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Relative momentum and energy efficiency

Physics Asked by sosterta on January 19, 2021

I have a rocket with fuel. The fuel weighs 10 and the rocket weighs 1. I expend all the fuel to accelerate the rocket. I observe the velocity of the expended fuel to be -10, and the velocity of the rocket to be 100, such that the momentum is conserved. I observe the kinetic energy of the fuel to be 1000, and that of the rocket to be 10000. Assuming there is no wasted energy, the potential energy was 11000, and so the efficiency of converting potential energy to rocket energy is 10/11.

Is this correct? Does a rocket become less energy efficient as it expends fuel? Is this somehow taken into account when designing rockets, or is this just a quirk of how the math works out?

One Answer

  • Initial mass of the rocket setup is 11 (The fuel is part of the rocket and is acquiring momentum along with the rocket)
  • A differential equation is to be set for fuel consumption (or the total rocket mass) with respect to time or height or any other quantity to solve for the values.

Answered by TKA on January 19, 2021

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