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If my rocket requires thrust $T$ to maintain a velocity $v$, the propulsive power can be calculated using

$$P=Tv$$

Assuming the same friction (so the required thrust to maintain the velocity doesn't change) but with double the velocity, I need twice the propulsive power.

I understand the formula but I don't understand the physical reasoning. What does "twice the propulsive power" mean exactly, and why would it need to double to maintain the higher velocity when the resistance to motion (friction) is unchanged?

I understand from wikipedia (https://en.wikipedia.org/wiki/Thrust#Thrust_to_propulsive_power) that a static-fire produces no propulsive power because none of the energy from the thrust goes into moving the rocket, but I don't understand the reasoning here.

Gerhard
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If my rocket requires thrust $T$ to maintain a velocity $v$

That would be unusual for a rocket. Rocket velocity isn't normally limited by friction. In most circumstances, any available thrust will increase the velocity. The equation is your current thrust and current velocity, not some theoretical maximum.

Assuming the same friction (so the required thrust to maintain the velocity doesn't change) but with double the velocity, I need twice the propulsive power.

Again, it's not what you need, it's what you are producing. If the engine produces constant thrust (common for rocket engines), then at double the velocity, it is developing double the power. This can seem odd at first if you think of what's happening to only the rocket. But if you look at what is happening to the energy of the exhaust, it all balances out.

There are several other questions that relate to this formula.
Does a rocket engine apply more power as the rocket's speed increases?

Where does the extra kinetic energy of the rocket come from?

BowlOfRed
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Assuming that the rocket is moving at a constant speed v upwards, the required power would be P1=Tv, as you've stated.

The required thrust would be T=mg+F, where F is the friction force.

If the speed doubles, but friction does not change, the required thrust will stay the same so the power will double: P2=T*2v.

Neglecting the loss of fuel, we can say that the rocket needs twice as much power because, each second, it needs to perform twice as much work against friction and against gravity, i.e., towards increasing its potential energy.

V.F.
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