Question

4. Matlab Solvers: A Case Study in Mechanics Suppose we have two objects orbiting in space, with masses 1 - and , rotating around each other. For example, think of the earth and the moon, where the moon moves around the earth at distance 1. (Of course, here both the masses and the distance are normalized.) A third object, which is relatively much smaller and does not affect the motion of the first two, is also orbiting in space. Think of this as a space-ship, or a meteorite. To simplify the analysis, we assume that all trajectories lie in the same plane. We choose the barycenter of mass as the origin of our coordinate system. Moreover, we adopt a rotating frame of coordinates. In this moving frame, the earth and the moon are always located at the points (y,y) = (-1,0), (YM,y) = (1 - 4,0), and their barycenter is at (0,0). We denote by (y(t), y(t)) the position of the space ship at time t, w.r.t. this rotating frame of coordinates. The equations of motion are: -91+ Y2 Y2 41 +242 - 2 (dm) ya y2-24-v de ((y1 + x)2 + y2)/2 dM ((41-v)+ y2)/2 = 0.012277471, = 1-. Notice that, in each of above ODEs, the first two terms account for the Coriolis force, due to the fact that the rotating frame is not an inertial frame. The last two terms account for the gravitational pull of the earth and the moon, respectively. The numbers de di

Answer 1

Solution:

According to the data, the following as the MAT hab Code. that involves Code step Code * An ordinary differential equation CODE) is an equation Some shdinary directives. of a function. * Please find belowe MAT Lab by step for the given ODE. % ODE equation in matlab function f f = funicty) f-t*y/sqrt(2-y92); [tvi fi óde23c'funľ, [0,5], 1); [tva f2] = ode 45 ('funt', [o 5], 1); Plot citvifi'-, tv2,62,'--') tu C'8' = -1; /sst(2-82 - 8(9)= tín, 51 gridarisc[o 5 0 1]) ) % initial values. y! - - 28. ty, +54, te-at +2e-at- Cos(3t) -39, -29-843 38,+34 +242 + cascat) 9,60)=1, Y00)=-1, 9(0)=0

t in [o, pil2]. % F(t, Y) for any given ty, I, and y, and name it funsys.m function Fu = funsys (t, ross FVCI,1)= 2 * Y (1) + Y (2) +5 * Y(3)+exp(-2*t). FV (2,1)=-3* Y(1)-2 *Y(2)- 8*Y(3) +2 *exp(-2xt] - Cos (3*t) FV (3,1) = 3* YC1 + 3 *Y(2) +2**(3)+COS(3*t) % Eulers method h=0.1; % step's size N=2400 ; % number of Steps YCI)=1; tva.fi for n=1N y entuz yen)+h*(-6* yon)); 220+1) = n*h; end plot (ay) % 4th order Runge kutta method cl clf clear all

a to solve dy/dt = f(t, y). In general it can be function of both % define fct, f) fenstr='y-t^2 +l; f=inline (fenstr, It's 'y' ); % to initial time to=0; % Yo, corresponding Valle of y at to yo-0.5 -70 tf, upper boundary of timet (end time) tf=2; %on, number of steps to take n=5" 4 % title Display disp (sprintf ("In In The 4th order Runge- kulta Method of soling ordinary differeth Equations' he(tf-to]/n; disp (sprintf("b = (tf-t.)/n's) disp (sprintf (' disp (sprintf (' = -log', h)) (%9-%og) Vlog,'tf, to no)

% Adding step size disp sprintf = f(49,79), fact, ya) tali) = to s ya (1)= you for i=lan diep (sprintf ("\n Step login disp (speintf ('. _'. %. taciti = ta cioth; % Calculating ki, k2, 63 and k4 ki =f(ta(i), yaci) k2= f(ta (1) +0.5*b, yacito.5 tkih)) K3 = f(ta (1) +0.5*h, yaci) +0.5*k2 *h); k4= f(ta li lth, yaci+k3 *h); using 4ith order Runge- kutta formula ya (i+1) = ya ci)+'/* (kit2**2 +2*k3+k4-)th; disp (1) Find kl and ke using Step information.') disp (Sprintf fkl= f(49g, 77.9) 1-1, 1-13) C' %g& in', KID) the previous =% disp (speintf('

%g in 'ska) disp cspscent 7 (23 € (19.8 +0.5 th ,9%+ 0.5xkixh) 1-1, 1-19) dispespeíutt ('=f(%8+0.5*79,79 +0.5*"ng* %8)) tali), h, yaci), ki, h)) disp (sprintf ('=f(log, tog), tacis +0.5th, yali) +0.5*k1th)) disp (sprintf C disp(sprintf ("K3 =f(%9 +0.5*h, 97.9 +0:54k2 #h), 1-1, 1-1) disp (sprintf('=f(79+0.5*%],"1.940.5*1.98 staci, b, yali), k2, h)) disp Esprintf (=46%9,99) ta cito:6*b, yali) to:5* K2 * h) disp (sprintf(' disp Espriutf (" k4 = f(x%.gth, 98.9 +23* h3, 1-11-1) dispe spriutt(' = f(log, lg), tacith, yali)+k3*h)) disp (sprintf(" =%g |h, ki)) displspeintt (12) Apply the Runge-kutta 4th Sider metood to estimate y %g ? dispe sprint f C 978-999+/6* (kl+2*k2 +20 kg +4)*h, 1,-1)) disp (sprintf(' = %g + %g* 7. g*%g' facia/6, (kl+ 2 * k2t2 #k3tk4), h) Log In, k3)

Witaciti) 1 *')) disp (sprintf(! - login', ya (i+19)) disp (spréutf(' at trg=%g!, Ai end disp (sprintf (Inin *** Results ** % The following finds what is Exact solution tspan = [to tf]; t, y] = ode45 Cf, Espanya) [yfi dummy) = Size(y): yf = y Cyfi); called the plotting te eract and approximate solution of the ODE. hold on alable ('x'), glabel (y); title ('. Exact and Approximate solution of the ODE by the 4th older Runge-kutta metrach plot it, y, --', 'hine Width', 2) color', [o o 0 plot (ta, ya, -, LineWidth, 2, 'Color', [ oo] legend ("Exact', 'Approximation');

dont represent the exact) disp('The figure window that now appears shows the approximate solutions dispe pre celúse Continuous Straight lines The blue line disp ( Solution. In this case "exact refers to the Scullution obtained by the's disptsprintf('Matlab function ode45. \n') disp('Note error at the approximate valle obtained as well as the true value and') disp(' relative true our desired Point 7cxf'.) disp (sprintf ('\n Approximate-zg', yaln+i))} dés (sprintf ("Exact = %g'-gf) dis (sprintf Cin True Error - Eract - Approximci désp (sprintf ('='lg - yog', yf, ya cnti ))) disp (sprintf(" = %g!, If - ya(n+1))) disp (sprintf ('ln Absolute Relative True Error Percentage')) disp (sprintf(' | CExact - Approximate) /Exact *100')) disp (sprintf(' - 1%91%9*100 yıf-yam+i) yf)

- disp (sprintf ľ 179 18.91*100 yf- ya (n+1), yf) disp (sprintf ( % g'abs (lyf-Ya(n+1)/ yt )*100) dispe sprintf(' In The approximation can be more accurate if we made one our Step')) disp (sprīntt('size smaller (hat is, increasing the number of steps ). In In's)

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