I was thinking about this problem:

How much work is required to hold an object stationary in a gravitational field?


How much energy is required to keep an object stationary in a gravitational field, i.e. how many J/s are required?

and I realized I could not come up with a satisfactory answer.

Let's setup an example to limit our scope: First of all let's consider an mini-copter of mass m=10 kg on earth which uses its' rotor blades to reach altitude of h=1 meter. The work required to get it there is roughly: mgh = 9.82 m/s * 10 kg * 1 m.

Which means we spent 98.2 Joules of energy getting it there. So if this work was done under the time t= 2s the effect of the mini-copter engine would be roughly $mgh/t$ ~ 98.2 Joules / 5s ~ 19.64 watt.

Now, obviously the engine would have to keep on running to keep the quadrocopter at the height of 1m, roughly how much effect would be required?

So this is kind of a high-school level question but I can't seem to wrap my head around it.

  • $\begingroup$ Depends on the size of the rotor blades. The larger the blades, the less power is required. Can you work out why? $\endgroup$
    – CuriousOne
    Dec 23 '14 at 14:06
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    $\begingroup$ For a mini-copter, it would be better to consider the work done on the rotor blades and the copter's main body separately: you want to keep the main body stationary, so no displacement, so main body does no work. You want the rotor blades to move (rotate) to supply the lifting force. The rotor blades do work: they do work by the driving torque, and do work against air resistance. The works for and against should cancel, so the rotor blade KE is constant. The impacting of the blades with the air causes the lifting force, and blades need to be fast enough for a strong enough lift. $\endgroup$
    – Involute
    Dec 23 '14 at 14:34
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    $\begingroup$ Be careful: a helium balloon expends no energy to stay aloft. (To you naysayers who want to work in a vacuum: all your helicopters won't stay up in a vacuum either :-) ) $\endgroup$ Dec 23 '14 at 15:25
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    $\begingroup$ For a craft of mass $M$, and the gravitational field strength at a certain point is $g$. The weight acting on the craft is $Mg$. Now let's say you are expelling particles at velocity $v$, and at a mass flow rate of $\dot m$. Now, $\dot m = ρAv$ where $A$ is area of nozzle, $ρ$ is particle density. The rate of change of momentum, i.e the lift force is therefore $ρAv^2$, so $v = $\sqrt {\frac{Mg}{ρA}}$ $\endgroup$
    – Involute
    Dec 23 '14 at 16:32
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    $\begingroup$ @JimmyPettersson Any speed at all, so long as you launch the right number per unit time! $\endgroup$ Dec 23 '14 at 16:40

You do not need to spend any energy to keep an object stationary in a gravitational field, but you do need a force that is opposite to gravity, so that the object can be at rest.

For instance, you can put the object on a table, or "hang" it, and the object will stay there without any energy requirement. So the answer is zero, you do not need to do any work. The fact that you might need to spend energy to generate the force does not change that.

  • $\begingroup$ Yes, in this case the rotor blades will be generating this force by pushing air downwards, the force will be equivalent to mg. How would you estimate the effect of the copters engine with a simplified model? $\endgroup$ Dec 23 '14 at 15:05
  • $\begingroup$ @JimmyPettersson I am not sure what is the question, what do you specifically mean by "the effect". $\endgroup$
    – user65081
    Dec 23 '14 at 15:07
  • $\begingroup$ To me effect = watt = J/s (this might be a misstranslation on the term "effect" on my part), but I think effect in US/UK you instead say power? $\endgroup$ Dec 23 '14 at 15:10
  • $\begingroup$ @JimmyPettersson yes, it might be a translation issue, the power you need is arbitrary, it can go from zero in the case of a table, to arbitrarily high, if you have a very inefficient machine that wastes a lot of energy. There is no "right" (a specific number) answer. $\endgroup$
    – user65081
    Dec 23 '14 at 15:15
  • $\begingroup$ @JimmyPettersson In your initial estimate, you've ignored the whole issue for "getting to 1m height" - you assumed a perfect engine. Within the same assumption, keeping the copter at the same height costs exactly zero energy. With a simple rotor-blade helicopter, this isn't the case, of course - you're generating the force by moving masses of air, and that's the energy you're spending (plus losses to heat etc. as usual). Additionally, this is heavily dependent on height - near the ground, you can exploit energy-saving measures. $\endgroup$
    – Luaan
    Apr 10 '16 at 23:24

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