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