Question

A rocket has an initial mass of 28500 kg, of which 20% is the payload (i.e., the mass of the rocket minus the fuel). It burns fuel at a rate of 195 kg/s and exhausts its gas at a relative speed of 1.5 km/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

a) Thrust is the force exerted by the exiting engine gases
= mass flow rate * velocity of exhaust
= 195 kg/s * 1.5 * 1000 m/s = 292500 N

b) Of total load of 28500 kg 80% is fuel = 22800 kg
burning at 195 kg/s it will burn for
t = 22800 kg / 195 kg/s = 117 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 = -195, then
m = -195t + C, C being the constant of integration.
Now m = 28500 when t = 0, hence C = 28500, or
m = 28500 - 195t

If F is the constant thrust then the varying acceleration
a = - g + F / (28500 -195t)

Then if h is the height at time t, we have
h" = a = - g + F/(28500-195t)

Integrating with t we get velocity
v = h' = - gt - (F/195) ln(28500-195t) + D,
where D is the constant of integration.

Since velocity is 0 at t = 0
D = (F/195) ln28500, or

v = - gt + (F/195) ln (28500 / (28500-195t))

Simply substitute t found earlier to get rocket's final velocity when all the fuel is burnt. The rocket will of course decelerate after that due to gravity and its speed will fall, but that's another matter.

v = 1271.5 m/s