# How much power does it take to keep a massive particle suspended in a gravitational field?

For instance if I have a rocket of mass $m$ in a uniform gravitational field $g$, and I want to keep it floating in the air via thrust alone, then how much power in the form of (say) chemical energy would it expend?

This is a simple question, but I can't seem to find an answer to it. The answer shouldn't be 0, however, applying the definition of work

$$W = \int \textbf{F}\cdot \textbf{dr}$$

means that the work done is 0, since there is no displacement. Also, since

$$P = \int \textbf{F}\cdot \textbf{dv}$$

also means that the Power is 0 since keeping the rocket held in space means its speed is constantly 0.

• My first idea is that the second integral should be applied to the ejected material, which is accelerated by the rocket from rest (in the rocket's frame) to some final v. Mar 22, 2015 at 20:01
• Both integrals should be applied to the ejected material! Indeed the rocket body stays still and is not gaining potential energy so the total work on it is necessarily zero. Jan 15, 2016 at 21:42

The equation you need is that force is equal to the rate of change of momentum. The force is the weight of the rocket, $Mg$, and the rate of change of momentum is the mass ejected from the exhaust per second multiplied by the exhaust velocity.
$$Mg = v\frac{dm}{dt}$$
So choose your exhaust velocity $v$, and you can work out the required $dm/dt$. The power is then just the change in kinetic energy per second so:
$$W = \tfrac{1}{2}v^2\frac{dm}{dt}$$