AE3613 HW 6 Problem 2

function AE3613HW6P2()
%AE3613 Homework #6
%Christopher Bates
%Due 27 Oct 2016

%Problem Number 2
%Setting variables
rpd = 3.14159/180; % radians per degree
minperhr = 60;
secpermin = 60;
mu = 3.986 * 10^5;
%%%%%%From HW5 answer%%%%%%
TOFhr = 36.976952; %Time of flight in hours
TOFmin = TOFhr * minperhr;
TOFsec = TOFmin * secpermin;
a = 58714.9036; % km
e = 0.88456; % e = eccentricity
asd = 288.493; % Dr.P solution in degrees
DeltaV = 3.380; % km/s

n = (mu/(a^3))^(.5); % n = 4.4376e-05

% Using MATLAB's Solve function
        %E1 = eccentric anomaly
syms E2
eqn = ((E2-e*sin(E2))==n*TOFsec); %Kepler's Eqn

solE2 = solve(eqn,E2); %Solution in Radians = 5.08251

%Solution in degrees
solutiondeg = solE2/rpd;  % = 291.2069 degrees

%Calculating error using Dr.P's values

error = (asd-solutiondeg)/asd;

%Percent error = -9.4070e-01% error
i = error*100;
j = solE2;

fprintf('\nThe calculated eccentric anomaly is: %12.4e\n',j);
fprintf('The calculated percent error is: %16.4e %% \n',i);


%%%%%Part 2%%%%%
satmass = 50; %kg
I_sp = 400; % sec
mp= satmass*(1-e^(DeltaV/I_sp)); % .0518 kg

fprintf('The calculated mass of propellant is: %11.4e kg \n',mp);
%%%OUTPUT%%%
%The calculated eccentric anomaly is:   5.0825e+00
%The calculated percent error is:      -9.4070e-01 %
%The calculated mass of propellant is:  5.1799e-02 kg


%Assuming the propellant mass was calculated properly, the impact this will
%have on the smallsat design is probably small. This amount of propellant is
%half of a percent of the weight of the smallsat. The weight of the
%propellant used is just over a pound, so this weight can possibly even be
%recovered by reducing the weight of something else or by cutting the
%amount of fuel in the maneuvering thrusters.
end