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# Tag Archives: apogee altitude

## 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');