Rocket Propulsion Problems and Solutions2

 Problem#1

Rocket Science. A rocket has total mass M_i = 360 kg, including 330 kg of fuel and oxidizer. In interstellar space it starts from rest, turns on its engine at time t = 0, and puts out exhaust with relative speed ve = 1 500 m/s at the constant rate k = 2.50 kg/s. The fuel will last for an actual burn time of 330 kg/(2.5 kg/s) = 132 s, but define a “projected depletion time” as T_p = M_i/k = 144 s. (This would be the burn time if the rocket could use its payload and fuel tanks as fuel, and even the walls of the combustion chamber.) (a) Show that during the burn the velocity of the rocket is given as a function of time by

v(t) = –v_eln[1 – (t/T_p)]

(b) Make a graph of the velocity of the rocket as a function of time for times running from 0 to 132 s. (c) Show that the acceleration of the rocket is

a(t) = v_e/(T_p – t)

(d) Graph the acceleration as a function of time. (e) Show that the position of the rocket is

x(t) = v_e(T_p = t)ln[1 – (t/T_p)] + v_et

(f) Graph the position during the burn.

Answer:

(a) we use

v_f – v_i = v_eln(M_i/M_f)

v_f = –v_eln(M_f/M_i)

Now, M_i = M_i – kt, so

v_f = –v_eln[(M_i – kt)/M_i]

v_f = –v_eln[(1 – kt/M_i)]

With the definition T_p = M_i/k, this becomes

v_f = –v_eln[(1 – kt/T_pk)]

v(t) = –v_eln[(1 – t/T_p]

(b) given: T_p = 144s, v_e = 1500 m/s, then

v(t) = –(1500 m/s)ln[1 – t/144s]

(c) the acceleration of the rocket is given by

a = dv/dt = d{–v_eln[(1 – kt/T_pk)]}/dt

a(t) = v_e/(T_p – t)

(d) given: T_p = 144s, v_e = 1500 m/s, then

a(t) = (1500 m/s)/(144s – t)

(e) we use
x(t) = v_i + ∫v(t)dt

x(t) = ∫₀^t{–v_eln[(1 – t/T_p]}dt

= v_eT_p[(1 – t/T_p)ln(1 – t/T_p) – (1 – t/T_p)]₀^t

x(t) = v_e(T_p – t)ln(1 – t/T_p) + v_et

(f) given: T_p = 144s, v_e = 1500 m/s, then

x(t) = (1.50 km/s)(144s – t)ln(1 – t/144s) + v_et

Problem#2
An orbiting spacecraft is described not as a “zero-g,” but rather as a “microgravity” environment for its occupants and for on-board experiments. Astronauts experience slight lurches due to the motions of equipment and other astronauts, and due to venting of materials from the craft. Assume that a 3 500-kg spacecraft undergoes an acceleration of 2.50 µg = 2.45 x 10^-5 m/s² due to a leak from one of its hydraulic control systems. The fluid is known to escape with a speed of 70.0 m/s into the vacuum of space. How much fluid will be lost in 1 h if the leak is not stopped?

Answer:
The thrust acting on the spacecraft is

∑F = Ma = (3500kg)(2.45 x 10^-5 m/s²) = 8.58 x 10^-2 N

Then, from

F_thrust = (dM/dt)v_e =

8.58 x 10^-2 N = (∆M/3600s)(70.0 m/s)

∆M = 4.41 kg 

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