Rocket Propulsion

Classical Mechanics Level pending

A space probe has a total mass of $1.30\times10^{3} kg$ , $5.20\times10^{2} kg$ of which is fuel.

The probe fires its rocket thrusters to make a final course correction,

burning the fuel at a constant rate until its gone. The burn lasts $1.00\times10^{2} s$ and produces a constant thrust of 11000 $N$ .

Estimate the speed of the probe 80.0 seconds into the burn.

Details : Assume the probe is so far away from celestial bodies so that the only significant force acting on the probe

is the thrust from the rocket.

Hint: conservation of linear momentum.

Thanks to Steven Chase for pointing out my error!

1 solution

Amal Hari
Feb 24, 2019

We use conservation of linear momentum ,

Since mass and velocity are variables,

$\dfrac{dP}{dt}=\dfrac{dM}{dt} v +\dfrac{dv}{dt} M$

Given F= $\dfrac{dP}{dt}$ =11000 N

For the rocket,

$dP=dM v_{r} +dv M$

For the exhaust gas at any instant $dP=dM v_{er}$ , $v_{er}$ is the relative velocity of exhaust gas w.r.t rocket.

$\dfrac{ dP}{dt}=\dfrac{dM}{dt} v_{er}$

$\dfrac{dM}{dt} =\dfrac{520}{100} =5.20$

$v_{er}=\dfrac{11000}{5.2}$

$v_{er}=2115.3846 \Longrightarrow$ [1]

The momentum of exhaust gas expelled at any instant is equal and opposite to change in momentum of rocket at any instant.

From a stationary frame of reference,

$dM v_{r} +dv_{r} M=-dM v_{e}$

$dv_{r} M=-dM v_{e}-dM v_{r}$

$dv_{r} M=dM (-v_{e}-v_{r} )$

$v_{e}+v_{r} =v_{er}$ , because velocities are in opposite directions,

$-v_{e}-v_{r} =-v_{er}$

$-\dfrac{dv}{v_{er}} =\dfrac{dM}{M}$

Integrating both sides,

$-\dfrac{v_{f}-v_{i}}{v_{er}} =lnM_{f} -lnM_{i}$

$\dfrac{v_{f}-v_{i}}{v_{er}} =ln\dfrac{M_{i}}{M_{f}}$

$v_{f}-v_{i} =v_{er}ln\dfrac{M_{i}}{M_{f}}$

$v_{80}-0 =2115.3846\times ln\dfrac{1300}{884}$ =815.824