Lagrange Points

[jgrillo2002]

:sos: Hi I need to know first hand on how do you calculate the state vectors for reaching a L1 or any of the other points of a orbiting body using equation MFD. because Im proposing of making a space station there and im not sure how to get there

BUmp

 

[tony]

Can you just burn when you're lined up with L2, until your apogee is the distance of L1?

 

[jgrillo2002]

Can you just burn when you're lined up with L2, until your apogee is the distance of L1?

no. its a totally different procedure. you need to the position that the two bodies cancel each other

 

[tony]

That's where I'm trying to get you to. L1 is inbetween the Earth and Sun, about 1497000 km from Earth, in the direction opposite L2.

The Sun and Earth don't cancel each other there. That point is much closer to Earth. L1 is the point where Earth's pull decreases the Sun's gravity vector to the point that the spacecraft's orbital period is exactly 1 year, despite orbiting the Sun interior to the Earth, where its period would normally be shorter. Once perfectly positioned there, you will need to make station-keeping burns along the Earth-Sun line to prevent from drifting away from L1.

I tried it in Gravity Simulator. Here's a few screen shots:
http://orbitsimulator.com/orbiter/toL1a.GIF
http://orbitsimulator.com/orbiter/toL1b.GIF

The Sun pulls you pretty hard at the end, so you can actually burn until your apogee is a little shy of L1, and the Sun's pull will make up the difference.

The distance to L1 can be approximated with the Hill Sphere formula. But I think the exact distance can only be computed numerically (at least that's the only way I know to do it).

 

[jgrillo2002]

ill give that a shot. what about L3 and L4. those were used by some missions I know

 

[ijuin]

Lunar L3 and L4 perhaps--Sun-Earth L3 is on the opposite side of the Sun from the Earth--nearly 300 million km away, and L4/L5 are sixty degrees ahead of and behind Earth in its orbit.

 

[tblaxland]

ill give that a shot. what about L3 and L4. those were used by some missions I know

I am not aware of any to either L3, 4, or 5. L3 would be a difficult place to stage a mission, being on the opposite side of the sun.

 

[n0mad23]

This might be helpful in terms of helping visualization:

 

[n0mad23]

This one, too:

 

[jgrillo2002]

Tried that but I was slowly returning to earth. any way I can calculate the exact spot to stop. I already used the formula. went to that distance but just came back down

 

[tony]

Here's a calculator I made to find the exact spot. It does it numerically, starting with the Hill Sphere as an estimate, then zeroing in on L1 by computing the orbital period of an object being tugged one way by the Sun and the other way by the Earth. When it finds a distance where the period is exactly 1 year, it reports the answer.

Even if you get it perfect, you'll still fall off the L1 point eventually for the same reason you can't balance a pencil on its tip. Any drift you have in the direction of your orbit will correct itself. Any drift you have in the z-axis (up down with respect to the ecliptic) will correct itself. Purposely putting drift on these axes gives you Lissijous and Halo orbits around L1. But will always have to do station-keeping burns along the Earth-Sun line to keep from drifting towards or away from Earth.


*** Edit, ignore the following "exe" link and code and use this javascript I made instead : http://orbitsimulator.com/formulas/LagrangePointFinder.html , The javascript is more accurate, especially when the secondary's mass is appreciable compared to the primary's.
*** Edit

http://orbitsimulator.com/orbiter/L1Finder.exe

Private Sub btnComputeL1_Click()
    Dim Pi As Double, G As Double, Ms As Double, Mt As Double, a1 As Double, a2 As Double
    Dim accs As Double, acct As Double, accx As Double, acctot As Double, accnet As Double
    Dim hs As Double, v As Double
 
    Pi = Atn(1) * 4 'define Pi
    G = 6.6727004301751E-11 'define gravitational constant
    Ms = Val(txtM1.Text) ' Get mass of primary body from the GUI
    Mt = Val(txtM2.Text) ' Get mass of secondary body from the GUI
    a1 = Val(txtSMA.Text) ' Get semi-major axis of secondary body from the GUI
    p1 = 2 * Pi * Sqr(a1 ^ 3 / (G * (Ms + Mt))) 'period of Earth
    hs = a1 * (Mt / (3 * Ms)) ^ (1 / 3) 'compute the hill sphere as a starting point
    lb = hs * 0.98: ub = hs * 1.02 'as an upper bound and a lower bound, choose 98% and 102% of the hill sphere
 
    While Abs(ub - lb) > 1 'While |ub - lb| is less than 1 gives a precision of 1 meter
        a2 = (ub + lb) / 2 'semi-major axis of the L1 particle with respect to the secondary body
        accs = G * Ms / (a1 - a2) ^ 2 'acceleration by the primary body on the L1 satellite
        acct = G * Mt / (a2) ^ 2 'acceleration by the secondary body on the L1 satellite
        'accx = G * Mt / (2 * a1) ^ 2 'acceleration by an Earth-mass object in Earth's L3 point on the L1 satellite
        acctot = accs - acct + accx 'compute combined acceleration
        accnet = acctot / accs 'multiplier to convert primary mass into simulated primary mass
        p2 = 2 * Pi * Sqr((a1 - a2) ^ 3 / (G * Ms * accnet)) ' period of satellite around the simulated primary mass
        If (p1 < p2) Then 'set new upper or lower bound
            lb = a2
        Else
            ub = a2
        End If
    Wend
    txtL1.Text = a2 'output the answer to the GUI
    v = 2 * Pi * (a1 - a2) / p1 'compute the orbital velocity of the L1 particle
    txtVelocity.Text = v 'output the orbital velocity to the GUI
End Sub

 

[agentgonzo]

Here's a calculator I made to find the exact spot. It does it numerically...

Can't you do it analytically?
I haven't read through it in detail, but the following gives functions for finding the all 5 Lagrange points.
http://www.physics.montana.edu/faculty/cornish/lagrange.pdf

 

[tony]

I'm not sure it can be done analytically. In that paper it states "We are unable to find closed-form solutions to equation 10 for general values of alpha, so instead we seek approximate solutions valid in the limit alpha << 1."

When I plug numbers in their formula, L1 is off by nearly 1 million km. The numerical solution I posted will get you within 1 meter, assuming your masses and sma are accurate.

 

[ar81]

More on lagrange points
http://en.wikipedia.org/wiki/Lagrange_point