Force on the rocket in the rocket problem

[dqqd]

 I recently learnt the rocket problem: how to derive rocket  equation. I accidently saw the problem about the force that is exerted on rocket part (rocket and unused fuel). 

My problem is: F(t) fore force that is exerted on rocket part (rocket and unused fuel) at time t should be F(t)=mdv/dt+vdm/dt, where m is the mass of rocket part at time t and v is the velocity of rocket part at time t, according to F=dp/dt=mdv/dt+vdm/dt which is true as the foundation of established physics, right? 
If we use the velocity function with respect to mass derived from the rocket equation v=v0+ve*ln(m0/m) and assume rocket expels gas mass at a constant mass flow rate R: m(t)=m0-Rt, the final result of force F(t)=mdv/dt+vdm/dt=ve*R+R*(v0+ve*ln(m0/m0-Rt)) where mdv/dt=ve*R which is called the thrust of rocket in many materials, and vdm/dt=R*(v0+ve*ln(m0/m0-Rt)). Thus, the force on the rocket is not only just the thrust but acutally the thrust plus vdm/dt=R*(v0+ve*ln(m0/m0-Rt)), right? 

[Genady]

Consider a bucket of water with mass \(m\) and the water evaporating at rate \(\frac {dm}{dt}\). No force is applied to the bucket.

Consider now the same bucket in a reference frame where it moves with velocity \(v\). If we apply the formula "F=dp/dt=mdv/dt+vdm/dt" in this reference frame, we get a ridiculous result that there is a force \(F=v \frac {dm}{dt}\) on the bucket.

Conclusion: this is a wrong way to apply the above formula.

Edited Thursday at 02:14 PM by Genady

[swansont]

45 minutes ago, Genady said:

Consider a bucket of water with mass m and the water evaporating at rate dmdt . No force is applied to the bucket.

Yes, there will be. A molecule of mass M that leaves at speed v has a momentum of Mv. The residual mass will have a momentum of Mv in the opposite direction, because momentum is conserved. Since m>>M the momentum will be small.

Since force is the time rate of change of momentum, the force will be v dm/dt

An extreme case of this would be water boiling and steam channeled through a nozzle.

[Genady]

58 minutes ago, swansont said:

Yes, there will be. A molecule of mass M that leaves at speed v has a momentum of Mv. The residual mass will have a momentum of Mv in the opposite direction, because momentum is conserved. Since m>>M the momentum will be small.

Since force is the time rate of change of momentum, the force will be v dm/dt

An extreme case of this would be water boiling and steam channeled through a nozzle.

Remove the bucket and consider a ball of water in free fall. The molecules leave in all directions. The total force on the ball is zero.

[swansont]

2 hours ago, Genady said:

Remove the bucket and consider a ball of water in free fall. The molecules leave in all directions. The total force on the ball is zero.

Yes. But there is a bucket. The molecules leaving in one direction (on average) is an important part of the scenario.

Much like an explosion in freefall vs a rocket

[Genady]

14 minutes ago, swansont said:

Yes. But there is a bucket. The molecules leaving in one direction (on average) is an important part of the scenario.

Much like an explosion in freefall vs a rocket

There is no rocket in my scenario. My scenario demonstrates that by applying the quoted formula wrongly, we get a ridiculous result that a force on the bucket is proportional to velocity with which an observer runs by it.

[swansont]

26 minutes ago, Genady said:

There is no rocket in my scenario.

Except there is, since evaporation would exert a force.

26 minutes ago, Genady said:

My scenario demonstrates that by applying the quoted formula wrongly, we get a ridiculous result that a force on the bucket is proportional to velocity with which an observer runs by it.

The v is the velocity of the ejected mass relative to the bulk mass. It doesn’t change in the other frame

If that was your point, it was not at all clear to me.

[studiot]

I feel the OP question was better laid out in the PMs I have been receiving.

Quote

a problem about the force that is exerted on rocket part (rocket and unused fuel) and I saw you answered a lot about the problem.

I would like to consult you about the problem: F(t) fore force that is exerted on rocket part (rocket and unused fuel) at time t should be

F(t)=mdv/dt+vdm/dt,

where m is the mass of rocket part at time t and v is the velocity of rocket part at time t,

according to F=dp/dt=mdv/dt+vdm/dt which is true as the foundation of established physics, right?

If we use the velocity function with respect to mass derived from the rocket equation v=v0+ve*ln(m0/m)

and assume rocket expels gas mass at a constant mass flow rate R: m(t)=m0-Rt, the final result of force F(t)=mdv/dt+vdm/dt=ve*R+R*(v0+ve*ln(m0/m0-Rt))

where mdv/dt=ve*R which is called the thrust of rocket in many materials,

and vdm/dt=R*(v0+ve*ln(m0/m0-Rt)).

