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Bi-Elliptic Hohmann Transfer

In this example we calculate the total change in speed required for a bi-elliptic Hohmann transfer from a geocentric circular orbit of 7200 km radius to circular orbit of 125000 km radius. The apogee of the first transfer ellipse is 190000 km.

 clc;
 clear all;

 R_i  = 7200;     % [km]
 R1_a = 190000;   % [km]
 R_f  = 125000;   % [km]

mu   = 398600;   % [km^3/s^2] Earth’s gravitational parameter
% For initial circular orbit
V_i = (mu/R_i)^0.5;
% Speed at apogee and perigee for the first transfer ellipse
V1_a = (2*mu*R_i/(R1_a*(R1_a+R_i)))^0.5;
V1_p = (2*mu*R1_a/(R_i*(R1_a+R_i)))^0.5;
% Semimajor axes of the first transfer ellipse
a1 = (R_i + R1_a)/2;
% Speed at apogee and perigee for the second transfer ellipse
V2_a = (2*mu*R_f/(R1_a*(R1_a+R_f)))^0.5;
V2_p = (2*mu*R1_a/(R_f*(R1_a + R_f)))^0.5;
% Semimajor axes of the second transfer ellipse
a2 = (R_f + R1_a)/2;
% For target circular orbit
V_f = (mu/R_f)^0.5;

% For bi-elliptic maneuver the total speed change required
dV =  abs(V_i - V1_p)+ abs(V1_a - V2_a) + abs(V_f - V2_p); %[km/s]

% Time required for transfer
t_bi = pi/(mu)^0.5*(a1^1.5+a2^1.5);     %[s]

fprintf('Total speed change = %4.4f [km/s]\n',dV);
fprintf('Time required for transfer = %4.2f [hours]\n',t_bi/3600);
Total speed change = 3.9626 [km/s]
Time required for transfer = 129.19 [hours]

 

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1 Comment

  1. ck says:

    thank you. 🙂

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