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');
Small Satellites 