Lagrange points calculations

[cristiapi]

I wrote an MFD that displays the distance of the ship from the 5 Lagrange points.
I used the formulas found at the link
https://en.wikipedia.org/wiki/Lagrangian_point#Mathematical_details
rewritten for the 3D space.

Now I would improve the calculations but it’s not clear to me what happens in a Lagrange point.
For example, consider the instant 2017-10-05 18:15:07 UTC when Sun, Earth and Moon are almost in a straight line (their angular distance is about 0.00967 deg, this is to make the calculations easier).

For L2 point of the Earth-Moon system, I get a distance = 436158.875344 km from the Earth and 62681.581136 km from the Moon.

The Moon’s angular speed for the above instant is 13.998 deg/day and it’s the same for L2; hence, the centripetal acceleration at L2 is 3.48743 mm/s².

The accelerations at L2 exerted by the celestial bodies are:
Moon: 1.24786 mm/s²
Earth: 2.09531 mm/s²
Sun: 5.89786 mm/s²
the sum is 9.23615 mm/s².

I thought to find centripetal acceleration = gravitational acceleration.
Please, could anyone explain what I’m doing wrong?

 

[martins]

Are you trying to extend your Lagrange point calculation to 3-body systems? This will probably run into problems quite quickly, since the relative position of the 3 bodies, as seen from the Lagrange point, is continuously changing, so it can't be reduced to a a virtual point mass as in the 2-body case.

In your L2 calculation it looks like you were only considering Earth and Moon, so it is not surprising that the addition of the Sun will produce a perturbation. For the time point considered, the Sun will exert a gravitational force aligned with that of Earth and Moon, so it will increase the mass of the virtual body experienced at L2. Since you want to keep the angular velocity constant, this means you need to push the L2 point further out. However, the net grativational force changes as the Moon and the L2 point are orbiting Earth, distorting the orbit of L2.

If the direction of the sun's perturbation were invariant, I suspect the effect would be the distortion of the L2 orbit into an ellipse, so it might still be possible in theory to place an object there in a stable configuration.

However, since the Earth-Moon system orbits the Sun, the Sun's gravitational force vector is also rotating, which probably means that the L2 elliptical orbit is also rotating. So I don't think that this can be solved without active station keeping.

 

[perseus]

if you choose a reference system Earth, since in an instant you can consider it with constant speed. You should not consider the effects of the Sun.

3,48743 mm / s 2 =~ (Moon: 1.24786 mm/s2) +(Earth: 2.09531 mm/s2)

approximation error

If you take as reference systems you must add to the speed at the point of Lagranje the speed of the Earth and with this calculate the centrifugal force

ω=(13.998 deg/(day* 24*3600 s) + V Earth/ 149.6 E+9)

ac = ω^2 (R Earth + R Lagranje)



the centripetal acceleration at L2

 

[dgatsoulis]

Have a look here. (Elliptic Restricted 3 Body Problem).

Also LagrangeMFD's documentation.

 

[cristiapi]

Are you trying to extend your Lagrange point calculation to 3-body systems?

Not only to 3-body systems; I’m trying to extend the calculation to n-body systems (consider, for example, Jupiter or Pluto systems, where the perturbations are very big).
My problem is that I don’t know whether I’m right when I say that the centripetal acceleration must be equal to the gravitational acceleration; in other words, the vector sum of all the accelerations must be zero. Is that right?

---------- Post added at 00:25 ---------- Previous post was at 00:07 ----------

if you choose a reference system Earth, since in an instant you can consider it with constant speed. You should not consider the effects of the Sun.

But the Sun is there and it moves the Lagrange points; I need to consider all the celestial bodies to improve my calculations.

If you take as reference systems you must add to the speed at the point of Lagranje the speed of the Earth and with this calculate the centrifugal force

ω=(13.998 deg/(day* 24*3600 s) + V Earth/ 149.6 E+9)

ac = ω^2 (R Earth + R Lagranje)

the centripetal acceleration at L2

I use the ac you wrote in my calculations, but I'm not sure to understand (R Earth + R Lagranje). I just wrote
ac = ω² * R_Lagrange
where R_Lagrange is the distance of the Lagrange point from the main body and ω is the instantaneous angular velocity of the smaller body wrt the main body.

[I use several of your ships in my scenarios, thank you ]

 

[perseus]

But the Sun is there and it moves the Lagrange points; I need to consider all the celestial bodies to improve my calculations.

Yes, but.. If it includes the Sun, it also you includes the radius of rotation and speed of the Earth-Moon System.

And the speed of this with respect to the Sun.

Any object in orbit of the Earth or the Moon, will have this speed with respect to the Sun. and therefore in the points of Lagranje. The velocities are relative to the reference system,

even in classical mechanics.

use the ac you wrote in my calculations, but I'm not sure to understand (R Earth + R Lagranje). I just wrote
ac = ω2 * R_Lagrange
where R_Lagrange is the distance of the Lagrange point from the main body and ω is the instantaneous angular velocity of the smaller body wrt the main body.

ac = ω2 * (R_Lagrange + R Earth-Moon System)

The Earth with the Moon makes one revolution per year and has speed with respect to the Sun.

and ω=(ω Earth-Moon System + 13.998 deg/day)

ω Earth-Moon System = 2 Π /(365.6*24*3600) rad/s

R Earth-Moon System = 149.6 E+9 meters

ac Earth-Moon System = ω2 * (R_Lagrange + R Earth-Moon System)
ac Earth-Moon System =5.919 mm/s

Moon: 1.24786 mm/s2
Earth: 2.09531 mm/s2
Sun: 5.89786 mm/s2
the sum is 9.23615 mm/s2. ]

5.919 mm/s + the centripetal acceleration at L2 is 3.48743 mm/s2. = 9.40643 mm/s2

the error 0.2 mm/s2 may be due to the approximate radius of the system Earth-Moon .

If he does not take this into account, it is as if the Moon, the Earth, and any object at the Langranje points are falling to the Sun.

Since it does not take into account its velocities with respect to the Sun.

 

[cristiapi]

Yes, but.. If it includes the Sun, it also you includes the radius of rotation and speed of the Earth-Moon System.

And the speed of this with respect to the Sun.

You're right, I didn't think.

Now I do all the calculations wrt the solar system barycenter and the differences are reasonable.
For example, let's consider Jupiter-Callisto L2 point for 2017-10-01 00:00 UTC:

the centripetal acceleration at the L2 is (with unrealistic accuracy) [m/s²]:
0.0363447951, ax= -0.0218807237508152, ay= -0.0289954426102335, az= -0.00120098524773781

the gravitational acceleration at L2 is:
0.0365373, ax= 0.0220006199243, ay= 0.0291459844992, az= 0.00120687919858

notice the sign of the components.

In my understanding, the small difference comes from the formula that I use to calculate the L2 point which is for the restricted 3-body problem (the link I posted).
Now I just need to move L2 (I'll use the binary search) until centripetal_acc - gravitational_acc < small_vector.

Thanks a lot!