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Relative Motion of Satellites, Numerical Simulation
This example shows how to use the state vectors of spacecraft A and B to find the position, velocity and acceleration of B relative to A in the LVLH frame attached to A. We numerically solve fundamental equation of relative two-body motion to obtain the trajectory of spacecraft B relative to spacecraft A and the distance between two satellites. For more details about the theory and algorithm look at Chapter 7 of H. D. Curtis, Orbital Mechanics for Engineering Students,Second Edition, Elsevier 2010
clc; clear all; % Initial State vectors of Satellite A and B RA0 = [-266.77, 3865.8, 5426.2]; % [km] VA0 = [-6.4836, -3.6198, 2.4156]; % [km/s] RB0 = [-5890.7, -2979.8, 1792.2]; % [km] VB0 = [0.93583, -5.2403, -5.5009]; % [km/s] mu = 398600; % Earth’s gravitational parameter [km^3/s^2] t = 0; % initial time dt = 15; % Simulation time step [s] dT = 4*24*3600; % Simulation interval [s] % Using fourth-order Runge–Kutta method to solve fundamental equation % of relative two-body motion F_r = @(R) -mu/(norm(R)^3)*R; VA = VA0; RA = RA0; VB = VB0; RB = RB0; ind = 1; while (t <= dT) % Relative position hA = cross(RA, VA); % Angular momentum of A % Unit vectors i, j,k of the co-moving frame i = RA/norm(RA); k = hA/norm(hA); j = cross(k,i); % Transformation matrix Qxx: QXx = [i; j; k]; Om = hA/norm(RA)^2; Om_dt = -2*VA*RA'/norm(RA)^2.*Om; % Accelerations of A and B,inertial frame aA = -mu*RA/norm(RA)^3; aB = -mu*RB/norm(RB)^3; % Relative position,inertial frame Rr = RB - RA; % Relative position,LVLH frame attached to A R_BA(ind,:) = QXx*Rr'; % A Satellite k_1 = dt*F_r(RA); k_2 = dt*F_r(RA+0.5*k_1); k_3 = dt*F_r(RA+0.5*k_2); k_4 = dt*F_r(RA+k_3); VA = VA + (1/6)*(k_1+2*k_2+2*k_3+k_4); RA = RA + VA*dt; % B Satellite k_1 = dt*F_r(RB); k_2 = dt*F_r(RB+0.5*k_1); k_3 = dt*F_r(RB+0.5*k_2); k_4 = dt*F_r(RB+k_3); VB = VB + (1/6)*(k_1+2*k_2+2*k_3+k_4); RB = RB + VB*dt; R_A(ind,:) = RA; R_B(ind,:) = RB; time(ind) = t; t = t+dt; ind = ind+1; end r_BA = (R_BA(:,1).^2+R_BA(:,2).^2+R_BA(:,3).^2).^0.5;
close all; figure(1); hold on; plot3(R_A(:,1),R_A(:,2),R_A(:,3),'r'); plot3(R_B(:,1),R_B(:,2),R_B(:,3),'y'); title('Satellites orbits around earth'); legend('Satellite A','Satellites B'); xlabel('km');ylabel('km'); % Plotting Earth load('topo.mat','topo','topomap1'); colormap(topomap1); % Create the surface. radius_earth=6378; [x,y,z] = sphere(50); x =radius_earth*x; y =radius_earth*y; z =radius_earth*z; props.AmbientStrength = 0.1; props.DiffuseStrength = 1; props.SpecularColorReflectance = .5; props.SpecularExponent = 20; props.SpecularStrength = 1; props.FaceColor= 'texture'; props.EdgeColor = 'none'; props.FaceLighting = 'phong'; props.Cdata = topo; surface(x,y,z,props); hold off; figure(2); plot3(R_BA(:,1),R_BA(:,2),R_BA(:,3),'k'); title('The trajectory of spacecraft B relative to spacecraft A'); xlabel('km');ylabel('km');zlabel('km'); figure(3); plot(time/3600,r_BA); title('Distance between two satellites'); xlabel('hour');ylabel('km') min_r = min(r_BA); max_r = max(r_BA); fprintf('Max distance between two satellites %6.4f km \n',max_r); fprintf('Min distance between two satellites %6.4f km \n',min_r);
Max distance between two satellites 13850.3054 km Min distance between two satellites 262.0271 km
Two Body Problem Numerical Solution,Satellite – Earth, R & V after dT
The initial position and velocity of an earth orbiting satellite in earth centered inertial frame is known. In this example we will numerically solve fundamental equation of relative two-body motion to find the distance of the satellite from the center of the earth and its speed after 24 hours.
