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https://www.mathworks.com/matlabcentral/answers/499567-plot-the-orbit-of-a-satellite#comment_1188443

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.

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.

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.

%

% Last modified: 2018/01/27 M. Mahooti

%

%--------------------------------------------------------------------------

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

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