Question

Calculate the force exerted on a rocket when the propelling gases are being expelled at a rate of 1300 kg/s with a speed of 4.5 × 104 m/s.

Answer 1

Considering the rocket+gases as an isolated system, the variation of momentum of the rocket should be equal to the variation of momentum of the gases, which is
$\Delta p = \Delta m \cdot v$
since the speed v at which the gas is expelled is constant.
If we divide both terms per the time interval, $\Delta t$, we get
$\frac{\Delta p}{\Delta t}= \frac{\Delta m}{\Delta t} v$
But $\frac{\Delta p}{\Delta t}$ is equal to the force F exerted on the gas by the rocket (and for Newton's third law, this is equal to the force exerted by the gas on the rocket), while $\frac{\Delta m}{\Delta t}$ is the rate at which the gas is expelled, 1300 kg/s. Therefore, the force exerted on the rocket is
$F= \frac{\Delta m}{\Delta t} v = (1300 kg/s)(4.5 \cdot 10^4 m/s)=5.85 \cdot 10^7 N$