Is a Helicopter on a Geostationary Orbit?

[Master Wayne]

Is a helicopter which is stationary relative to the surface on a geostationary orbit? Why, or why not?

 

[Drakkith]

No, it is not in an orbit at all. An orbit is the gravitationally curved path of one object around a point or another body.

 

[Master Wayne]

No, it is not in an orbit at all. An orbit is the gravitationally curved path of one object around a point or another body.

But if it stays still on the air for a whole day, it's trajectory will be a circle around, and slightly above, the surface of the earth. I can't see the difference from a geostationary satellite.

 

[Drakkith]

But if it stays still on the air for a whole day, it's trajectory will be a circle around, and slightly above, the surface of the earth. I can't see the difference from a geostationary satellite.

The difference is that a satellite is in free fall in its orbit around the Earth. A helicopter is not. As an example, let's replace the helicopter with a blimp floating stationary in the air. The air is a fluid, just like water, so a blimp hovering in the air is similar to a submarine underwater. Neither are in free fall under the influence of gravity, so neither are in an orbit.

Edit: Note that there is no real difference for this discussion whether our helicopter/blimp/sub is in the air or underwater. The surface of the Earth is an arbitrary choice for deciding what a satellite is orbiting around. Just think of Jupiter or other gas giants which have no solid surface.

 

[Master Wayne]

The difference is that a satellite is in free fall in its orbit around the Earth. A helicopter is not. As an example, let's replace the helicopter with a blimp floating stationary in the air. The air is a fluid, just like water, so a blimp hovering in the air is similar to a submarine underwater. Neither are in free fall under the influence of gravity, so neither are in an orbit.

Edit: Note that there is no real difference for this discussion whether our helicopter/blimp/sub is in the air or underwater. The surface of the Earth is an arbitrary choice for deciding what a satellite is orbiting around. Just think of Jupiter or other gas giants which have no solid surface.

I see your point. Thanks for the reply.

But now look at it this way. What is the centripetal force which keeps the satellite on a circular trajectory around the planet? Its weight.

And what is the centripetal force which keeps the helicopter on a circular trajectory around the planet? Also its weight. The aerodynamic forces (or hydrodynamic, for that matter) which keep the helicopter (or submarine) suspended are not exactly equal in magnitude to its weight. There's a little weight "left", and that acts as the centripetal force.

So when you look at it, both the satellite and the helicopter are spinning around the planet in a circle due to their weight. The fact that there are other forces acting on the helicopter doesn't really matter, as I see it.

 

[phinds]

To come at it a slightly different way from Drakkith's excellent answer, you seem to think that "orbit" has to do with somethings position relative to a spot on the surface of the Earth. It does not. It has to do with the way that the object is interacting with all of earth. According to your understanding everything on the Earth that isn't moving relative to some spot on the ground is in a geostationary orbit. You've got the "geostationary" part right but not the "orbit" part.

 

[Drakkith]

But now look at it this way. What is the centripetal force which keeps the satellite on a circular trajectory around the planet? Its weight.

And what is the centripetal force which keeps the helicopter on a circular trajectory around the planet? Also its weight.

All of which is irrelevant since an orbit is the path due solely to gravity. A helicopter is being held up by lift and cannot take the path that gravity wants it to.

The aerodynamic forces (or hydrodynamic, for that matter) which keep the helicopter (or submarine) suspended are not exactly equal in magnitude to its weight. There's a little weight "left", and that acts as the centripetal force.

That is incorrect. If the weight of an object is not countered exactly by something such as lift then the object will accelerate downward and fall.

So when you look at it, both the satellite and the helicopter are spinning around the planet in a circle due to their weight. The fact that there are other forces acting on the helicopter doesn't really matter, as I see it.

The fact that there are other forces acting on the helicopter is the exact reason that it isn't in orbit. Note that a helicopter follows none of the laws of orbital mechanics, laws which were developed specifically to describe objects in space that orbit with no other forces other than gravity acting on them.

 

[Master Wayne]

All of which is irrelevant since an orbit is the path due solely to gravity. The fact that there are other forces acting on the helicopter is the exact reason that it isn't in orbit.

Alright, now I got it. It all comes down to the definition of orbit.

That is incorrect. If the weight of an object is not countered exactly by something such as lift then the object will accelerate downward and fall.

If the weight of the object is countered exactly by something such as lift, then the net force on that object is zero. If the net force is zero, then how come the object is traveling in a circle?

