# orbital elements of satellite

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ali hassan on 7 Dec 2020
Edited: Walter Roberson on 26 May 2021
pakistan has launched a missile on india and israel ? which missile will follow which orbit? i.e polar, equatorial, etc? is it easy for pakistan to launch a missile on india or for india to launch a missile on pakistan? prove mathematically.
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ali hassan on 7 Dec 2020
actually, this is a sort of theoretical question based on background knowledge on astromy. i am not good at it but any body like sir MEG NOAH will take at max 5 mins to answer it. it involves no coding but some great background and if anybod helps to catch one end of rope, the other end could easily be found out. i hope its clear but still you feel something wrong in question i am plaesed to answer it sir.and i will like to appreciate you for the help you did in my project TDOA localization.wil be always appreciated sir.
BEST REGARDS

Walter Roberson on 7 Dec 2020
If the Earth is rotating the target around to meet you, then you typically need less fuel.
If the Earth is rotating the target around away from you, then you need longer flight times, but that also potentially gives you more time to do course corrections. Most often, the time for course corrections is not a significant consideration. However, remember that India and Pakistan are literally adjacent to each other, so for a sufficiently small and close target that is rotating to meet you, the Earth might rotate the target past your starting point while you are still in the launch phase, faster than you had a chance to aim, so there is a range of distances for which the extra time of flight as the target rotates away from you is beneficial.
None of the scenarios you describe involve orbits.

### More Answers (1)

Meysam Mahooti on 26 May 2021
Edited: Walter Roberson on 26 May 2021
%--------------------------------------------------------------------------
%
% Elements: Computes orbital elements from two given position vectors and
% associated times
%
% Inputs:
% GM Gravitational coefficient
% (gravitational constant * mass of central body)
% Mjd_a Time t_a (Modified Julian Date)
% Mjd_b Time t_b (Modified Julian Date)
% r_a Position vector at time t_a
% r_b Position vector at time t_b
% Outputs:
% Keplerian elements (a,e,i,Omega,omega,M)
% a Semimajor axis
% e Eccentricity
% i Inclination [rad]
% Omega Longitude of the ascending node [rad]
% omega Argument of pericenter [rad]
% M Mean anomaly [rad]
% at time t_a
%
% Notes:
% The function cannot be used with state vectors describing a circular
% or non-inclined orbit.
%
%
%--------------------------------------------------------------------------
function [a,e,i,Omega,omega,M] = Elements(GM,Mjd_a,Mjd_b,r_a,r_b)
% Calculate vector r_0 (fraction of r_b perpendicular to r_a) and the
% magnitudes of r_a,r_b and r_0
pi2 = 2*pi;
s_a = norm(r_a);
e_a = r_a/s_a;
s_b = norm(r_b);
fac = dot(r_b,e_a);
r_0 = r_b-fac*e_a;
s_0 = norm(r_0);
e_0 = r_0/s_0;
% Inclination and ascending node
W = cross(e_a,e_0);
Omega = atan2(W(1),-W(2)); % Long. ascend. node
Omega = mod(Omega,pi2);
i = atan2(sqrt(W(1)^2+W(2)^2),W(3)); % Inclination
if (i==0)
u = atan2(r_a(2),r_a(1));
else
u = atan2(+e_a(3),(-e_a(1)*W(2)+e_a(2)*W(1)));
end
% Semilatus rectum
tau = sqrt(GM)*86400*abs(Mjd_b-Mjd_a);
eta = FindEta(r_a,r_b,tau);
p = (s_a*s_0*eta/tau)^2;
% Eccentricity, true anomaly and argument of perihelion
cos_dnu = fac/s_b;
sin_dnu = s_0/s_b;
ecos_nu = p/s_a-1;
esin_nu = (ecos_nu*cos_dnu-(p/s_b-1))/sin_dnu;
e = sqrt(ecos_nu^2+esin_nu^2);
nu = atan2(esin_nu,ecos_nu);
omega = mod(u-nu,pi2);
% Perihelion distance, semimajor axis and mean motion
a = p/(1-e^2);
n = sqrt(GM/abs(a^3));
% Mean anomaly and time of perihelion passage
if (e<1)
E = atan2(sqrt((1-e)*(1+e))*esin_nu,ecos_nu+e^2);
M = mod(E-e*sin(E),pi2);
else
sinhH = sqrt((e-1)*(e+1))*esin_nu/(e+e*ecos_nu);
M = e*sinhH-log(sinhH+sqrt(1+sinhH^2));
end