The following first order linear differential equation is a form of Tsiolkovsky\

Question

The following first order linear differential equation is a form of Tsiolkovsky's rocket equation that includes terms for external forces due to weight and atmospheric drag. dv dt Assume m(t) mo-At is the mass of the rocket, mo is the initial mass, ? is the rate at which fuel is burned and t is time in seconds. Assume v(t) is the velocity in meters/second, k is the atmospheric drag coefficient, g is a constant acceleration due to gravity (9.8 m/s) and R is a constant term measuring the force of the rocket's thrust. 1. Show the steps needed to rewrite the above equation in the form =-g + (2-r) u +- (Equation 2) dt 2. Suppose mo -200 kg, R 2000 N, and v(oj -0 meters/sec. Substitute these values into the rocket equation (Equation 2) and solve for velocity as a function of time. Integrate the velocity function to find a function h(t) for the height of the rocket over time. Use the computer to generate separate graphs for both the velocity and height functions on the 3. 4. interval Osts60. Use appropriate vertical scales in each graph and label each axis. Print your graphs and staple to your work. Suppose the mass of the fuel is 50 kg so that at time t-50 seconds all of the fuel has been consumed (rocket engine burnout) and the rocket's mass becomes a constant m 150 kg. Assume that, because the fuel is spent, dm-0 (the mass of the rocket is constant) and R·0 (zero rocket thrust S Find the rocket's velocity and height at burnout. 6 Assuming g 9.8 m/s2 and k 3, use Equation 1 to write a differential equation for the velocity of the rocket after burnout. ONLY need solution to #6 #

Explanation / Answer

for the given situation

from rocket equation

dv/dt = -g + (lambda - k)v/(mo - lambda*t) + R/(mo - lambda*t)

m(t) = mo - lambda*t

lambda = -dm/dt = constant

mo = 50 kg

R = 2000 N

lambda = 1 kg/s

g = 9.8 m/s/s

k = 3 kg/s

v(o) = 0

hence

dv/dt = -9.8 - 2v/(50 - t) + 2000/(50 - t)

solving

v(t) = c1(50 - t)^2 - 9.8t^2/(50 - t) + 20t/(50 - t) - 25500/(50 - t)

v(0) = c1(50)^2 - 510 = 0

c1 = 0.204

hence

v(t) = (-9.8t^2 - 0.204(50 - t)^3 - 20t + 25500)/(50 - t)

hence

5. at t = 50 s

v(50) = 1000 m/s

now, dh/dt = v(t)

dh/dt = v(t) = (-9.8t^2 - 0.204(50 - t)^3 - 20t + 25500)/(50 - t)

integrating

for h(0) = 0

h(t) = t^2(15.1 - 0.068t)

hence at t = 50 s

h(50) = 29250 m = 29.25500km

6. g = 9.8 m/s/s

after burnout

50*dv/( 30.2*50 - v) = dt

integrarting

50*ln(30.2*50 - v) = -t + c

at t = 0, v = 1000 m/s

hence

311.720536 = c

50*ln(30.2*50 - v) = 311.7205362 - t

v = 30.2*50 - e^((311.72 - t)/50)

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