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NEWTON-RAPSON METHOD FOR KEPLER’S ECCENTRICITY

##Constants and initializations
ecc=0.1; ## Eccentricity
M=24.851090; ## Mean anomaly(degrees)
Mr=M*pi/180; ## Mean anomaly (rads)
f=[]; ## The fundamental eqn we want to solve
## Kepler's Equation's zero function f(E)=E-esinE-M=0
fp=[]; ## 1st derivative of f(E). fp(E)=1+ecosE
Ed=[]; ## Eccentric anomaly(degrees)
Er=[]; ## Eccentric anomaly(rads)
Ed(1)=M+0.85*sign(sin(Mr))*ecc; ## The eccentric anomaly initial
## guess for Danby's method(degrees)
Er(1)=Ed(1)*pi/180; ## rads
for i=1:100
f(i)=Ed(i)-ecc*sin(Er(i))-M;
fp(i)=1+ecc*cos(Er(i));
Ed(i+1)=Ed(i)-f(i)/fp(i); ## The iteration for an updated eccentric anomaly E(i+1)
## based on a current value E(i) by Newton-Rapson Method
Er(i+1)=Ed(i+1)*pi/180;
endfor
save -text NEWTONRAPSON.dat
printf('We find E to be %f. This is the answer of Q1(a)\n',Ed(i+1));

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NEWTON-RAPSON METHOD FOR HEAT FLOW

##Constants and initializations
a=5.67E-8; ## Stefan-Boltzman constant[Watt/meter^2Kelvin^4]
e=0.8; ## Rod surface emissivity [Dimensionless]
h=20; ## Heat transfer coefficient of air flow [W/m^2-K]
Tinf=Ts=25; ## Temperature of air and the walls of the close[Celcius]
D=0.1; ## Diameter of the rod[meter]
I2R=100; ## Electric power dissipated in rod (Ohmic Heat)[W]
T=[]; ## Temperature of the rod[*C]
T(1)=25; ## Initial guess of the temperature of the rod[*C]
Q=[]; ## Heat function [W]
Qp=[]; ## First derivative of Q wrt T [W/C*].
for i=1:100
Q(i)=pi*D*(h*(T(i)-Tinf)+e*a*(T(i)^4-Ts^4))-I2R;
Qp(i)=pi*D*(h+4*e*a*T(i)^3);
T(i+1)=T(i)-Q(i)/Qp(i); ## Newton-Rapson Method
endfor
printf('The steady state temperature is %f\n',T(i+1))
save -text HeatFlowTemp.dat
## The plot
t=1:100; ##temperature
for n=1:100
H(n)=pi*D*(h*(t(n)-Tinf)+e*a*(t(n)^4-Ts^4))-I2R;
endfor
plot(t,H)
xlabel('T(Celcius)');
ylabel('Q(Watt)');
legend('Q(T)');
title('Heat flow vs Temperatu…
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