clc; clear all; R0 = [6750 0 0]; %[km] V0 = [0 10.5 0]; %[km/s] mu = 398600; % Earth’s gravitational parameter [km^3/s^2] t = 0; % initial time dt = 10; % time step [s] dT = 24*3600; % Time interval [s] % Using fourth-order Runge–Kutta method to solve fundamental equation % of relative two-body motion F_r = @(R) -mu/(norm(R)^3)*R; Rd = V0; R = R0; i = 1; while (t <= dT) Rv(i,:) = R; tv(i) =t; k_1 = dt*F_r(R); k_2 = dt*F_r(R+0.5*k_1); k_3 = dt*F_r(R+0.5*k_2); k_4 = dt*F_r(R+k_3); Rd = Rd + (1/6)*(k_1+2*k_2+2*k_3+k_4); Vv(i,:) = Rd; R = R + Rd*dt; t = t+dt; i = i+1; end rn = (Rv(:,1).^2+Rv(:,2).^2+Rv(:,3).^2).^0.5; % Radius Vector vn = (Vv(:,1).^2+Vv(:,2).^2+Vv(:,3).^2).^0.5; % Speed Vector fprintf('Distance from Earth center = %4.2f [km] \n',norm(R)); fprintf('Satellite speed= %4.4f [km/s] \n',norm(Rd));
Distance from Earth center = 77061.78 [km] Satellite speed= 1.5791 [km/s]
Plots
figure(1); hold on;grid on; plot(tv/3600,rn); ylabel('Altitude [km]'); xlabel('Time [hour]'); title('Distance variation of the satellite from the center of the Earth'); figure(2); hold on;grid on; plot(tv/3600,vn); ylabel('Speed [km/s]'); xlabel('Time [hour]'); title('Satellite speed variation');
Two Body Problem Numerical Solution,Satellite – Earth
Given initial position and velocity of an earth orbiting satellite at a given instant.In this example we will numerically solve fundamental equation of relative two-body motion to find the maximum altitude(apogee altitude) reached by the satellite.
clc; clear all; R0 = [3102 5369 2625]; %[km] V0 = [-6.426 0.7735 5.943]; %[km/s] r = norm(R0); % Initial radius [km] v = norm(V0); % Initial speed [km/s] mu = 398600; % Earth’s gravitational parameter [km^3/s^2] a = mu/(2*mu/r - v^2); % Semimajor Axis [km] T = 2*pi*a^1.5/mu^0.5; % Orbital period [s] R_earth = 6378; % Earth radius [km] dt = T/1000; % time step [s] t = 0; % initial time % Using fourth-order Runge–Kutta method to solve fundamental equation % of relative two-body motion F_r = @(R) -mu/(norm(R)^3)*R; Rd = V0; R = R0; i = 1; while (t <= T) Rv(i,:) = R; tv(i) =t; k_1 = dt*F_r(R); k_2 = dt*F_r(R+0.5*k_1); k_3 = dt*F_r(R+0.5*k_2); k_4 = dt*F_r(R+k_3); Rd = Rd + (1/6)*(k_1+2*k_2+2*k_3+k_4); R = R + Rd*dt; t = t+dt; i = i+1; end rn = (Rv(:,1).^2+Rv(:,2).^2+Rv(:,3).^2).^0.5; Alt = rn - R_earth; % Altitude Vector Alt_max = max(Alt); % [km] fprintf('Maximum Altitude = %4.2f [km] \n',Alt_max);
Maximum Altitude = 6249.10 [km]
Ploting the altitude variation during one orbit
figure(1); hold on;grid on; plot(tv/3600,Alt); ylabel('Altitude [km]'); xlabel('Time [hour]'); title('Satellite altitude variation during one orbit');
Small Satellites 