This code is a MATLAB script that can be used to design and analyze Sun-synchronous orbits. A Sun-synchronous orbit is a geocentric orbit which combines altitude and inclination in such a way that an object in this orbit has an a nodal regression rate which is equals to Earth’s orbital rotation speed around the Sun. The object in this orbit constantly illuminated by the Sun. Input: a = Mean semimajor axis, e = Eccentricity

Output: i = Mean orbital inclination

clc; clear all; mu = 398600.440; % Earth’s gravitational parameter [km^3/s^2] Re = 6378; % Earth radius [km] J2 = 0.0010826269; % Second zonal gravity harmonic of the Earth we = 1.99106e-7; % Mean motion of the Earth in its orbit around the Sun [rad/s] % Input a = 7378; % Mean semimajor axis [km] e = 0.05; % Eccentricity

h = a*(1 - e^2); % [km] n = (mu/a^3)^0.5; % Mean motion [s-1] tol = 1e-10; % Error tolerance % Initial guess for the orbital inclination i0 = 180/pi*acos(-2/3*(h/Re)^2*we/(n*J2)); err = 1e1; while(err >= tol ) % J2 perturbed mean motion np = n*(1 + 1.5*J2*(Re/h)^2*(1 - e^2)^0.5*(1 - 3/2*sind(i0)^2)); i = 180/pi*acos(-2/3*(h/Re)^2*we/(np*J2)); err = abs(i - i0); i0 = i; end fprintf('Mean orbital inclination %4.5f [deg] \n',i);

Mean orbital inclination 99.43667 [deg]

Always in the sun? Is that certain? From my understanding, the nodal regression is about ~1 deg/day. Therefore, the beta angle you launch into would be about constant. If you launched into a beta angle that was close to ~70 to 90 (dusk or dawn), then I can see not having an eclipse. But if you launch at Noon…then you would have a 0 beta angle…and your max eclipse time of close to 30 minutes (~1/3 of an orbit). Or thereabout.

Thoughts?

(P.S. I don’t know, but I would think that you would only be able to achieve max beta anlge close to 90 at certain time of the year. My though is that would be around the Spring/Fall Equinox.)