How Does Power Affect Force in Space Travel?

[K^2]

Still switch this case to a rocket "hovering" above the moon (no atmosphere), the rocket isn't moving, so no work done on the rocket, only on the propellent expelled from the rocket. This is a similar case to the hovering helicopter (the heli doesn't move, only the air). I think it's a reasonable comparason.

Yeah, for hovering, the equations are similar. Though, mass flow rate has to be factored in separately.

So let's see...

$$F = mg = -\dot{m} V_p$$

$$P = -\frac{1}{2} V_p^2 \dot{m}$$

$$\dot{m} = -\frac{mg}{V_p}$$

$$P = \frac{1}{2} mgV_p$$

$$m(t) = m_0 e^{-\frac{gt}{V_p}}$$

$$P(t) = \frac{1}{2} m_0 gV_p e^{-\frac{gt}{V_p}}$$

Ok, maybe not so similar, but the mg term is the same.

 

[norrrbit]

That should be "m-dot": ṁ_p, the mass flow per unit time (not mass flow per second, not important here, but the time it takes to reach terminal velocity is not needed to be known).

Ok rcgldr, your remarks are understood, but I just wanted to make it as pictorial as possible for someone out of physics community.

 

[rcgldr]

That should be "m-dot": ṁ_p, the mass flow per unit time (not mass flow per second

Ok rcgldr, your remarks are understood, but I just wanted to make it as pictorial as possible for someone out of physics community.

My main concern was the impression from your previous post that the expelled propellent would accelerate to terminal velocity in 1 second. Using mass flow rate eliminates having to specify how long it takes for the propellent to reach terminal velocity.