Understanding Newton's Second Law for Rockets: F=dp/dt Explained

  • Context: Graduate 
  • Thread starter Thread starter RedX
  • Start date Start date
  • Tags Tags
    Rocket
Click For Summary

Discussion Overview

The discussion centers on the application of Newton's Second Law, specifically the equation F=dp/dt, in the context of rocket propulsion and mass ejection. Participants explore the implications of changing mass in a system, the role of relative velocity, and the interpretation of forces acting on a rocket as it expels fuel to achieve orbit.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • Some participants argue that F=dp/dt can be misleading when applied to rockets, suggesting that the equation is more suited for point particles rather than systems with changing mass.
  • Others propose that the force acting on a rocket is better represented by F=m*a, where a is the acceleration of the center of mass, but question how this applies when mass is ejected.
  • A participant clarifies that the velocity v in the equation F=d(mv)/dt refers to the velocity with respect to an inertial frame, not the relative velocity between the rocket and the ejected mass.
  • Another viewpoint suggests that the force depends on the frame of reference, and that using relative velocity in calculations may yield valid results for rocket problems.
  • Concerns are raised about how the force equation accounts for the mass loss of the rocket and whether it can accurately predict the dynamics involved in the ejection process.
  • One participant draws an analogy to a conveyor belt system to illustrate similar principles of mass change and force application without net acceleration.

Areas of Agreement / Disagreement

Participants express differing views on the applicability of F=dp/dt to rocket dynamics, with no consensus reached regarding the interpretation of velocity and the nature of forces acting on the system. The discussion remains unresolved with multiple competing perspectives.

Contextual Notes

Participants highlight the complexity of applying classical mechanics to systems with variable mass, noting the dependence on definitions and the need for careful consideration of reference frames. The discussion reflects uncertainty regarding the assumptions underlying the equations used.

RedX
Messages
963
Reaction score
3
I was looking through this forum, and I noticed one of those problems where a rocket ejects mass to get into orbit.

F=dp/dt

Now the arguments I saw went something like this:

p=mv

F=d(mv)/dt=v(dm/dt)+m(dv/dt)

But that equation is for a point particle I believe. All you can say for a rocket is:

F=m*a

where a is the acceleration of the center of mass.

For one thing, in the equation

F=d(mv)/dt=v(dm/dt)+m(dv/dt)

what is v? It's the relative velocity between the ejected mass and the rocket right? How does F=dp/dt "know" information about how the mass is going to be ejected? You could chuck the fuel with a small velocity or high velocity.

I think when you write

F=d(mv)/dt

you do that because relativity says m is a function of v (which is a function of t). But we are still assuming a point particle, and I don't think this expression can honestly be used for a rocket problem. However, I see everyone doing that, so I'm not too sure of myself on this one.
 
Physics news on Phys.org


Originally posted by RedX
I was looking through this forum, and I noticed one of those problems where a rocket ejects mass to get into orbit.

F=dp/dt

Now the arguments I saw went something like this:

p=mv

F=d(mv)/dt=v(dm/dt)+m(dv/dt)

But that equation is for a point particle I believe. All you can say for a rocket is:

F=m*a

where a is the acceleration of the center of mass.

For one thing, in the equation

F=d(mv)/dt=v(dm/dt)+m(dv/dt)

what is v? It's the relative velocity between the ejected mass and the rocket right? How does F=dp/dt "know" information about how the mass is going to be ejected? You could chuck the fuel with a small velocity or high velocity.

I think when you write

F=d(mv)/dt

you do that because relativity says m is a function of v (which is a function of t). But we are still assuming a point particle, and I don't think this expression can honestly be used for a rocket problem. However, I see everyone doing that, so I'm not too sure of myself on this one.


Well if you know a little calculus u would know that ΣF = ma = dp/dt = dmv/dt = m(dv/dt) = ma. ==> if mass is constant.
 


Originally posted by RedX
For one thing, in the equation

F=d(mv)/dt=v(dm/dt)+m(dv/dt)

what is v? It's the relative velocity between the ejected mass and the rocket right?
No. v is the velocity with respect to an inertial frame.
 


Originally posted by krab
No. v is the velocity with respect to an inertial frame.

Now I thought that the laws of physics are the same for different inertial frames. If you have

F=v(dm/dt)+ma

then the force depends on v, it depends on your frame of reference.

If you put the relative velocity (between the ejected fuel and the rocket) in the calculations, it works for rocket problems. However, I don't think it's legitimate physics to say that it works because of this equation:

F=v(dm/dt)+ma

and Newton meant for v to be the relative velocity.

