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We have a rocket in Space that accelerates with 10m/s². Our engine requires one Unit of Energy each second, while expelling neglible amounts of mass¹. Well assume the rocket has a mass of 100kg aswell as an inital speed of 0.

Scenario A: After 1 second the rocket has achieved a speed of 10 m/s while using 1 Unit of Energy.
Scenario B: After 10 seconds the rocket has achieved a speed of 100 m/s while using 10 Units of Energy.

Now introducing the formula for kinetic energy:
kinetic energy is 0.5 * mass * velocity squared $$ KE = \frac{1}{2}m\cdot v^2 $$

Applying this to our Scenarios:
Scenario A: $$ 0.5 \cdot 100\mathrm{kg}\cdot (10\frac{\mathrm{m}}{\mathrm{s}})^2 = 50\mathrm{kg}\cdot 100\frac{\mathrm{m}^2}{\mathrm{s}^2} = 5000\frac{\mathrm{kg}\cdot\mathrm{m}^2}{\mathrm{s}^2} $$

Our rocket now has 5 000 Joules of kinetic Energy, which was achieved using one Unit of Energy. Thats 5000 Joules per Unit of Energy!

Scenario B:
$$ 0.5 \cdot 100\mathrm{kg}\cdot (100\frac{\mathrm{m}}{\mathrm{s}})^2 = 50\mathrm{kg}\cdot 10.000\mathrm{m}\frac{\mathrm{m}^2}{\mathrm{s}^2} = 500.000\frac{\mathrm{kg}\cdot\mathrm{m}^2}{\mathrm{s}^2} $$ Our rocket now has 500 000 Joules of kinetic Energy, which was achieved using one Unit of Energy. Thats 50 000 Joules per Unit of Energy!

My Problem is the discrepancy in how many Joules one Unit yields between the two Scenarios. After (unsuccessfully) trying to research the topic, I think that my mistake must lie within the first part describing my understanding of how a rocket works.

Though if i try to transfer the kinetic energy formula onto my understanding of the rocket, i have to assume an exponential increase in the energy required to double the speed of an object. The problem is that this would be at odds with how real world rockets appear to work. It is not that their TWR drops while they easily go Hypersonic. Trying to explain this with the use of reaction mass is also not helpful because the fuel consumption is linear.
Thus i found myself here, asking the question of what i could be missing.

[1] e.g. an ion drive.

Oblivious_Squid
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1 Answers1

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f you have a constant power output then the acceleration will not be constant. It will actually decrease as the rocket speeds up. This is because $P=Fv$ (for motion in one direction). As the velocity $v$ increases, the force $F$, and hence the acceleration, must decrease.

BioPhysicist
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  • Doesnt this render the concept of Δv in rocketry useless because one can no longer rely on 100m/s of Δv being able to change a rockets velocity by 100m/s when not burning along the same vector? – Oblivious_Squid Apr 14 '21 at 15:23
  • @Oblivious_Squid I am not sure I fully understand what you are asking. The momentum of the system has to be conserved; it is not necessarily the case that the changes in velocity have to be equal. Momentum depends on both mass and velocity – BioPhysicist Apr 14 '21 at 15:44