AEE 4263 HW 10

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AEE 4263-02 Space Flight Mechanics Homework HW10 Fall 2023 1. (15 points) A spacecraft has the following inertia tensor: I = [ 500 − 8 − 7 − 8 300 − 5 − 7 − 5 400 ] kg − m 2 a. (5 points) What are the minimum, intermediate, and maximum principal moments of inertia to the nearest kg − m 2 ? b. (10 points) What is the transformation matrix from body to principal coordinates (i.e., C b p ) to four decimal places? Solution: a. Minimum MOI= 501 kg − m 2 Intermediate MOI= 400 kg − m 2 Maximum MOI= 299 kg − m 2 b. C b p = -0.9970 -0.0416 -0.0652 0.0380 -0.9978 0.0551 0.0674 -0.0525 -0.9963 MATLAB: %% Homework 10 Problem 1 close all clear all clc I= [500 -8 -7; -8 300 -5; -7 -5 400] %kg-m^2 [eigenVectors, eigenValues] = eig(I) I_pricipal= [-eigenVectors(:,3), eigenVectors(:,1), eigenVectors(:,2)] I_pricipal'*I*I_pricipal; MOI= diag(eigenValues) 1

2. (15 points) A spacecraft is spinning about the z-axis of its principal frame at 2 π rad / s . The principal moments of inertia about the center of mass are: I xx = 370 kg − m 2 I yy = 420 kg − m 2 I zz = 530 kg − m 2 The nutation damper has the following properties: R = 1 m μ = 0.01 m = 10 kg k = 10000 N / m c = 150 N − s m Use the Routh Hurwitz stability criteria to assess the spacecraft’s stability as a: a. (7 points) major axis spinner b. (7 points) minor axis spinner c. (1 point) why do you not need to evaluate the intermediate axis? Solution: a. Major axis = Stable r1_maj =1538460 r2_maj =23859450 r3_maj =1588940954.55 r4_maj =134666.21 r5_maj =7361318631.66 b. Minor axis = Unstable r1_min =2203740 r2_min =34177050 r3_min =2276680293.88 r4_min =-149591.50 r5_min =2970492893.22 c. Intermediate axis: Did not need to evaluate because motion is not stable by there being different signs on the r’s. Spin is unstable about the intermediate axis. r1_int =1941390 r2_int =29964450 r3_int =1996576766.15 r4_int =-231592.44 r5_int =-2584430101.53 MATLAB: %% Homework 10 Problem 2 close all clear all clc R= 1; %m mu= 0.01; m= 10; %kg k= 10000; %N/m c= 150; %N-s/m w0= 2*pi; % Major 2

