[timey (OP)]

What will the helicopter's clock say?

If a helicopter, fitted out with long range fuel tanks (chuckle) were to hover at elevation from the equator maintaining its position in space while the take off point spins away at 1,040mph - as per the clock on the ground at sea level, 24 hours will pass before the take off point is under the helicopter again, but the clock on the helicopter will show that more than 24 hours has passed due to its altitude and the fact that it is not moving at 1,040mph like the clock on the ground is.

However, the Earth is not only spinning on its axis but also belting around the sun at a staggering 18.5 miles/second (as measured by the ground clock), which is 66,600mph.
Clearly the helicopter will not be able to maintain an exact position of altitude above the equator because the helicopter is not able to achieve speeds of 66,600mph.
If it tried to take off into the inline motion and maintain a position in space at a constant altitude from the Earth, the helicopter not being able to travel at 66,600mph would cause the Earth to crash into the helicopter.
Alternatively, if the helicopter took off away from the inline motion then the Earth would be receding away from the helicopters position in space and the helicopter would end up being left behind.

Let's tell the helicopter to take off away from the inline motion to maintain its position in space and wait for the Earth to complete its circumnavigation of the Sun.
Not withstanding the difficulties the pilot would have in landing the helicopter onto the Earth hurtling towards it at 66,600mph, the clock at sea level at the equator would have recorded that a years worth of time had past, but what would the clock on the helicopter have to say?

However, the Solar System is travelling with the rotation of the galaxy at a whopping 515,000mph, as measured by the clock on Earth at sea level equator, so the helicopter maintaining its position in space would watch the solar system receding away from its position.

Let's tell the helicopter to hover at that position for 230 million years, as measured by the clock on Earth at sea level equator, until the Solar System comes back to that position.  The clock on Earth is saying that 230 million years has passed by, but what will the clock on the helicopter have to say?
From it's position of take off on the equator Earth the helicopter has reduced its motion in space by 515,00mph+66,600mph+1,040mph=182,640mph, so SR motion related time dilation will be somewhat reduced.  Whereas GR altitude time dilation will be rather hard to determine.

However, the Galaxy is also moving through space, and if the helicopter were to maintain its position in space it would observe the Galaxy receding away at the phenomenal rate of 1.3 million mph, as measured by the clock on Earth at sea level equator.
But what will the clock on the helicopter say?
By now the SR motion related time dilation effects will be minimal, and what are the GR gravity potential time dilation calculations saying?

How fast is the helicopter's clock running?

[Kryptid]

I kind of want to calculate this myself, but I'm unsure what effect the clock on the ground being in an accelerated reference frame would have. Moving at 182,640 miles per hour relative to an observer apparently has a different effect depending on whether that motion is a constant linear velocity or an orbital acceleration. I've read that even has something to do with how to resolve the twin paradox. For what it's worth, I calculated that an observer would measure the clock of an object with a linear velocity of 182,640 miles per hour relative to themself to be ticking 99.999996334% as fast as their own. For 1.3 million miles per hour, it's 99.999814%.

[Colin2B]

I kind of want to calculate this myself, but I'm unsure what effect the clock on the ground being in an accelerated reference frame would have. Moving at 182,640 miles per hour relative to an observer apparently has a different effect depending on whether that motion is a constant linear velocity or an orbital acceleration. I've read that even has something to do with how to resolve the twin paradox.

The twin paradox doesn't require acceleration to explain the effect because you can start the twins in 2 spaceships already travelling at speed and get the same effect. The result is due to which reference frame you end up making the final measurements in.
The problem with the twin paradox is that like the example above it starts with the moving twin (or helicopter) starting at rest relative to earth. This assumes acceleration in order to reach the speeds quoted relative to the earth starting point. To make the full calculation involves GR, but what you have done is OK as long as you assume the acceleration phase is over and the 'copter is now travelling at full speed relative to earth, just accept it's not the full story - to do that you need to sit down and work out the space-time diagrams for each phase or do the GR maths.

[timey (OP)]

The problem with the twin paradox is that like the example above it starts with the moving twin (or helicopter) starting at rest relative to earth. This assumes acceleration in order to reach the speeds quoted relative to the earth starting point. To make the full calculation involves GR, but what you have done is OK as long as you assume the acceleration phase is over and the 'copter is now travelling at full speed relative to earth,

Well actually Colin, I have devised the OP to be an inverse of the twin paradox because the helicopter is asked to maintain the same position in space.

This means that the helicopter:
Is not spinning with Earth's rotation on its axis = minus 1,040 mph
Is not circumnavigating around the Sun = minus 66,600mph
Is not spinning with the Galaxy's rotation = minus 115,000mph
And is not moving through space with the Galaxy = minus 1.3 million mph.

The helicopters speed has been reduced by 1, 482 ,640mph which will increase the helicopter's clock's rate of time.
The helicopter's gravity potential has drastically increased due to the mass of the Galaxy travelling away from the helicopter at 1.3 million mph, which will increase the helicopter's clock's rate of time.

All measurements of speed and time are being held relative to the clock on Earth at sea level Equator.

The question is:
How much faster than the clock on Earth will the helicopter's clock be running?

[timey (OP)]

The answer to this question is that the helicopter, in order to maintain it's position in space will have to do work against the gravitational fields of the masses it seeks 'not' to be moving with.
This is the same work that the helicopter would have to do if it were to try to achieve the same motion it seeks to reduce as additional speed.

