- #1

- 16

- 0

mathematically analyse, using diagrams, rockets burning in terms of momentum conservation and the rate of fuel consumption R and derive the acceleration for a rocket during its launch stage.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter lebbo
- Start date

- #1

- 16

- 0

mathematically analyse, using diagrams, rockets burning in terms of momentum conservation and the rate of fuel consumption R and derive the acceleration for a rocket during its launch stage.

- #2

cristo

Staff Emeritus

Science Advisor

- 8,122

- 74

- #3

- 22

- 0

- #4

- 261

- 2

Also, the Internet is your friend. We're not going to just tell you how to do the problem, but nothing's stopping you from hunting down explanations online. That way, you'll learn by working through the derivations on your own.

- #5

- 16

- 0

we know that rockets momentum final= R

then the F= R/t

were R is the rate of fuel consumption and F is the total thrust

am i gettin there ?

- #6

- 145

- 0

If we imagine the rocket sitting in outer space it can make the thought process simpler.

If we assume initially the rocket is stationary its initial momentum will be zero, after it starts accelerating the total momentum of the system will still be zero (1st clue).

Secondly... Momentum = Mass * Velocity, however, in this case the mass of the rocket is not constant! Try working out the Thrust via differentiation.

- #7

- 145

- 0

Why do you describe "R" as both the rate of fuel usage and the change of momentum?

- #8

- 16

- 0

and therefore F=(R)(Vex)

Vex is the velocity when the rocket loses the fuel because it goes faster with less weight

total momentum intial= (M+m)V=0

total momentum final=M(V+v)=0

therefore total momentum: (M+m)V=M(V+v)

m : this is delta m is the fuel (before and after)

but wat do i do with the fuel that is expelled

- #9

- 145

- 0

When you write F=R*Vex, if as you also say R=M/T

This is the thrust. However you then describe Vex as something other than the exhaust velocity (then it is not the thrust). Also, its very unclear what you mean by M and m.

Before you try solving this for the general case in which R might vary take R to be constant. Lets also assume that the exhaust velocity is constant.

I say this because your momentum equations look rather confused and we should sort that out first.

Vex = exhaust vel.

Mf = mass of Fuel.

M = Initial mass of rocket.

V = final mass of rocket

Once the rocket has used all its fuel...

Vex*Mf = (M-Mf)*V

Which is the bog standard cons. of mom. eqn.

If your ok with this then we can set up the "rocket equation".

This is the thrust. However you then describe Vex as something other than the exhaust velocity (then it is not the thrust). Also, its very unclear what you mean by M and m.

Before you try solving this for the general case in which R might vary take R to be constant. Lets also assume that the exhaust velocity is constant.

I say this because your momentum equations look rather confused and we should sort that out first.

Vex = exhaust vel.

Mf = mass of Fuel.

M = Initial mass of rocket.

V = final mass of rocket

Once the rocket has used all its fuel...

Vex*Mf = (M-Mf)*V

Which is the bog standard cons. of mom. eqn.

If your ok with this then we can set up the "rocket equation".

Last edited:

- #10

- 16

- 0

yea i get wat ur doin

ur sayin momentum before=momentum after

ur sayin momentum before=momentum after

- #11

HallsofIvy

Science Advisor

Homework Helper

- 41,847

- 969

Share: