Question

1. (Coding problem) Consider the satellite equations GMx (x2 + y2)3/2 GMY (x2 + y2)3/2 with the initial condition x(0) = 0,

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Answer #1

(1) The system can be first converted into a system of four ordinary differential equations and then can be solved using ode45 (Runke-Kutta 4th order) of MATLAB by the following code:

G=6.67408*10^(-11);
M=5.974*10^(24);
R=6378.1*10^3;
v=10000;
f = @(t,x) [x(2);(-G*M*x(1))/((x(1))^2+(x(3))^2)^(3/2);x(4);(-G*M*x(1))/((x(3))^2+(x(3))^2)^(3/2)]; % The system %
[t,xa] = ode45(f,[0 10],[0 v 20000+R 0]);
distance=sqrt((xa(85,1))^2+(xa(85,3))^2) % distance of the satelite from the centre of earth %

plot(xa(:,1),xa(:,3))
xlabel('x(t)'), ylabel('y(t)')

You can see the distance we get is 6398880 which is larger than R, i.e., 6378100. The plot (x,y) is given below :
6.3981 X 106 6.3981 6.3981 6.3981 6.3981 y(t) 6.3981 6.3981 6.3981 6.3981 6.3981 0 1 2 3 4 6 7 8 9 10 5 X(t) 104

(b) To determine the v_{min}, one can change v in the above MATLAB code and check the distance for each of this v. Then see for which it is less than R.

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1. (Coding problem) Consider the satellite equations GMx (x2 + y2)3/2' GMY (x2 + y2)3/2" with...
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