A rocket’s mass decreases as it burns fuel. The equation of motion for a rocket

Question

 A rocket’s mass decreases as it burns fuel. The equation of motion for a rocket in vertical flight can be obtained from   Newton’s law, 

 d/dt(mv)=F-mg Where, m=m0(1-(rt/b) g=g0(R/(R/(R+x))^2 

 In the above equations,   t = time  x = the rocket’s altitude above the earth’s surface (x = 0 when t = 0)  v = upward vertical velocity (v = 0 when t = 0)  R = radius of the earth  

 Note that m and g are NOT constants. F is the rocket’s thrust, its mass is a function of time, and  gravitational acceleration decreases with altitude. The rocket’s initial mass is m 0 , the burn time is  b, and r is the fraction of the total mass accounted for by the fuel.  

 Constants:  F = 48,000 N  m 0  = 2200 kg  r = 0.8  g 0  = 9.81 m/s 2 

 b = 60 s  R = 6.4 x 10 6  m   

 Write a Matlab program to solve the above differential equation and answer the questions below.   HINT: you may need to reformulate the problem. 

Explanation / Answer

%ode for rocket in vertical flight function dxdt=frocket(t,x) global F r b m0 g0 R dxdt = zeros(2,1); %2x1 array for dxdt dxdt(2)=x(1); m=m0*(1-((r*t)/b)); g=g0*(R / (R+x(2)))^2; % v=0; %t=0; %x=0; %X2=v % dxdt(2)=(F/(m0*(1-((r*t)/ b))))-(g0*((R /(R+x(2)))^2)) + ((29.33*x(2))/(m0*(1-((r*t)/b)))); % dxdt(1)=(F/m)-(g) + (29.33)*(x(1))/(m); dxdt(1)=(F/(m0*(1-((r*t)/ b)))) return

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