Ixx_maj= 370; %kg-m^2 Iyy_maj= 420; %kg-m^2 Izz_maj= 530; %kg-m^2 MOI_maj= [Ixx_maj Iyy_maj Izz_maj]; a4_maj= (1-mu)*m*Ixx_maj*Iyy_maj; a3_maj= c*Ixx_maj*(Iyy_maj+ (1-mu)*m*R^2); a2_maj= k*(Iyy_maj+(1-mu)*m*R^2)*Ixx_maj + (1-mu)*m*((Ixx_maj- Izz_maj)*(Iyy_maj-Izz_maj)-(1-mu)*Ixx_maj*m*R^2)*w0^2; a1_maj= c*((Ixx_maj-Izz_maj-(1-mu)*m*R^2)*(Iyy_maj-Izz_maj))*w0^2; a0_maj= k*((Ixx_maj-Izz_maj-(1-mu)*m*R^2)*(Iyy_maj-Izz_maj))*w0^2 + ((Iyy_maj-Izz_maj)*(1-mu)^2)*m^2*R^2*w0^4; r1_maj= a4_maj r2_maj= a3_maj r3_maj= a2_maj-((a4_maj*a1_maj)/a3_maj) r4_maj= a1_maj- (a3_maj^2 *a0_maj)/((a3_maj*a2_maj) - (a4_maj*a1_maj)) r5_maj=a0_maj % Minor Ixx_min= 530; %kg-m^2 Iyy_min= 420; %kg-m^2 Izz_min= 370; %kg-m^2 MOI_min= [Ixx_min Iyy_min Izz_min]; a4_min= (1-mu)*m*Ixx_min*Iyy_min; a3_min= c*Ixx_min*(Iyy_min+ (1-mu)*m*R^2); a2_min= k*(Iyy_min+(1-mu)*m*R^2)*Ixx_min + (1-mu)*m*((Ixx_min- Izz_min)*(Iyy_min-Izz_min)-(1-mu)*Ixx_min*m*R^2)*w0^2; a1_min= c*((Ixx_min-Izz_min-(1-mu)*m*R^2)*(Iyy_min-Izz_min))*w0^2; a0_min= k*((Ixx_min-Izz_min-(1-mu)*m*R^2)*(Iyy_min-Izz_min))*w0^2 + ((Iyy_min-Izz_min)*(1-mu)^2)*m^2*R^2*w0^4; r1_min= a4_min r2_min= a3_min r3_min= a2_min-((a4_min*a1_min)/a3_min) r4_min= a1_min- (a3_min^2 *a0_min)/((a3_min*a2_min) - (a4_min*a1_min)) r5_min=a0_min % Intermediate Ixx_int= 370; %kg-m^2 Iyy_int= 530; %kg-m^2 Izz_int= 420; %kg-m^2 MOI_int= [Ixx_int Iyy_int Izz_int]; a4_int= (1-mu)*m*Ixx_int*Iyy_int; a3_int= c*Ixx_int*(Iyy_int+ (1-mu)*m*R^2); a2_int= k*(Iyy_int+(1-mu)*m*R^2)*Ixx_int + (1-mu)*m*((Ixx_int- Izz_int)*(Iyy_int-Izz_int)-(1-mu)*Ixx_int*m*R^2)*w0^2; a1_int= c*((Ixx_int-Izz_int-(1-mu)*m*R^2)*(Iyy_int-Izz_int))*w0^2; a0_int= k*((Ixx_int-Izz_int-(1-mu)*m*R^2)*(Iyy_int-Izz_int))*w0^2 + ((Iyy_int-Izz_int)*(1-mu)^2)*m^2*R^2*w0^4; 3

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r1_int= a4_int r2_int= a3_int r3_int= a2_int-((a4_int*a1_int)/a3_int) r4_int= a1_int- (a3_int^2 *a0_int)/((a3_int*a2_int) - (a4_int*a1_int)) r5_int=a0_int 4

3. (20 points) A single stage rocket is launched vertically. Its properties are: Launch mass: m 0 = 62100 kg Mass ratio: n = 7.2 Specific impulse: I sp = 420 s Thrust: T = 942 kN Neglecting drag and assuming a constant gravitational acceleration of g 0 = 9.81 m / s , compute: a. (5 points) Total burn time in seconds b. (5 points) Altitude at burnout in km c. (5 points) Burnout speed in km/s d. (5 points) Maximum altitude in km Solution: a. tb0 = 233.89 sec b. hb0 =388.52 km c. vb0 =5.84 km/s d. hmax =2126.30 km MATLAB: %% Homework 10 Problem 3 close all clear all clc m0= 62100; %kg n= 7.2; Isp= 420; %s T= 942000; %N g0= 9.81; %m/s % a. Total burn time in seconds mf= m0/n; me= T/(Isp*g0); tb0= (m0-mf)/me % b. Altitude at burnout in km c= Isp*g0; hb0= ((c/me)*(mf*log(mf/m0)+m0-mf) - 0.5*((m0-mf)/me)^2*g0)/1000 % c. Burnout speed in km/s vb0= (c*log(m0/mf) - g0*((m0-mf)/me))/1000 % d. Maximum altitude in km hmax= ((0.5*c^2/g0)*log(n)^2 - ((c*m0)/me)*((n*log(n) - (n-1))/n))/1000 5