Rocket and external forces

1. Oct 1, 2007

Oblio

I have this equation:

dP = P(t+dt) - P(t) = mdv +dmv(ex)

(ex=external)

I have to show that that will equal

ma = -ma + F(ex)

(Although in the book, acceleration is shown as the derivative of v, (dotted v)).

In the example I was reading, there are no external forces, so my first equation simply boiled down to mdv=-dmv.
I'm not sure how to substitute in ALL 3 of those momentums above....

Thanks!

2. Oct 1, 2007

Oblio

Sorry. V(ex) is exhaust speed at which the rocket is ejecting spent fuel, relative to the rocket.

3. Oct 1, 2007

Oblio

In further reading, I'm not sure about the earlier stage of the first equation I posted...

A rocket is traveling in the positive x with mass m,and velocity v, and ejecting spent fuel at exhaust speed of v(ex) relative to the rocket. Since the rocket is ejecting mass, the rockets mass m is steadily decreasing. At time t, the momentumis P(t) = mv. A short time later at t+dt the rockets mass is m+dm where dm is negative, and its momentum is (m+dm)(v+dv). The fuel ejected in the time dt has mass (-dm) and velocity v-v(ex) relative to the ground. thus the total momentum (rocket plus the fuel just ejected) at t+dt is

P(t+dt)= (m+dm)(v+dv) - dm(v-v(ex)

If dm is already negative, why is it subtracted again?

4. Oct 1, 2007

Staff: Mentor

The mass of propellant being expelled is the source of the force/momentum applied to the rocket, but it also means less mass to be accelerated or subjected to that force.

The remaining propelleant is a mass, in addition to the payload mass, that must be accelerated.

5. Oct 1, 2007

Oblio

Ah ok.
So the first (m+dm) is the 'cargo' propellant, and the '-dm(v-v(ex)) is my external force?

6. Oct 1, 2007

Oblio

In one way I think it has to be F (ext) but dm is just v... theres no acceleration in those terms...

7. Oct 1, 2007

Oblio

when the book has -mdv =dmv (ex) the simplify by dividing both by dt....

but I have -mg = mdv + dmv (ex).

I can't find a way to use seperation of variables here...

8. Oct 2, 2007

Oblio

Although this is done and handed in, I'm still not sure about:

P(t+dt)= (m+dm)(v+dv) - dm(v-v(ex)

is dm(v-vex) my external force?

9. Oct 2, 2007

learningphysics

Can you write the question word for word? I don't understand what you need to prove.

10. Oct 2, 2007

Oblio

With the help of the book I completed the question, but didn't understand steps along the way that the book made.

Anyways,

Consider a rocket traveling in a straight line subject to an external force F(ext) acting along the same line. Show that its equation of motion is m$$\dot{v}$$ = -$$\dot{m}$$v(ex) + F(ext).
It says to review the derivation of this:

m$$\dot{v}$$=-$$\dot{m}$$v (ex) (eqn 3.6)

from

P(t+dt) = (m+dm)(v+dv)-dm(v-v(ex)) = mv+dmv + dmv(ex) (he neglects 'the doubly small product dmdv)

which turns into:

dP=P(t+dt) - P(t) = mdv+dmv(ex) (eqn 3.4) and this turns into 3.6

So that in my case, I am keeping the external force, where they got rid of it to get 3.6.

So I set it up to be:

-mg=mdv +dmv(ex)

In the very beginning,
am I correct in saying that: -dm(v-v(ex)) is my external force?

11. Oct 2, 2007

D H

Staff Emeritus
No. The external force is something generated by some external agent (hence the name), such as the atmosphere. Forces are additive, so you can add the effective force from thrusting and the external force to get the total force.

12. Oct 2, 2007

learningphysics

The external force just equals $$\frac{dP}{dt}$$. in your book example external force is 0, hence dP/dt is 0. But in this case external force is nonzero.

13. Oct 3, 2007

Oblio

P(t+dt) = (m+dm)(v+dv)-dm(v-v(ex))

what IS -dm(v-vex) then?

P=mv, and that exists before this part of the equation..

14. Oct 3, 2007

learningphysics

-dm(v-vex) is the momentum of the exhaust.

P(t+dt) - P(t) = F(ext)*dt

Don't get vex confused with "external"... vex is the velocity of the exhaust relative to the rocket.