Thus, the force on the rocket is not only just the thrust, but acutally the thrust plus vdm/dt=R*(v0+ve*ln(m0/m0-Rt)), right?

Thank you for your help and time and I look forward to your reply.

I am going to be away till later on Saturday so if I can't come up with anything tonight, any input from me will have to wait until later in the weekend.

[Genady]

4 hours ago, swansont said:

The v is the velocity of the ejected mass relative to the bulk mass. It doesn’t change in the other frame

But in the OP, v is not "the velocity of the ejected mass relative to the bulk mass," but rather (my emphasis),

 

9 hours ago, dqqd said:

F(t)=mdv/dt+vdm/dt, where m is the mass of rocket part at time t and v is the velocity of rocket part

It depends on reference frame.

Edited Thursday at 11:19 PM by Genady

[dqqd]

On 11/21/2024 at 11:16 PM, Genady said:

But in the OP, v is not "the velocity of the ejected mass relative to the bulk mass," but rather (my emphasis),

 

It depends on reference frame.

According to F= dp/dt, the force is now defined by the change in momentum. For a variable-mass object, its change in momentum indeed does depend on the inertial frame you choose. So I think it's OK. 

[Genady]

1 hour ago, dqqd said:

According to F= dp/dt, the force is now defined by the change in momentum. For a variable-mass object, its change in momentum indeed does depend on the inertial frame you choose. So I think it's OK. 

Force, however, does not depend on the inertial frame you choose.

[dqqd]

2 hours ago, Genady said:

Force, however, does not depend on the inertial frame you choose.

If the force that you mean is defined by F=ma=mdv/dt for a constant-mass object, then the force does not depend on the inertial frame you choose because the change in velocity does not depend on the inertial frame you choose. However, for a variable-mass object, the force has a new definition: F=dp/dt, and the force now does depend on the inertial frame you choose because the change in momentum really depends on the inertial frame you choose.  I think the main point is that the definition that we use here to define force does matter, since a conclusion or claim about the force should be derived from its definition.

[Genady]

4 minutes ago, dqqd said:

the force has a new definition

I disagree: the force has not been redefined.

In the formula \(F=\frac {dP}{dt}\), \(P\) is the momentum of the entire system. If the system is composed of parts, \(P=P_1+P_2+...\), then \(F=\frac {d(P_1+P_2+...)}{dt}=m_1 \frac {dv_1}{dt}+v_1 \frac {dm_1}{dt}+m_2 \frac {dv_2}{dt}+v_2 \frac {dm_2}{dt}+...\), where no mass crosses the boundary of the composed system.

[swansont]

37 minutes ago, dqqd said:

If the force that you mean is defined by F=ma=mdv/dt for a constant-mass object, then the force does not depend on the inertial frame you choose because the change in velocity does not depend on the inertial frame you choose. However, for a variable-mass object, the force has a new definition: F=dp/dt, and the force now does depend on the inertial frame you choose because the change in momentum really depends on the inertial frame you choose.  I think the main point is that the definition that we use here to define force does matter, since a conclusion or claim about the force should be derived from its definition.

As Genady has pointed out, the v in the equation is the velocity of the ejected mass relative to the parent mass, which is not frame dependent.

The change in momentum will also not depend on the frame, since any velocity term from a frame change that’s included in the calculation cancels out — it’s added to both values.

frame 1: the change in momentum is mv2- mv1

In frame 2, moving at u wrt frame 1, it’s m(v2+u) - m(v1+u) = mv2- mv1

[Genady]

@dqqd, this analysis in Wikipedia is quite clear: Variable-mass system - Wikipedia

[dqqd]

17 hours ago, swansont said:

The change in momentum will also not depend on the frame, since any velocity term from a frame change that’s included in the calculation cancels out — it’s added to both values.

frame 1: the change in momentum is mv2- mv1

In frame 2, moving at u wrt frame 1, it’s m(v2+u) - m(v1+u) = mv2- mv1

Yes, for a constant mass object, the change in momentum does not depend on the frame, as your formula proves. But what about a variable mass object:

Frame 1: the change in momentum is m2v2- m1v1.

In frame 2, moving at a velocity u relative to frame 1: the change in momentum is m2(v2+u) - m1(v1+u) = m2v2-m1v1+u(m2-m1)

where the mass of the object changes from m1 to m2. You can then see that the change in momentum depends on u, the relative velocity of the frame you have chosen.

