(a) A rocket is travelling out of the earth’s atmosphere. During flight it burns fuel at a rate of 80kg/s producing a constant thrust of 500kN which drives the rocket in a vertical trajectory. It has a mass of 3500Kg when it reaches a speed of 1220m/s. At this point in the rocket’s trajectory the gravitational pull of the earth is negligible and air resistance can be also be neglected.

Question

(i) Determine the rocket’s velocity 10s after it has reached a speed of 1220m/s

(ii) What would the error in the velocity calculation have been if the mass of the rocket had been assumed constant over the 10s?

Answer 1

Let time t=0 when mass is 3500 and speed is 1220 m/s.
then
F(t)=500 kN
m(t)=3500-80t
a(t)=(500*1000/m(t))=acceleration
=>
(i)
v(10)=v(0)+∫(500000/(3500-80t))dt
for t=0 to 10.
=1220-500000[ln(3500-80t)]/80
=1220+1621.945
=2841.945 m/s
(ii) using initial mass
v(10)=v(0)+(500000/3500*10)
=1220+1428.57
=2648.57 m/s
So error of simplification = (2648.57-2841.945)/2941.945= -6.8%