If you are sitting in a chair in a room for a whole day, then you will have traveled a full circle after 24 hours. There must be a net force on you for you to do that. The way I have always seen it is that the chair's force on you is slightly smaller than your weight, leaving a small net force pointing towards the center of the planet. Same would go for the helicopter. Don't you agree?

 

[Drakkith]

If the weight of the object is countered exactly by something such as lift, then the net force on that object is zero. If the net force is zero, then how come the object is traveling in a circle?

That's a good question. I actually can't answer it at the moment.

 

[Master Wayne]

That's a good question. I actually can't answer it at the moment.

Thank you very much for your help!

 

[ehild]

Alright, now I got it. It all comes down to the definition of orbit.



If the weight of the object is countered exactly by something such as lift, then the net force on that object is zero. If the net force is zero, then how come the object is traveling in a circle?

If you are sitting in a chair in a room for a whole day, then you will have traveled a full circle after 24 hours. There must be a net force on you for you to do that. The way I have always seen it is that the chair's force on you is slightly smaller than your weight, leaving a small net force pointing towards the center of the planet. Same would go for the helicopter. Don't you agree?

You are right.

ehild

 

[abitslow]

You probably should start a new thread. Posting a series of questions makes it difficult to see what you do and do not understand. How fast are YOU traveling right now? If you answered ANYTHING except "Relative to what?" then you fail physics. You don't "get it". Relative to the CMB radiation? Relative to the center of mass of the Virgo Supercluster? Relative to the path of the Local Cluster? Relative to Sagittarius A*? Relative to the Sun? Relative to the Earth's center? Relative to The Surface of the Earth? You are spinning. Your motion around the center of the Earth depends on your latitude, but is thousands of miles per hour, and I know you know this. Centripetal and centrifugal force are created in a non-inertial rest frame. That is, they exist relative to the surface of the Earth (which is also spinning). OK, too advanced for you I think... you seem to think that the method used to suspend you at a certain height matters. Think about it. Whether it is the ground supporting you, or a rocket engine, or whirling blades lifting air, you are being supported. You have angular momentum. You will tend to follow the surface of the Earth around not just because of friction, but because angular momentum is conserved. It could be that someday you will get into space and experience microgravity (aka weightlessness) first hand. Perhaps if you've learned the physics, you will understand that right now (if your are sitting down) you can feel the force of gravity (which is actually the chair pushing you up) and that being in a gravitational field does not make you feel weight. Unless that gravity is counteracted, you are weightless. So, don't blame gravity for your weight; blame the electromagnetic repulsion (and pauli exclusion) for preventing you from doing what comes naturally (falling to the center of the Earth). There are two (actually at least three) Conservation Laws you must consider Linear Momentum is conserved and Angular Momentum is conserved. It gets a bit more complicated once tidal forces need to be considered, but that's too advanced a subject for now.
Your motion (or lack of it, relative to the surface of the Earth) is quite accurately explained just by considering the conservation of momentum (for you).

 

[Master Wayne]

You probably should start a new thread. Posting a series of questions makes it difficult to see what you do and do not understand. How fast are YOU traveling right now? If you answered ANYTHING except "Relative to what?" then you fail physics. You don't "get it". Relative to the CMB radiation? Relative to the center of mass of the Virgo Supercluster? Relative to the path of the Local Cluster? Relative to Sagittarius A*? Relative to the Sun? Relative to the Earth's center? Relative to The Surface of the Earth? You are spinning. Your motion around the center of the Earth depends on your latitude, but is thousands of miles per hour, and I know you know this. Centripetal and centrifugal force are created in a non-inertial rest frame. That is, they exist relative to the surface of the Earth (which is also spinning). OK, too advanced for you I think... you seem to think that the method used to suspend you at a certain height matters. Think about it. Whether it is the ground supporting you, or a rocket engine, or whirling blades lifting air, you are being supported. You have angular momentum. You will tend to follow the surface of the Earth around not just because of friction, but because angular momentum is conserved. It could be that someday you will get into space and experience microgravity (aka weightlessness) first hand. Perhaps if you've learned the physics, you will understand that right now (if your are sitting down) you can feel the force of gravity (which is actually the chair pushing you up) and that being in a gravitational field does not make you feel weight. Unless that gravity is counteracted, you are weightless. So, don't blame gravity for your weight; blame the electromagnetic repulsion (and pauli exclusion) for preventing you from doing what comes naturally (falling to the center of the Earth). There are two (actually at least three) Conservation Laws you must consider Linear Momentum is conserved and Angular Momentum is conserved. It gets a bit more complicated once tidal forces need to be considered, but that's too advanced a subject for now.
Your motion (or lack of it, relative to the surface of the Earth) is quite accurately explained just by considering the conservation of momentum (for you).