17 hours ago, Genady said:

In the formula F=dPdt , P is the momentum of the entire system. If the system is composed of parts, P=P1+P2+... , then F=d(P1+P2+...)dt=m1dv1dt+v1dm1dt+m2dv2dt+v2dm2dt+... , where no mass crosses the boundary of the composed system.

If you mean P the momentum of the entire system (including the rocket part and all expelled fuel part to ensure no mass crosses the boundary of the composed system), F here you write will be the net force on the whole system, the sum of the force on the rocket part and the force on the expelled fuel part. Is my understanding right?

[swansont]

31 minutes ago, dqqd said:

Frame 1: the change in momentum is m2v2- m1v1.

In frame 2, moving at a velocity u relative to frame 1: the change in momentum is m2(v2+u) - m1(v1+u) = m2v2-m1v1+u(m2-m1)

where the mass of the object changes from m1 to m2. You can then see that the change in momentum depends on u, the relative velocity of the frame you have chosen.

No, that’s not correct. A change in momentum means the difference in momentum at two different points in time. That’s what the 1 and 2 refer to in my equation. If you have two masses, they each have a velocity at each time.

For a mass M that ejects a mass m, the difference in their momenta is not the change in momentum. The change in momentum of that system is zero, since there is no external force. 

[dqqd]

14 minutes ago, swansont said:

No, that’s not correct. A change in momentum means the difference in momentum at two different points in time. That’s what the 1 and 2 refer to in my equation. If you have two masses, they each have a velocity at each time.

For a mass M that ejects a mass m, the difference in their momenta is not the change in momentum. The change in momentum of that system is zero, since there is no external force. 

If I understand correctly, the system you mean here will includes both the rocket part and all the expelled fuel part, right? But my question is to consider only the rocket part and work out the force exerted on the rocket part, whose mass varies with time.

17 minutes ago, swansont said:

No, that’s not correct. A change in momentum means the difference in momentum at two different points in time. That’s what the 1 and 2 refer to in my equation. If you have two masses, they each have a velocity at each time.

For a mass M that ejects a mass m, the difference in their momenta is not the change in momentum. The change in momentum of that system is zero, since there is no external force. 

And m1 and m2 in my formula are not two mass of two objects but the mass of the same object at two different moment: at t1, the mass of the rocket part is m1; at t2, the mass of the rocket part is m2, because the rocket part mass varies with time, right? 

[swansont]

19 minutes ago, dqqd said:

If I understand correctly, the system you mean here will includes both the rocket part and all the expelled fuel part, right? But my question is to consider only the rocket part and work out the force exerted on the rocket part, whose mass varies with time.

And m1 and m2 in my formula are not two mass of two objects but the mass of the same object at two different moment: at t1, the mass of the rocket part is m1; at t2, the mass of the rocket part is m2, because the rocket part mass varies with time, right? 

The differential equation being used does not have m1 and m2 in it. You can’t mix and match these equations and methods. 

dm is very small, so from one instant to the next the rocket’s mass is approximately constant. 

[studiot]

On 11/21/2024 at 1:51 PM, dqqd said:

I recently learnt the rocket problem: how to derive rocket  equation. I accidently saw the problem about the force that is exerted on rocket part (rocket and unused fuel). 

It seems to me that the current discussion is going round and round in circles, probably due to not fully appreciating the underlying Physics and Maths.

There are several rocket equations, depending upon the circumstances, both concerning the rocket, its surroundings and what you want to know about it.

It is not a good idea to mix up these circumstances.

There are also several possible routes to solving the problem.

Mathematically any equation that can be presented (and solved) as a differential equation can also be presented and solved as an integral equation.
Some treatments do it this way and use what is known as an impulse.

 

It would be very helpful if you could describe your interest further so that we can determine the appropriate response.

Are you primarily in the Physics or the Applied Maths  ?

If we go right back to basics, the rocket problem is a sub class of the dynamics of variable mass systems, both for loosing mass (as in the rocket) or gaining mass (as in the raindrop) or even a bit of both.

A general equation can be derived for all these circumstances, and appropriate values put in for each sub class.

The systems described are described under Galilean/Newtonian relativity, not Einstinian relativity.

As such the common variable linking the parts of the problem together is time.

Differentiation is performed with respect to time. (use t)

The problem is set up considering the chages to the variables during a period  from t to (t+δt)

and the solution found by multiplying out the algebra, discarding the product of small quantities  as negilgible and finally taking the limit as δt tends to zero.