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THE LAGRANGE COEFFICIENTS
The task we are trying to solve is with a given initial position and velocity of an orbiting body at a given instant calculate the position and velocity at any later time. Assume an earth satellite moves in the xy plane of an inertial frame with origin at the earth’s center. Initial position and velocity of the satellite at that frame are
clc; clear all; R0 = [8000.2 -6900.5]; %[km] V0 = [0.450 8.9000]; %[km/s] % Compute the position and velocity vectors when the change in satellite % true anomaly equals to 100 [deg]. dfi = 100; %[deg] % Magnitude of R0 and V0 vectors r0 = norm(R0); %[km] v0 = norm(V0); %[km/s] % Radial component of V0 vr0 = R0*V0'/r0; %[km/s] % Magnitude of the constant angular momentum h = r0*(v0^2-vr0^2)^0.5; %[km^2/s] mu = 398600; % Earth’s gravitational parameter [km^3/s^2] r = h^2/mu*1/(1+(h^2/mu/r0 -1)*cosd(dfi)- h*vr0/mu*sind(dfi)); % Lagrange coefficients in terms of the change in true anomaly f = 1 - mu*r/h^2*(1-cosd(dfi)); g = r*r0/h*sind(dfi); fd = mu/h*(1 - cosd(dfi))/sind(dfi)*(mu/h^2*(1 - cosd(dfi)) -1/r0 - 1/r); gd = 1 - mu*r0/h^2*(1 - cosd(dfi)); % Resultant position and velocity vectors R = f*R0 + g*V0; V = fd*R0 + gd*V0; fprintf('R = %4.2f*i + %4.2f*j [km] \n',R(1),R(2)); fprintf('V = %4.4f*i + %4.4f*j [km/s] \n',V(1),V(2));
R = 3634.07*i + 6101.27*j [km] V = -7.6623*i + 7.5831*j [km/s]
Inertial components of the angular acceleration
Example 10.4, Orbital Mechanics for Engineering Students, 2nd Edition.
The inertial components of the angular momentum of a torque-free rigid body are
Hg = [320; -375; 450 ]; %[kg*m^2/s] IJK
% the Euler angles [deg] are
fi = 20;
theta = 50;
psi = 75;
% The inertia tensor in the body-fixed principal frame is
Ig = [1000, 0, 0;
0, 2000, 0;
0, 0, 3000]; %[kg*m^2]
Obtain the inertial components of the (absolute) angular acceleration Matrix of the transformation from body-fixed frame to inertial frame
QxX = [-sind(fi)*cosd(theta)*sind(psi) + cosd(fi)*cosd(psi), ...
-sind(fi)*cosd(theta)*cosd(psi) - cosd(fi)*sind(psi),sind(fi)*sind(theta);
cosd(fi)*cosd(theta)*sind(psi) + sind(fi)*cosd(psi),...
cosd(fi)*cosd(theta)*cosd(psi) - sind(fi)*sind(psi),-cosd(fi)*sind(theta);
sind(theta)*sind(psi), sind(theta)*cosd(psi), cosd(theta)
]
QxX =
0.0309 -0.9646 0.2620
0.6720 -0.1740 -0.7198
0.7399 0.1983 0.6428
Matrix of the transformation from inertial frame to body-fixed frame
QXx = QxX'
QXx =
0.0309 0.6720 0.7399
-0.9646 -0.1740 0.1983
0.2620 -0.7198 0.6428
Obtain the components of HG in the body frame
Hgx = QXx*Hg % [kg*m^2/s]
Hgx = 90.8616 -154.1810 643.0376
The components of angular velocity in the body frame
Ig_inv = inv(Ig); wx = Ig_inv*Hgx % [rad/s]
wx =
0.0909
-0.0771
0.2143
From Euler ’s equations of motion we calculate angular acceleration in the body frame.
alfa_x = - Ig_inv*cross(wx,Ig*wx) % [rad/s^2]
alfa_x =
0.0165
0.0195
0.0023
Angular acceleration in the inertial frame
alfa_X = QxX*alfa_x % [rad/s^2] IJK
alfa_X =
-0.0177
0.0060
0.0176
Published with MATLAB® 7.10
Small Satellites 