AEE 4263 HW 3

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Apr 3, 2024

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AEE 4263-02 Space Flight Mechanics Name: Homework HW3 Fall 2023 For all problems, use the following constants: Earth gravitational parameter: μ = 398600.4418 km 3 / s 2 Earth radius: R E = 6378.137 km 1. (15 pts) Given a satellite in an elliptical Earth orbit, with semi-major axis a = 8684.137 km , and eccentricity e = 0.214538 . a) What is the orbit’s specific angular momentum magnitude h in km 2 / s ? b) find the following parameters for true anomalies of 0, 30, 90, 120, 180, 240, 270, and 330 degrees: i. orbital radius r in km . ii. orbital speed v in km / s . iii. radial velocity component v r in km / s . iv. tangential (horizontal) velocity component v in km / s . v. flight path angle γ in degrees . c) What are the apogee and perigee altitudes? d) What is the symmetry between the parameters in part b) ? Solution: a) h = 57464 km^2/s b) i. r(0)=6821km, r(30)=6986km, r(90)=8284km, r(120)=9280km, r(180)=10548km, r(240)=9280km, r(270)=8284km, r(330)=6986km ii. v(0)=8.42km/s, v(30)=8.26km/s, v(90)=7.09km/s, v(120)=6.33km/s, v(180)=5.45km/s, v(240)=6.33km/s, v(270)=7.09km/s, v(330)=8.26km/s iii. vr(0)=0.00km/s, vr(30)=0.74km/s, vr(90)=1.49km/s, vr(120)=1.29km/s, vr(180)=0.00km/s, vr(240)=-1.29km/s, vr(270)=-1.49km/s, vr(330)=-0.74km/s iv. v (0)=8.42km/s, v (30)=8.23km/s, v (90)=6.94km/s, v (120)=6.19km/s, v (180)=5.45km/s, v (240)=6.19km/s, v (270)=6.94km/s, v (330)=8.23km/s v. γ (0)=0.00deg, γ (30)=5.17deg, γ (90)=12.11deg, γ (120)=11.76deg, γ (180)=0.00deg, γ (240)=-11.76deg, γ (270)=-12.11deg, γ(330)=-5.17deg c) ha = 4169.5km, hp = 442.53km d) There is symmetry between 30&330deg, 90&270deg, 120&240deg, for the orbital radius, orbital speed, and horizontal velocity. There is symmetry between 0&180deg, 30&330deg, 90&270deg, 120&240deg for the magnitude of radial velocity component and flight path but they are opposite in sign. MATLAB: close all clear all clc mu= 398600.4418; %km^3/s^2 Re=6378.137; %km a= 8684.137; %km e= 0.214583; %a) What is the orbit’s specific angular momentum magnitude
h= sqrt(mu*(1-e^2)*a) % b) find the following parameters for true anomalies of 0, 30, 90, 120, 180, 240, 270, and 330 degrees: % i. orbital radius r in km. % ii. orbital speed v in km/s. % iii. radial velocity component v_r in km/s. % iv. tangential (horizontal) velocity component v_ in km/s. % v. flight path angle γ in degrees. theta= [0 30 90 120 180 240 270 330]; %deg n=length(theta); %i for i=1:n r(i)= (h^2)/(mu*(1+(e*cosd(theta(i))))); fprintf( 'r(%g)=%0.0fkm,\t' ,theta(i),r(i)) end %ii for i=1:n v(i)= sqrt(mu*((2/r(i))-(1/a))); fprintf( 'v(%g)=%0.2fkm/s,\t' ,theta(i),v(i)) end %iii for i=1:n vr(i)= (mu*e*sind(theta(i)))/h; fprintf( 'vr(%g)=%0.2fkm/s,\t' ,theta(i),vr(i)) end %iv for i=1:n vperp(i)= (mu*(1+e*cosd(theta(i))))/h; fprintf( 'v (%g)=%0.2fkm/s,\t' ,theta(i),vperp(i)) end %v for i=1:n fp(i)= atand((e*sind(theta(i))/(1+e*cosd(theta(i))))); fprintf( 'γ(%g)=%0.2fdeg,\t' ,theta(i),fp(i)) end %c) What are the apogee and perigee altitudes? ha=(h^2/(mu*(1-e)))-Re %km hp=(h^2/(mu*(1-e)))-Re %km %d) What is the symmetry between the parameters in part b) ? disp( 'There is symmetry between 30&330deg, 90&270deg, 120&240deg, for the orbital' ) disp( 'radius, orbital speed, and horizontal velocity. There is symmetry between' ) disp( '0&180deg, 30&330deg, 90&270deg, 120&240deg for the magnitude of radial' ) disp( 'velocity component and flight path but they are opposite in sign.' )
2. (5 pts) Given a spacecraft departing Earth on a hyperbolic trajectory, with a semi-major axis a = 10,000 km and eccentricity e = 1.284 , find the following: a) True anomaly of the asymptote θ in degrees . b) Hyperbolic excess speed v in km / s . c) Aiming radius in km . Solution: a) θ 1 = 141.15deg, θ 2 = 218.85deg b) v = 6.31km/s c) = b = 8053.9km MATLAB: close all clear all clc mu= 398600.4418; %km^3/s^2 Re=6378.137; %km a= 10000; %km e= 1.284; %a) True anomaly of the asymptote θ_∞ in degrees. theta_inf1=acosd(-1/e) %deg theta_inf2=360-theta_inf1 %deg %b) Hyperbolic excess speed v_∞ in km/s. v_inf=sqrt(mu/a) %km/s %c) Aiming radius ∆ in km. b= a*sqrt(e^2-1) %km
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