# 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. – morrna Mar 22 '15 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. – Andrea Jan 15 '16 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:
$$W = \tfrac{1}{2}v^2\frac{dm}{dt}$$