Problem with KE and work equation (perfect rocket in space example)

AI Thread Summary
The discussion centers on calculating the kinetic energy of a rocket using the Tsiolkovsky Δv equation and confirming it with the total work equation. A key issue identified is a flawed approach in deriving the formula for force, as it does not adequately account for the momentum change due to both thrust and mass flow from the fuel. The importance of considering the entire system, including the fuel, is emphasized to avoid miscalculations. Additionally, it is suggested that the empty rocket's mass should be included to prevent singularities when the fuel is depleted. The clarity and organization of the work presented are praised, aiding in troubleshooting the assumptions made.
Jrs580
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Homework Statement
This isn't a formal HW, but here's the problem basically: Calculate the kinetic energy of a rocket after time=t, based off the Tsiolkovsky Δv equation, then confirm it with the total work equation.
Relevant Equations
Δv=v_e ln (m/m_r), 1/2 mv^2, W=Int(F(t).v(t))dt, p=mv, F=dp/dt
Can someone help me by taking a look at the attachment and figuring out where I am making a bad assumption? It's driving me nuts.
 

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Jrs580 said:
Homework Statement: This isn't a formal HW, but here's the problem basically: Calculate the kinetic energy of a rocket after time=t, based off the Tsiolkovsky Δv equation, then confirm it with the total work equation.
Relevant Equations: Δv=v_e ln (m/m_r), 1/2 mv^2, W=Int(F(t).v(t))dt, p=mv, F=dp/dt

Can someone help me by taking a look at the attachment and figuring out where I am making a bad assumption? It's driving me nuts.
I dislike having to open PDF files in order to see the contents of a posting. The PDF has five pages, including a number of typeset equations. Not even a summary is presented here.

Let me go over it in outline.

You begin by defining variables. ##m_i## (initial mass), ##R## (mass flow rate), ##u## (exhaust velocity), ##t## (elapsed time), ##m(t)## (mass as of time ##t##), ##v(t)## (velocity as of time ##t##), ##f(t)## (force on vehicle at time ##t##) and ##K(t)## (kinetic energy of vehicle at time ##t##)

You present the Tsiolkovsky rocket equation and use it to get an expression for ##v(t)## as ##u \ln ( \frac{m_i}{m_i - Rt})##

You proceed to try to derive a formula for ##f(t)## by considering the momentum change of the vehicle. This approach seems flawed.

There are two contributions to the momentum change of the vehicle. One contribution is the thrust from the rocket motor. The other contribution is the mass flow as fuel is pumped (at negligible relative velocity) into the rocket motor and out of the system of interest.

I am going to stop here at the first discovered error.

One more thing. I applaud the organization, style and clarity of your work. You make it clear what you are doing and why. This is extremely valuable assistance in the troubleshooting process.
 
A problem is what you are doing seems to be with the derivative ##\frac{dp}{dt}##. The external force acts on the rocket, the fuel, and the ejecta. Just looking at the rocket velocity doesn't account for enough momentum. Several ways of doing the accounting are developed in:

https://www.physicsforums.com/threads/how-to-develop-the-rocket-equation.961707/post-6895695

Or the entire thread in general.

Also see the recent insight:
https://www.physicsforums.com/insights/how-to-apply-newtons-second-law-to-variable-mass-systems/
 
Last edited:
jbriggs444 said:
You proceed to try to derive a formula for ##f(t)## by considering the momentum change of the vehicle. This approach seems flawed.
I agree with this statement. You call ##f(t)## the "external force". What is the system to which this force is external? Specifically, does it include the fuel which at one moment is part of ##m(t)## and the next moment it is not? That issue is clarified in the insight article that @erobz recommended.

Also, it will be prudent to assume that the empty rocket has some mass, to avoid logarithmic singularities when the fuel runs out at which point the acceleration becomes zero.
 
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