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

In summary, the problem with the kinetic energy (KE) and work equation in the context of a perfect rocket in space arises from the misunderstanding of how work is done in a system where mass is expelled as fuel. In space, a rocket propels itself by expelling mass at high velocity, which generates thrust. However, the work-energy principle, which relates the work done on an object to its change in kinetic energy, must account for the changing mass of the rocket. This leads to complexities in applying the traditional KE and work equations, as they do not directly apply to systems where mass is not constant, necessitating a more nuanced approach using the rocket equation.
  • #1
Jrs580
20
4
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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  • #2
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.
 
  • #3
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/
 
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  • #4
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.
 

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

What is the work-energy principle in the context of a rocket in space?

The work-energy principle states that the work done on an object is equal to the change in its kinetic energy. In the context of a rocket in space, when the rocket's engines fire, they perform work by expelling exhaust gases, which changes the rocket's kinetic energy. This principle helps us understand how the rocket accelerates as it expends fuel.

How does the lack of air resistance affect the work done by a rocket in space?

In space, there is no air resistance, which means that all the work done by the rocket's engines directly contributes to its acceleration. Unlike on Earth, where air resistance can dissipate energy and reduce the effective work done, in the vacuum of space, the rocket can achieve higher speeds more efficiently because there are no opposing forces to counteract its thrust.

How is kinetic energy calculated for a rocket in space?

The kinetic energy (KE) of a rocket in space is calculated using the formula KE = 0.5 * m * v², where m is the mass of the rocket and v is its velocity. As the rocket accelerates, either by burning fuel or through other means, its velocity increases, which in turn increases its kinetic energy significantly due to the square relationship between velocity and kinetic energy.

What role does thrust play in the work done by a rocket?

Thrust is the force generated by the rocket's engines as they expel exhaust gases. This thrust performs work on the rocket, which is calculated as the product of thrust and the distance over which it acts. The work done by the thrust results in an increase in the rocket's kinetic energy, allowing it to accelerate in the vacuum of space.

Can a rocket in space ever reach a point where it requires no additional work to continue moving?

Yes, a rocket in space can reach a point where it requires no additional work to continue moving, due to the principle of inertia. Once the rocket is in motion and there are no external forces acting on it (like gravity or friction), it will continue to move at a constant velocity according to Newton's first law of motion. However, if the rocket needs to change speed or direction, it must perform additional work through its engines.

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