So would the helicopter's clock be positively affected by SR time dilation  with the work it takes for the reduced motion in the same way that it would under the remit of additional motion?

[timey (OP)]

The posts above outline an inverse representation of the twin paradox...

Would anyone care to coment?

[timey (OP)]

Ok then - in light of 'no response' perhaps I'll change the question...

Let's say that we ask the helicopter to take off from Earth at sea level equator, and by up-keeping a speed of 66,600mph it travels Earth's orbital radius of the Sun in the opposite direction that Earth is travelling, passes Earth once, to travel a complete orbit of the Sun and then lands back on Earth at the take off point.
The clock at the take off point shows that 1 year has passed by, but what will the helicopter's clock say?

The clock on the helicopter has reduced its speed by 1, 040mph compared to the clock spinning with the Earth's equatorial centripetal motion at the take off point...
If we then, with the help of an extended arm upon which we place the helicopter's clock, spin the helicopter so that the clock is moving at 1,040mph in addition to the 66,600mph it is orbiting the Sun at, the helicopter's clock and the Earth's clock should be equal according to SR.

But what about equality under GR considerations?
Ok well - the radius from Sun is equal for both helicopter and Earth so they are, or - considering the elliptical factor - will have been by the completion of the orbit, equally positioned in the gravity potential of the Sun.
Via the equation pe=mgh the helicopter's potential energy will be lesser than the Earth's but if one were (for some reason) seeking absolute equality pe/m would do the trick.

However, this illuminates the fact that the helicopter is a much smaller mass than the Earth, and bigger masses are supposed to be inherent with clocks that will run slower than clocks will on smaller masses...
So - in this instance, apart from differing mass sizes, everything is equal for the helicopter's clock and the clock on Earth where the speed is the same, and the overall positioning in the gravity potential of the Sun is the same.
Here we have mathematical description available to us that describes that the clocks will run at the same rate...

So where does the factor for the clock on the bigger mass running slower reside?

[timey (OP)]

Can anyone answer the question?

[Kryptid]

This is difficult for me to answer because the relative velocity between the two clocks will be constantly changing (half the time the clocks are moving away from each other and half the time they are coming towards each other). I don't know how to calculate that. All I know is that the time dilation factor will be very, very small. I'm guessing the difference at the end of the year would be less than a second.

[timey (OP)]

The question outlines the fact that although the helicopter and the Earth are orbiting the Sun in opposing directions, they are travelling at the same speed where the clock on the helicopter is also spinning at 1,040mph as the clock on Earth is.
The SR effects are thus equalised for both clocks.
Each clock by the time 1 year has passed will have moved away from, and closer to each other's gravitational field twice, but their collective position with regards to the Sun over the course of a year will be equal.

I designed this question in order that there is minimal acceleration or deceleration considerations.

All parameters are equal for each clock apart from the fact that the clock on Earth is on a vastly greater mass.
Greater masses are supposed to cause a clock to run slower than lesser masses.

On the basis that we have equalised both the GR gravity potential issues with regards to each clocks position in the Sun's gravity potential, and we have equalised SR time dilation considerations for both clocks as they are travelling through space at the same speed, the question is:

Where does the factor for the clock on the bigger mass running slower exist?

[timey (OP)]

Ok then - let's have a proper little think about this:

In the OP the helicopter is asked to take off from a position at the equator, maintain it's position in space and wait there for the Earth to circumnavigate the Sun where it lands on the Earth 1 year later as per what the Earth clock says...
The helicopter's clock is moving 66,600mph orbital speed+1,040mph centripetal motion speed slower than the Earth's clock so it should have a faster rate of time according to SR.
But although the helicopter appears to be not moving, it has had to do work to decelerate, and will also have to do work to accelerate in order to land.*

Q1:  Does this amount to the same deal as accelerating to move faster than the Earth, and then decelerating in order to land?

(*The helicopter would also, because it is not in orbit to the Sun, have to do work to maintain its position of altitude above the Sun, but we don't need to go there)

In post 6 the helicopter is asked to take off from the Earth and circumnavigate the Sun in the opposing direction.
The helicopter is already moving at 66,600mph+1,040mph with the Earth when it takes off but it needs to maintain the speed of 66,600mph orbital speed in the opposite direction that Earth is orbiting the Sun.
The helicopter will need to accelerate  from 66,600mph to escape Earth's gravity field.
Considering that the helicopter was already moving with the Earth at 66,600mph...

Q2:is this bit of acceleration all the work that the helicopter would need to do to maintain an orbital speed of 66,600mph?

Q3:  Or would the change in direction be a factor?

The helicopter will need to make an equal amount of deceleration to land back on the Earth.
Considering that the helicopter has accelerated and decelerated equally in both scenarios...

Q4:  Does this mean that we can cancel the accelerative and decelerative factors from our time dilation considerations?

In the second scenario, by placing the helicopter's clock on an extended arm, we also ask the helicopter to spin so that the clock is not only travelling at 66,600mph, but is also spinning at the speed of 1,040mph.
The clock on the helicopter is now travelling at the same speed as the clock on Earth, and over the course of the year will have maintained an equal altitude from the Sun.
The only difference between the conditions for both clocks is that the helicopter's clock is on a much smaller mass.  Everything else is equal.
Conventional physics states that the clock on the bigger mass will run slower than the clock on the smaller mass, so in consideration of all else being equal...

Q5:  Where does the factor for the clock on the bigger mass running slower exist?