Thanks for taking the time to write all of this. I'm aware of everything you said.

However, the question being asked here is really pretty simple: due to the rotation of the Earth, objects lying still on the planet's surface are always describing curved trajectories. For an object to describe a curved trajectory, the net force on it must be different from zero. What force is acting on all of these objects to cause them to move like this? I'm suggesting it's their weight, which is not exactly balanced by "the electromagnetic repulsion (and Pauli exclusion)". What do you suggest?

 

[rcgldr]

However, the question being asked here is really pretty simple: due to the rotation of the Earth, objects lying still on the planet's surface are always describing curved trajectories. For an object to describe a curved trajectory, the net force on it must be different from zero. What force is acting on all of these objects to cause them to move like this? I'm suggesting it's their weight, which is not exactly balanced by ... (upwards force from the Earth's surface)

It's a small difference. The net force on an object on the surface of the Earth is a tiny centripetal force, except at the north or south pole where the net force is zero and the object rotates but does not accelerate towards the center of the earth. At the equator, centripetal acceleration is about 0.03392 m / s^2.

 

[sophiecentaur]

If the weight of the object is countered exactly by something such as lift, then the net force on that object is zero. If the net force is zero, then how come the object is traveling in a circle?

The fact that it is traveling in a circle tells you that the net force is not zero. IF it were, it would carry on in a straight line. Its weight force is much more than enough to hold it onto the surface of the planet because the necessary centripetal force is so tiny. The rest of the weight force is canceled by the upward force of the ground.

The definition of Orbit must involve the notion of free fall. Once you provide any force (even a rocket engine) then the motion is no longer Orbital. At lease, that definition produced consistency.

 

[Drakkith]

Question: a centripetal force is required to cause an object to move in a uniform circular motion. Since both myself and the Earth's surface are rotating, then both myself and the ground feel the same centripetal force. So, relative to the ground, I am not accelerating at all, correct? And if so, then the force of my weight is exactly canceled by the ground pushing up on me with nothing left over (relative to the ground). Is all that correct, or have I misunderstood something?

 

[paisiello2]

I think you and the ground experience the same centripetal acceleration, not the same centripetal force since that depends on the relative masses of yourself and the ground.

Otherwise I think you have it right.

 

[Drakkith]

I think you and the ground experience the same centripetal acceleration, not the same centripetal force since that depends on the relative masses of yourself and the ground.

Otherwise I think you have it right.

Ah yes, thank you for the correction.

 

[paisiello2]

Now here is a bit of semantics for you:

If an orbit is defined as the path of an object subjected only to gravitational forces, then can anything really be 100% in orbit? Satellites in orbit experience orbital decay because of drag caused by the Earth's atmosphere, magnetic field, and/or tidal forces. Objects in space are going to be affected by the solar wind.

So the definition of when something is in orbit or not seems arbitrary.

 

[FactChecker]

The helicopter path is certainly "geostationary". So is my path as I sit on this chair. The definition of "orbit" in Webster's and other common dictionaries do not specify how the orbit path is maintained. In the context of flying vehicles, "orbit" is used to describe a flight path around a point, powered or otherwise. It is clear from the posts here that "orbit" has a more specific understood meaning in physics (at least where an entire planet is at the center).

 

[FactChecker]

One point that I do not see mentioned here is that the only geostationary orbit around the Earth is on the plain of the equator (22,236 mi above Earth's equator, moving in the direction of Earth rotation). Most of the discussion here has been about geosynchronous, not geostationary. For a helicopter to stay in geostationary position in the northern hemisphere, it must do more than provide lift to cancel the gravitational force, it must continually correct toward the North.

 

[sophiecentaur]

An alternative way of putting it would be that the upwards force from the ground will be equal to the weight force minus the centripetal force.

 

[paisiello2]

For a helicopter to stay in geostationary position in the northern hemisphere, it must do more than provide lift to cancel the gravitational force, it must continually correct toward the North.

Is this because any object in "orbit" and only affected by gravity must be in a plane that passes through the center of the earth? And therefore the only true geostationary satellites must be at the equator?

 

[FactChecker]

Is this because any object in "orbit" and only affected by gravity must be in a plane that passes through the center of the earth? And therefore the only true geostationary satellites must be at the equator?

Yes. And any plain through the center of the Earth that is tilted away from the plain through the equator would make an orbit that varies North and South.

