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# Orbit Properties Exercises

For the Earth remote sensing satellite Landsat 1/2 a Sun and Earth synchronous circular orbit has been selected at which the subsatellite points repeat after 18 days. The orbital period was 103.267 minutes.

clc;
clear all;close all;
T = 103.267*60;         % [s] Orbital period
J2 = 0.00108263;

1. How many orbits are completed within 18 days (round to integer value)?

num = round(18 * 24*3600/T);
fprintf('Number of Orbits(round) =  %d \n', num);
Number of Orbits(round) =  251

2. Calculate the semi-major axis and altitude above Earth’s surface.

mu = 398600;
R_earth = 6378;  % [km] Earth Radius
a = (T^2/(4*pi^2)*mu)^(1/3);
fprintf('Semi-major Axis =  %4.2f [km] \n', a);
H = a - R_earth;
fprintf('Altitude above Earths surface =  %4.2f [km]\n', H);
Semi-major Axis =  7291.23 [km]
Altitude above Earths surface =  913.23 [km]

3. What is the inclination of this Sun and Earth synchronous satellite? For Earth Synchronous Satellites ( 2*pi*m = n*dFi)

m = 18;     % Number of Days
dFi = 2*pi*m/(num);
Fi_earth = 2*pi*T/(86160);
RAAN =  Fi_earth - dFi ;
cosi = - RAAN*a^2/(3*pi*J2*R_earth^2);
inc = acos(cosi)*180/pi;       %Inclination in [deg]
fprintf('Inclination =  %f [deg] \n\n',inc);
dFi =  0.450587 [rad]
Inclination =  99.252960 [deg]

Consider a satellite in a circular orbit at 800 km altitude and 28.5 inclination. At the time that the satellite passes near a ground station in Los Angeles at a latitude of 34 and longitude of 242, the instantaneous longitude of the ascending node is 150 (longNode). Local smog limits the elevation angle at which one can communicate with the satellite to 8 above the horizon.

R = 800;        % Circular orbit [km]
inc = 28.5;     % Inclination [deg]
GS_lat = 34;    % Latitude  [de]
GS_lon = 242;   % Longitude [deg]

SS_lon = 150;   % Longitude of the ascending node,subsatellite point[deg]
SS_lat = 90-inc;
eps_min = 8;   % [deg]minimum necessary elevation of the ground station

% 1. Compute the follo wing parameters for its pass over the Los Angeles
% ground station

d = 2 *6378;     % [km] Earth visual diameter
D =(6378 + 800); % [km] Distance to the Earth center from Satelite
H = 800;         % [km] Altitude

% a) Earth angular radius [deg],
ro = asin(R_earth/(R_earth + H))*180/pi;
fprintf('a. Earth angular radius = %4.2f [deg] \n',ro);
% b) maximum nadir angle [deg],
xeta_max = asin(cosd(eps_min)*R_earth/(R_earth + H))*180/pi;
fprintf('b. Maximum nadir angle  = %4.2f [deg] \n',xeta_max);
% c) maximum Earth central angle [deg],
lambda_max = 90 - eps_min - xeta_max;
fprintf('c. Maximum Earth central angle  = %4.2f [deg] \n',lambda_max);
% d) maximum distance [km],
Dmax = R_earth* sind(lambda_max)/sind(xeta_max);
fprintf('d. Maximum range Dmax = %4.2f [km] \n',Dmax);
% e) minimum Earth central angle [deg],
lambda_min = asin(sind(SS_lat)*sind(GS_lat) + cosd(SS_lat)*cosd(GS_lat)*...
cosd(GS_lon - SS_lon + 90))*180/pi;
fprintf('e. Minimum Earth Central Angle =  %4.2f [deg] \n',lambda_min);
% f) minimum nadir angle [deg],
xeta_min = atan(R_earth/(R_earth + H)*sind(lambda_min)/...
(1 - R_earth/(R_earth + H)*cosd(lambda_min)))*180/pi;
fprintf('f. Minimum nadir angle  =  %4.2f [deg] \n',xeta_min);

lambda = lambda_min:0.5:lambda_max;
xeta = atan(R_earth/(R_earth + H)*sind(lambda)./...
(1 - R_earth/(R_earth + H)*cosd(lambda)))*180/pi;
figure(1);
hold on;
xlabel('Earth central angle \lambda [deg] ');
plot(lambda,xeta);
hold off;
eps = acos(sind(xeta)*((R_earth + H))/(R_earth))*180/pi;
figure(3);
hold on;
plot(xeta,eps);
ylabel('Spacecraft elevation angle  \epsilon [deg]');
chk_angle = xeta + eps + lambda;
figure(2);
hold on;
plot(xeta,'b');plot(eps,'r');plot(lambda,'k');
plot(chk_angle,'g');
legend('\eta','\epsilon','\lambda','\eta + \epsilon +\lambda');
xlabel('[deg]');
ylabel('[deg]');
% g) Minimum distance [km],
Dmin = R_earth* sind(lambda_min)/sind(xeta_min);
fprintf('g. Minimum range Dmin = %4.2f [km] \n',Dmin);
% h) maximum angular rate [deg/min],
% Period T the time in view is
T = 2*pi*sqrt((R_earth+H)^3/mu)/60;
omega = 360/T;
fprintf('h. Angular Rate = %4.2f [deg/min] \n',omega);
% i) Azimuth range for full ground station pass and [deg]
dFi =2*acos(tand(lambda_min)/tand(lambda_max))*180/pi;
fprintf('i. Azimuth range dFi = %4.2f [deg] \n',dFi);
% j) the time in view [min].
%fprintf('Period T =  %4.2f [min] \n',T);
T_view = T/pi*acos(cosd(lambda_max)/cosd(lambda_min));
fprintf('j. Time in view = %4.2f [min] \n\n',T_view);
a. Earth angular radius = 62.69 [deg]
b. Maximum nadir angle  = 61.63 [deg]
c. Maximum Earth central angle  = 20.37 [deg]
d. Maximum range Dmax = 2523.03 [km]
e. Minimum Earth Central Angle =  5.51 [deg]
f. Minimum nadir angle  =  36.46 [deg]
g. Minimum range Dmin = 1031.34 [km]
h. Angular Rate = 3.57 [deg/min]
i. Azimuth range dFi = 149.86 [deg]
j. Time in view = 11.01 [min]

For a Keplerian orbit with a period T = 205 min, eccentricity e = 0.4 and true anomaly = 60 [deg] determine how much time [s] has passed

%since the perigee passage.
T = 205*60; % [sec]
e = 0.4;
theta = 60; % [deg]
% Eccentric Anomaly E
E = 2 *atan(tand(theta/2)*sqrt((1 - e)/(1  + e)));
fprintf('Eccentric Anomaly E =   %4.2f  [rad] \n', E);
% Compute the mean anomaly M from the formula
M = E - e * sin(E);
fprintf('Mean Anomaly M =   %4.2f [rad]  \n', M);
t = T*M/(2*pi);
fprintf('time passed since the perigee passage =   %4.2f  [s] \n', t);
Eccentric Anomaly E =   0.72  [rad]
Mean Anomaly M =   0.46 [rad]
time passed since the perigee passage =   896.89  [s]

Published with MATLAB® 7.10