
Views
1K
  • · Replies 13 ·
Replies
13
Views
1K
  • · Replies 7 ·
Replies
7
Views
892
Replies
335
Views
17K