 

[SteamKing]

Yes. And any plain through the center of the Earth that is tilted away from the plain through the equator would make an orbit that varies North and South.

I think you mean 'plane' instead of 'plain'.

 

[FactChecker]

I think you mean 'plane' instead of 'plain'.

Ha! Sorry. I'm the worst speller I know.

 

[OCR]

... except at the north or south pole where the net force is zero and the object rotates but does not accelerate towards the center of the earth.

... does not accelerate towards the center of the earth.

If the "object" happened to be an accelerometer, then... would it read zero at the north or south pole?




OCR

 

[A.T.]

If the "object" happened to be an accelerometer, then... would it read zero at the north or south pole?

No, and neither would an accelerometer in a hovering helicopter, so neither one is "in orbit".

But in a classical inertial frame the hovering helicopter has centripetal coordiante acceleration, which an object at the pole doesn't.

 

[fandi.bataineh]

i think you are missing a fundamental point here.

when youre setting on your chair; your weight pulls you down, and the normal force exerted on you by the chair pushes you up, which means that; these two forces share a common line of action (in opposite directions), and if they are not equal in magnitude, their net force will also lie on the same line of action they lie on, i think you agree with me on this intuitive point.

so, let's assume that they (your weight "w" and the normal "n") are NOT EQUAL in magnitude, and your weight is slightly larger in magnitude than the normal force (w > n), as you claim, then the resultant net force (lets call it "f") will be in the direction of your weight, and with a nonzero magnitude (f ≠ 0), acctually its magnitude will be equal to the difference in magnitudes between your weight and the normal (f = w - n).

now, let's assume; for the sake of generalization, that youre setting on your chair, in your room, at any point "P" on the Earth's surface (that is your location) that DOES NOT lie on the equator, then youre ROTATING (NOT SPINNING) around the Earth's axis of rotation, which is the line connecting the north pole "N" and the south pole "S", in fact youre rotating around another point "X" which is the point on the Earth's axis of rotation that is closest to your position "P", and NOT around the Earth's center of mass (lets call it "O"; the origin).
please recall that your weight pulls you toward the Earth's center of mass "O", not toward "X", and these two points (X & O) are always distinct unless youre setting at some point on the equator.
now, i want you to use your imagination to make sure that; any nonzero (net) force (f ≠ 0) acting on you in the direction of your weight; or the radial direction; that is in the direction of the line connecting your current position "P" and the origin "O", WILL NOT make you rotate around the point "X", but rather around "O", which means that you will be moving on a GREAT CIRCLE on the Earth's surface, which obviously is not the case.
for you to rotate around "X", which is exactly the case; you need a nonzero (net) force to act on you in the direction of the line connecting your position "P" and the center of rotation "X", and toward the point "X", which is a NECESSARY CONDITION that any force (f) on the same line of action as your weight fails to satisfy.

if youre going to stick to your claim that (f ≠ 0) and (w ≠ n); and that the component of "f" in the direction of the line "PX" is responsible for your rotation around "X"; then think about the other component of "f" (the component perpendicular to the line "PX"), this component would drive you to dive in the ground and move toward the the plane that contains the Earth's equator on a line parallel to the Earth's axis of rotation (recall that the Earth's axis of rotation is the line "NS" which connects the north and south poles), also recall that the ground CANNOT counteract this force (the component of "f" parallel to Earth's axis of rotation) because its not perpendicular to it!
this cannot happen either.

this leaves us with the ultimate conclusion that (f = 0) and (w = n).

in fact; this is why geostationary orbits exist ONLY above the Earth's equator, or in the plain containing it (the equator), because this is the only plain that satisfies the two conditions for a geostationary orbit allowing the previously defined points (X & O) to meet in a single point.
its perpendicular to the Earth's axis of rotation (the first condition), and it contains the Earth's center of mass (the second condition).iam still thinking about the REAL force that makes you (in fact it makes all of us) rotate with the Earth's surface around its axis of rotation, maybe it has something to do with friction and air resistance, that is; maybe there is no centripetal force after all, but rather we are smoothly dragged by the Earth's surface (the ground) and its atmosphere, MAYBE!

 

[A.T.]

when youre setting on your chair; your weight pulls you down, and the normal force exerted on you by the chair pushes you up, which means that; these two forces share a common line of action (in opposite directions)...,

No, they don't always act along the same line. The gravitational force is not perpendicular to the surface (except at the poles and equator). The rest of your argument is based on this misconception.