Can matlab diffrentiate F with respect to θ , ϕ and ψ only ?.
Any help is apperciated
MATLAB: Symbolic Derivative in matlab
differential equationssymbolic
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This seems to work
syms rp alpha beta theta psi zA=cos(alpha)-cos(beta);B=sin(alpha)-sin(beta);C=(cos(beta)-1)/tan(beta)-(cos(alpha)-1)/tan(alpha);R=A*(cos(theta)-cos(psi))+B*cos(theta)*sin(psi);S=A*sin(theta)*sin(psi)-B*cos(theta)+C*cos(psi);phi=atan(-R/S);x=-rp*(cos(theta)*cos(phi)+sin(psi)*sin(theta)*sin(phi))*cos(alpha)-rp*(-cos(theta)*sin(phi)+sin(psi)*sin(theta)*cos(phi))*sin(alpha)+(rp/tan(alpha))*(cos(psi)*sin(phi)*(cos(alpha)-1)+cos(psi)*cos(phi)*sin(alpha));y=-cos(psi)*sin(phi)*rp;p=[x;y;z];T=Rot('y',theta)*Rot('x',psi)*Rot('z',phi);a1=T*[rp;0;0];a2=T*[rp*cos(alpha);rp*sin(alpha);0];a3=T*[rp*cos(beta);rp*sin(beta);0]; Con1=(p+a1)'*[0;1;0];Con2=(p+a2)'*[-sin(alpha);cos(alpha);0];Con3=(p+a3)'*[-sin(beta);cos(beta);0];disp([Con1; Con2; Con3])function output=Rot(axis,angle)syms x y zif axis==x output=[1,0,0;0,cos(angle),-sin(angle);0,sin(angle),cos(angle)];endif axis==y output=[cos(angle),0,sin(angle);0,1,0;-sin(angle),0,cos(angle)];endif axis==z output=[cos(angle),-sin(angle),0;sin(angle),cos(angle),0;0,0,1];endend K = 3*G*mE/R^5;It's been a while since I've studied any physics, but this looks odd to me.
RE = 6378.137e3 ; % m - radius of earth
RS = 500e3; % m - orbital altitude
R = (RE+RS); % m
mE = 5.972e24; % kg
G = 6.6743e-11; % m^3/kg s^2
By dimensional analysis, K has units of 
or 
times whatever the unit of 3 is. If that's supposed to be the gravitational force between a 3 kg satellite and the Earth, shouldn't that be over R^2 (which would give Newtons, 
) instead of over R^5?
or times whatever the unit of 3 is. If that's supposed to be the gravitational force between a 3 kg satellite and the Earth, shouldn't that be over R^2 (which would give Newtons, ) instead of over R^5?
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