You can show that the equation works when v is the relative velocity just by considering a mass dm changing its momentum from the rocket velocity to the velocity of the ejected mass, and divide this by dt to get the relative velocity times the mass flow rate.
 


Originally posted by RedX
I was looking through this forum, and I noticed one of those problems where a rocket ejects mass to get into orbit.

F=dp/dt

Now the arguments I saw went something like this:

p=mv

F=d(mv)/dt=v(dm/dt)+m(dv/dt)

But that equation is for a point particle I believe. All you can say for a rocket is:

F=m*a

where a is the acceleration of the center of mass.

For one thing, in the equation

F=d(mv)/dt=v(dm/dt)+m(dv/dt)

what is v? It's the relative velocity between the ejected mass and the rocket right? How does F=dp/dt "know" information about how the mass is going to be ejected? You could chuck the fuel with a small velocity or high velocity.

I think when you write

F=d(mv)/dt

you do that because relativity says m is a function of v (which is a function of t). But we are still assuming a point particle, and I don't think this expression can honestly be used for a rocket problem. However, I see everyone doing that, so I'm not too sure of myself on this one.

When a force acts on an object in SIMPLE, BASIC, classical mechanics, the force is applied onto the object's center of mass. When you draw a free-body diagram of the system, you don't draw the object, you represent the object simply as a point, because unless we are considering rotational motion here, the "size" of the object is irrelevant at this level.

In the above comment, you seem to be mixing the object in question. The system that is under consideration is the rocket. It has a mass m(t). It's mass is changing over time. It's velocity isn't, and this is measured based on some inertial reference frame. The stuff being spewed out of the rocket, as soon as it leaves the rocket, is no longer part of the "system" under consideration. The force F is the force that acts only on the rocket system, and not on the spewed gases, etc. The system here is just the rocket itself, and only the rocket.

You also can't say that

"F=m*a

where a is the acceleration of the center of mass."

because you need to specify the center of mass of what? The rocket? The rocket + ejected mass? Since the mass of the rocket is changing and being redistributed, is the location of the center of mass of the rocket also changing with time? Does this add an added "a" to the overall acceleration in the CM frame?

But what I don't quite understand is your statement:

"How does F=dp/dt "know" information about how the mass is going to be ejected? You could chuck the fuel with a small velocity or high velocity."

This is the issue of the chicken or the egg. When you push an object that is in contact with the floor, and you apply a force just so that it moves with constant velocity, did you know a priori just how much you need to push to get it to move this way? The question has been set up so that the conditions are: (i) it has no net acceleration and (ii) it is also losing mass. One can imaging of throttling the engine just enough to get to achive this, and so getting just the right level of rate of mass loss given the size of the mass "chunks" that the engine is spewing.

BTW, question like this isn't unique. Another popular scenario that is often encountered in intro physics is a conveyor belt moving with constant velocity, while a long length of chains with a constant mass per unit length is dropped onto it at a certain rate. Again, no net acceleration, but the conveyor belt (which is now THE system under consideration) is gaining mass. The treatment of this problem is identical to the rocket.

Zz.
 


Originally posted by RedX
Now I thought that the laws of physics are the same for different inertial frames.
Right. But you weren't using any inertial frame.
 
Okay I understand now. Thanks everyone.
 

Similar threads

Undergrad Meaning of the terms in the formula of the net external force
  • · Replies 17 ·
Replies
17
Views
2K
Insights How to Apply Newton’s Second Law to Variable Mass Systems
  • · Replies 0 ·
Replies
0
Views
2K
Undergrad Newton's Second Law - variable mass case
  • · Replies 14 ·
Replies
14
Views
3K
Undergrad Variable mass F=ma versus Rocket F=ma
  • · Replies 1 ·
Replies
1
Views
2K
Undergrad How to develop the rocket equation?
  • · Replies 47 ·
2
Replies
47
Views
5K
Graduate Momentum at time t and at a time t+dt for a rocket
  • · Replies 8 ·
Replies
8
Views
2K
Graduate Derivation of Rocket Equation Using Relative Velocity
  • · Replies 1 ·
Replies
1
Views
3K
Undergrad Doubt about variable mass systems
  • · Replies 30 ·
2
Replies
30
Views
3K
Undergrad The rocket equation, one more time
  • · Replies 41 ·
2
Replies
41
Views
4K
Undergrad Effective mass from the Lagrangian
  • · Replies 3 ·
Replies
3
Views
2K