Rocket Acceleration: Understanding Fuel Burn & Kinetic Energy

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SUMMARY

This discussion centers on the relationship between fuel burn and kinetic energy in rocket acceleration, specifically addressing the misconceptions surrounding constant acceleration and fuel consumption. Key concepts include the Tsiolkovsky ideal rocket equation and the Oberth Effect, which explains that a rocket's efficiency increases with speed. Participants clarify that the kinetic energy of both the rocket and the ejected reaction mass must be considered, emphasizing that energy is conserved in the rocket-exhaust system. The discussion concludes that acceleration increases over time with a constant burn rate, challenging the notion of constant fuel consumption.

PREREQUISITES
  • Understanding of Hamiltonian and Lagrangian mechanics
  • Familiarity with the Tsiolkovsky ideal rocket equation
  • Knowledge of the Oberth Effect in rocketry
  • Basic principles of conservation of energy and momentum
NEXT STEPS
  • Study the derivation of the Tsiolkovsky ideal rocket equation
  • Research the implications of the Oberth Effect on rocket design
  • Explore the relationship between fuel burn rate and acceleration in rocketry
  • Examine conservation laws in rocket-exhaust systems
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Aerospace engineers, physics students, and anyone interested in the mechanics of rocket propulsion and fuel efficiency.

ShamelessGit
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This isn't a homework problem, and I'm not sure if I'm putting this in the right section, so I apologize in advance if I'm doing something wrong.

So far I just learned hamiltonian and Lagrangian mechanics, but I was never taught about how fuel burns when a rocket accelerates, and I'm having a conceptual problem now.

It seems to make sense that there is potential energy in the rocket fuel which is being burned in order to accelerate the rocket. So assuming a rocket in space (no other forces) starts at rest, you would say that the energy is U0 = U + T, where U is the energy in the fuel and T is the kinetic energy.

It seems intuitive that the rocket would accelerate constantly, so you would say v = a*t and therefore T = (1/2)m*(at)^2. It seems strange to me that the kinetic energy, and therefore the rate at which fuel has been burned up, is proportional to a square. You can derive the energy equation by time and get dU/dt = - m*a^2*t, which means the rate at which fuel burns goes up linearly with time. That doesn't make sense to me. It seems to say that the rocket has to burn fuel faster when it is up at higher speeds, which would seem to mean that the rate at which it burns fuel is dependent on the reference frame in which you are measuring it, which does not make sense.

I did some math where I assumed only U0 = U + T and that dU/dt was constant, and got that acceleration was proportional to t^-(1/2). Is this wrong? Something about this whole situation seems wrong to me. It seems like very basic physics say that the acceleration of the rocket and the rate at which it burns fuel cannot both be constant.
 
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ShamelessGit said:
It seems intuitive that the rocket would accelerate constantly, so you would say v = a*t and therefore T = (1/2)m*(at)^2.

That's the kinetic energy of the rocket only. You forgot the kinetic energy of ejected reaction mass.
 
You have to be careful when applying Newtonian mechanics, in any form, to a rocket. The rocket's total energy is not a conserved quantity. A rocket works by ejecting mass from the rocket. Another way to look at it: The rocket is transferring momentum (and hence energy) to the external environment.

Yet another way to look at it: Energy is conserved in the rocket+exhaust system, as are linear momentum and mass.

What you should find is that acceleration increases over time given a constant burn rate. You should also be able to derive the Tsiolkovsky ideal rocket equation.
 
thanks guys
 
Actually you are correctly on to a non-intuitive phenomenon known as the Oberth Effect. Basically, the faster you are going, the more energy you get out of the same rocket fuel. This occurs because the rocket generates the same force regardless of relative velocity.
 

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