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Escape velocity,Sun and Earth surface

April 6, 2013 2:43 pm / Leave a comment

In this example we calculate the acceleration of gravity and the escape velocity at the Sun and Earth surface.

clc;clear all;
G = 6.67384E-11;                %Gravitational constant,[m^3kg-1s-2]

Sun

Ms = 1.98855e30;                %Solar mass [kg]
Rs = 6.955e8;                   %Solar radius [m]
V_esc = (2*G*Ms/Rs)^0.5/1000;   %Escape velocity [km/s]
g = G*Ms/Rs^2;                  %Acceleration of gravity[m*s-2]
fprintf('Acceleration of gravity[m*s-2] %4.2f \n',g);
fprintf('Escape velocity [km/s] %4.2f \n\n',V_esc);

Acceleration of gravity[m*s-2] 274.36 
Escape velocity [km/s] 617.76

Earth

Me = 5.98e24;                   %Earth mass [kg]
Re = 6378000;                   %Earth radius [m]
V_esc = (2*G*Me/Re)^0.5/1000;   %[km/s]
g = G*Me/Re^2;                  %Acceleration of gravity[m*s-2]
fprintf('Acceleration of gravity[m*s-2] %4.2f \n',g);
fprintf('Escape velocity [km/s] %4.2f \n',V_esc);%% Earth

Acceleration of gravity[m*s-2] 9.81 
Escape velocity [km/s] 11.19

Two Body Problem Numerical Solution,Satellite – Earth, R & V after dT

January 16, 2013 11:44 pm / Leave a comment

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');