Question

A rocket has an initial mass of 29500 kg, of which 20% is the payload (i.e., the mass of the rocket minus the fuel). It burns fuel at a rate of 190 kg/s and exhausts its gas at a relative speed of 2.5km/s.

(a) Find the thrust of the rocket.
kN
(b) Find the time until burnout.
s
(c) Find its final speed assuming it moves upward near the surface of the earth where the gravitational field g is constant.
km/s

Answer 1

.Thrust is the force exerted by the exiting engine gases
= mass flow rate * velocity of exhaust
= 90 kg/s * 2.5 * 1000 m/s = 225000 N

Of total load of 29500 kg 80% is fuel = 23600 kg
burning at 90 kg/s it will burn for
t = 23600 kg / 90 kg/s = 262.22 s

Now the rockets motion is more complex: even though the thrust is constant, the rockets mass is getting less and less, and this needs calculus, as follows:

If m is the mass at time t, then dm/dt = -90, then
m = -90t + C, C being the constant of integration.
Now m = 29500 when t = 0, hence C = 29500, or
m = 29500 - 90t

If F is the constant thrust then the varying acceleration
a = - g + F / (29500 -90t)

Then if h is the height at time t, we have
h" = a = - g + F/(29500-90t)

Integrating with t we get velocity
v = h' = - gt - (F/90) ln(29500-90t) + D,
where D is the constant of integration.

Since velocity is 0 at t = 0
D = (F/90) ln29500, or

v = - gt + (F/90) ln (29500 / (29500-90t))

V= -9.8*262.22+(225000/90) ln(29500/(29500-90*262.22))

V= 1453.75 m/s