Advanced Question Earth-Moon Gravitational Pull Question

[ghostrunner01]

Is it possible in orbiter and in real life where you are in a spacecraft between the moon and the earth that looks like this:

Earth Spacecraft Moon
O -> O

The Gravitational pull is even so you won't go toward the moon or earth.

Is that Possible? :hmm:


This one never Gets old :compbash:

 

[tblaxland]

You are describing the L1 Lagrange point. Yes, it is possible, but it is not a stable position due to the negative curvature of the gravitational potential at that point. The L4 and L5 points, however, have positive curvature of the gravitational potential and it is possible to orbit them in a stable manner (for example, trojan asteroids do this).

 

[ghostrunner01]

Is it possible to ge to those points in orbiter?

 

[tblaxland]

Is it possible to ge to those points in orbiter?

Sure, though it is no easy navigation task. CJP made this MFD to help though:
[ame="http://www.orbithangar.com/searchid.php?ID=3644"]Lagrange MFD 0.7[/ame]

 

[ghostrunner01]

Sweet

 

[RisingFury]

The stock Orbiter features a DG in one of those points if I remember correctly.

 

[martins]

There is a nice document about the Lagrange points here:
http://map.gsfc.nasa.gov/media/ContentMedia/lagrange.pdf
The interesting thing is that when plotting the potential fields for the stationary solutions in the rotating frame, the L4 and L5 points look unstable (they are maxima), but in fact they are stable, because as soon as a particle starts to wander off, it picks up speed, which means that the Coriolis term in the potential equation (which vanishes for the stationary solution) kicks in, thereby dynamically changing the effective potential field distribution, and the particle begins to orbit the Lagrange point. (Plotting the dynamic potentials as a function of time for an orbiting particle may make an interesting movie). The other points are saddle points and truly unstable (although the L3 point is apparently sufficiently weakly unstable for a particle to remain there for quite a long time (excluding other sources of perturbation).


-----Post Added-----


Ok, instead of doing useful work , I started to play around with this and have now implemented the potential and Lagrange point calculations described in the paper. If you want to try it, here is the Matlab code:

function lagrange

% Plot the effective stationary potential field of a restricted 3-body system
% in the rotating frame

G = 6.67259e-11;
M1 = 10;      % mass of body 1
M2 = 1;       % mass of body 2
mu1 = G*M1;
mu2 = G*M2;
alpha = M2/(M1+M2);
beta  = M1/(M1+M2);
r1 = [-alpha 0 0]; % position of body 1
r2 = [beta 0 0];   % position of body 2
R = alpha+beta;

omega = sqrt((mu1+mu2)/R^3); % angular velocity of rotating frame
Omega = [0 0 omega];

grid=[200 200];
range=[[-1.5*R -1.5*R];[1.5*R 1.5*R]];
X = ([1:grid(1)]-1) * (range(2,1)-range(1,1))/(grid(1)-1)+range(1,1);
Y = ([1:grid(2)]-1) * (range(2,2)-range(1,2))/(grid(2)-1)+range(1,2);

U = zeros(grid);
for i=1:grid(1)
    for j=1:grid(2)
        p = [X(i) Y(j) 0];
        % Gravitational potentials
        R1 = norm(p-r1);
        R2 = norm(p-r2);
        U1 = -mu1/R1;
        U2 = -mu2/R2;
        % Centrifugal potential
        oa = cross(Omega,p);
        Uc = 1/2 * dot(oa,oa);
        U(i,j) = U1+U2-Uc;
    end
end

contourf(X,Y,rot90(max(U,-1.5e-9)),20);
axis equal tight
hold on

% The following plot of the Lagrange points 1-3 requires the Optimisation
% toolbox. Otherwise you need to replace the fzero function with your own
% root finder.

for L=1:5
    [x,y] = CalcLagrange(L);
    plot(x,y,'*')
end

    function [x,y] = CalcLagrange(which)
        if which <= 3
            switch which
                case 1
                    s0=-1;
                    s1=1;
                case 2
                    s0=1;
                    s1=1;
                case 3
                    s0=-1;
                    s1=-1;
            end
            u=fzero(@lagpoly,0);
            x=R*(u+beta);
            y=0;
        else
            x=R/2*(M1-M2)/(M1+M2);
            y=sqrt(3)/2*R;
            if which==5
                y=-y;
            end
        end    
        
        function of=lagpoly(u)
            of = alpha*(s0+2*s0*u+(1+s0-s1)*u^2+2*u^3+u^4) - u^2*((1-s1)+3*u+3*u^2+u^3);
        end

    end

end

Enjoy!

Edit: Note that the simple zeroth order approximations given in the paper for the L1-L3 points (equations 11) do not work for this example, because the M1 >> M2 condition is not satisfied (M1=10 M2). Instead you have to solve for the roots of the 4th order polynomial equation (Eq. 10 in the paper). The L4 and L5 points are exact however. They always form an equilateral triangle with the two bodies, regardless of mass ratio.

 

[ghostrunner01]

Thanks guys for all of this info! I never knew about the lagrange points.

 

[martins]

Found another interesting article here: http://arxiv.org/abs/astro-ph/0612508 which discusses generalised potentials and Lagrange points in eccentric systems. They seem to conclude that for particular mass ratios and eccentricities, the L2 point can become stable too. This is fascinating stuff!

 

[ghostrunner01]

Whoa, I never knew that a simple question could keep growing and growing!

 

[SPASE_1976]

off topic

i have this question about auto capture

Mars moons Deimos and Phobos are believed to be captured asteroids but their orbits are almost perfect circles
How is this possible ? asteroids dont have thrusters and cockpits!
So my question is what happened there and if we can do that effect to orbit the moon without wasting fuel to reduce speed.

 

[Linguofreak]

off topic

i have this question about auto capture

Mars moons Deimos and Phobos are believed to be captured asteroids but their orbits are almost perfect circles
How is this possible ? asteroids dont have thrusters and cockpits!
So my question is what happened there and if we can do that effect to orbit the moon without wasting fuel to reduce speed.

Basically, they reached their current orbits through tidal interactions over very long periods of time. The timeframes involved are too long to be useful for spacecraft.

 

[spacekid1998]

Basically, they reached their current orbits through tidal interactions over very long periods of time. The timeframes involved are too long to be useful for spacecraft.

But the Earth is only 2011 years old?