Just my imagination - 2

In easy way to see that it is $B$ is to imagine you are orbiting a tiny marble a mile away. By the time you orbit once, your reflection has only gone around the tiny marble, while you have gone over 6 miles in the same amount of time.

Thus, the speed of you is higher than of your reflection.

Reflection executes a smaller circle compared to your flying machine, in the same time interval.

well I already know that

$v=ω\times{r}$

where

v:linear velocity

ω:angular velocity

r: radius

But since this is geometry so lets go back one step by integration

$s=\theta \times{r}$

where

s: circumference distance , $\theta$ : rotational angle , r: radius

so if you change the radius "S "will change as well in a direct relationship ,then at any given\certain moment" $\theta$ " if we took two arbitrary points "a",and "b" one on the surface and the other above it (respectfully), with the condition that they always lie on the same line and it must go through the center of the sphere (the reflection)

Point "b" the one above must be always faster than point "a" the one on surface to fulfill that condition (reflection)

or you can simply differentiate the "S" equation with respect to time to obtain the "V" equation first one above, noting that ω is constant

you can also imagine two track runners one in the nearest lane the other in the farthest lane in circular track with yourself as an observer at the center and they're always perfectly aligned to you

Let their be a circle resembling earth. Let their be another concentric circle bigger than it resembling the route taken by helicopter. Let their be two points anywhere on the bigger circle resembling the start point A and end point B of the helicopter. construct a radius out of the two previous points A and B. Let the intersection of the two radiuses with the smaller circle (earth) be two points named a' and b' . The chord AB is bigger than chord a'b'. Therefore, Distance cut by helicopter d(X) is bigger than distance cut by reflection d(Y). Since the time taken by the helicopter to cut chord AB (T1) is the same as the time taken by the reflection to cut chord a'b' (T2). Therefore, $\frac{d(Y)}{T2}$ < $\frac{d(X)}{T1}$ . Since v=d/t, Therefore, the speed of reflection is less than that of the helicopter.

Depending on the height of your helicopter, the height of the image varies between r-h (when your height h above Earth's surface is zero) to r/2 when your height is at infinity and r is the radius of the Earth (discounting "spherical aberration", Wikipedia it). So for practical helicopters, the image height is something like r-h. And the helicopter height is r+h. And the image and object travel at the same angular velocity, Instantaneous velocity is angular velocity * radius, so the greater radius (object) has the greater velocity. I have drunk quite a few beers and gluwine (new years eve) so apologies in advance for any lack of claret-y...

The speed of light may seem instant, but it isn't. It takes time for the reflection to catch up with the changes you make to yourself.

Helicopter is source for reflaction here, so speed of reflaction will always less than holicopter, however that differnce in speed is Negligible...

I don't know. Ha ha ..........