Work performed by a stationary object in a gravitational field “on Earth” [duplicate]

This question already has an answer here:

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

or:

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.

marked as duplicate by knzhou, John Rennie energy StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Sep 24 '18 at 13:47

• Depends on the size of the rotor blades. The larger the blades, the less power is required. Can you work out why? – CuriousOne Dec 23 '14 at 14:06
• 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. – Involutius Dec 23 '14 at 14:34
• 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 :-) ) – Carl Witthoft Dec 23 '14 at 15:25
• 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}}\$ – Involutius Dec 23 '14 at 16:32
• @JimmyPettersson Any speed at all, so long as you launch the right number per unit time! – Carl Witthoft Dec 23 '14 at 16:40