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Assumptions:

  • A rocket engine requires a constant mass flow rate to maintain a given force
  • F = ma

Dilemma:

Let's say I take a 1 Newton rocket engine and attach it to a rocket with a mass including the starting fuel and engine itself of 1 kilogram. During a burn, both will experience an acceleration of 1 m/s^2.

For the first 10 seconds, the average velocity will be 5 m/s. The distance covered will be 50 m, so the work done will be 50 J.

For the second 10 seconds, the average velocity will be 15 m/s. The distance covered will be 150 m, so the work done will be 150 J.

For the third 10 seconds, the average velocity will be 25 m/s. The distance covered will be 250 m, so the work done will be 250 J.

How does the work done increase if the mass flow rate of the fuel, and therefore the input energy flow rate, is constant?

Caveats:

In a real rocket, the mass is lost over time, so the mass flow rate decreases or the acceleration increases over time. For a rocket with enough mass and a high-enough energy fuel to burn for more than a few hundred seconds, this effect is negligible relative to the linear work increase over time above.

Qmechanic
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TheEnvironmentalist
  • 2,794

2 Answers2

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Work done is $W=\int_0^x F\,dx$, and in your case $F$ is constant. If you want to calculate this in smaller intervals you should choose space intervals and not time intervals (10s in your case). In each space interval $\Delta x$ the work done is $\Delta W=F\Delta x$ (same for all space intervals).

Clearly, as the rocket accelerates it takes less and less time to cover one such interval. In other words: over time intervals of equal length the work done must increase.

The derivation of the kinetic energy is as always: $$ E=\int_0^xF\,dx=\int_0^xma\,dx=\int_0^xm\frac{dv}{dt}\,dx=\int_0^vmv\,dv=\frac{1}{2}mv^2. $$

A more realistic rocket that looses mass due to fuel consumption is discussed here where also a reference is given.

Kurt G.
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Very simple to explain with words. The amount of energy may remain constant over the intervals you have chosen, but the work is the sum of all of the energies across all of the intervals considered.

JRL
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