What power develops rocket's engine if the rocket of mass M is hanging over the surface of earth, flow rate of gas is u. I solved this problem using work-energy theorem, and got the right answer. But how to solve it using dynamical approach? Is this possible?
I want to know how to use it because this problem appears in chapter on linear momentum.
Answer : mgu/2
P.S.
One may notice inappropriate words used for description of the problem, and it's no surprise - I translated it from russian))). I'm not that good at it.

Question

What power develops rocket's engine if the rocket of mass M is hanging over the surface of earth, flow rate of gas is u. I solved this problem using work-energy theorem, and got the right answer. But how to solve it using dynamical approach?  Is this possible?
I want to know how to use it because this problem appears in chapter on linear momentum.
 Answer : mgu/2
P.S.
One may notice inappropriate words used for description of the problem, and it's no surprise - I translated it from russian))). I'm not that good at it.

anonymous · Answer

@Vincent-Lyon.Fr

anonymous · Answer

For variable mass, dynamical equation of motion is as following:
$$\bar F_g+ \bar u \frac {dm}{dt}=m\frac{dv}{dt}$$
since the rocket isn't moving it's net acceleration is zero, so
$$\bar F_g+ \bar u \frac {dm}{dt}=0$$
Second  quantity on left-hand side accounts for the force of the rocket's engine. 
$$\bar u \frac {dm}{dt}=-\bar F_g=-mg$$
Now, the power of the rocket is
$$P = dW/dt$$
$$dW = -mg*dl$$
$$P = -mg \frac{dl}{dt}=mgu$$
$$\frac{dl}{dt}=-u$$
Because here we are considering the negative of the work done to the gas outflowing from the rocket(or what is the same the work done to the the rocket by the outflowing gas). What's the problem here? Why I'm getting mgu instead of mgu/2 obtained through energy approach?

anonymous · Answer

@apoorvk

anonymous · Answer

@UnkleRhaukus

unklerhaukus · Answer

$$KE=\frac {mv^2}2$$/

unklerhaukus · Answer

is  $\text dl$  the vertical distance infinitesimal?

anonymous · Answer

Yes dl is vertical distance covered by gas during dt.

anonymous · Answer

@UnkleRhaukus

anonymous · Answer

The question in general is can we use energy approach equivalently with dynamical approach, are there any limits of using them? Except the inertial reference frames condition.

unklerhaukus · Answer

this question is a little bit beyond me, i am not a rocket scientist yet this article looks helpful though http://www.philsrockets.org.uk/physics.pdf

anonymous · Answer

We went a bit deeper into boundaries of physics I feel weakness too, anyway thanks for the help.

unklerhaukus · Answer

are you a rocket scientist in training?

anonymous · Answer

Nope) It's simply entry exam for master's degree in our university.

anonymous · Answer

I mean question.

unklerhaukus · Answer

What is the masters degree  in?

anonymous · Answer

I mean I want to get study in a university to get master's degree, so it's just preparation questions.

anonymous · Answer

Oh sorry the question is about, what area?
Theoretical physics.

unklerhaukus · Answer

cool, by degree that im about halfway through is a
 bachelor of science (nuclear science technologies)

unklerhaukus · Answer

by*my

anonymous · Answer

It's quite close to my future area of study, but you know I'm experiencing some difficulties for now, cause I've graduated university with technical speciality, not scientific.

anonymous · Answer

I mean all subjects were taught on technical level not scentific.

apoorvk · Answer

@ArchiePhysics I am so sorry, this is master level preparation stuff, but I haven't even begun my bachelors yet. Don't think I'd be able to help here. "(

anonymous · Answer

It's ok man!)

apoorvk · Answer

I guess the dynamics solution you are asking for will involve a lot of complex calculus, so the work-energy theorem is the best suited to such problems. I think @jamesj may be able to help here.

anonymous · Answer

Oh thanks I hope he would help me!