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- 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
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