Escaping the Cape, Inertia of Serbia

'net velocity = velocity of the airplane with respect to the ground + velocity of the ground'.

'net velocity' relative to what? The fixed space the Earth lies in? In which case it is probably better to talk about 'speed' since they are not travelling in a straight line.

Secondly, given that air resistance can be ignored, then why would the rotation of the Earth affect the speeds of the planes relative to the fixed space the Earth lies in?

In fact, the rotation of the Earth decreases the speeds of the planes from Tina's point of view (which is what we're concerned with). Which would reverse the answer?

It seems that this explanation will hold, if Tina's position in 3D space was fixed.

But Tina in the airport is also moving along the surface of the earth.

So, it seems that the "velocity of the ground" should not be considered here.

What do you think? @Terry Chadwick

I am shocked that you would change the problem statement! Add to it, Ok. Add a version B where cannons are used and airplanes remain in version A.

You didn't "move the goal posts", you completely changed the field of play. Shame!

Surely it was meant that velocity is the same relative to the earth!?

If distance is based on the distance the airplane travels. I was assuming it meant straight-line distance.

The problem statement is ambiguous and incomplete. When is Tina's position equidistant from the other two (at takeoff, or landing?) and what control do the pilots have over engine power (must both planes apply the same power?). By varying these parameters any answer is possible.

Several proposed solutions state that one plane had a higher ground speed and therefor arrived first, but there was no adequate discussion of the flight path. The optimal flight path is a straight line ignoring ground motion (no air resistance).

If the landing point is at 45° latitude on the same longitude as Adam's position then, In my opinion, the best solution if for both planes to follow the same strategy (as described for Terry's Southbound flight in several solutions): fly a straight line North or South (respectively), with the North bound plane applying a small extra westward thrust to dissipate the initial eastward velocity. With this scenario each plain follows an equal path and takes equal time. It also allows each pilot to choose how much forward thrust to apply. Thus the outcome depends on the cooperation of the pilots.

If the landing position is not on the same longitude as Adam's then Adam has to fly a longer distance and will arrive later, unless he applies much more thrust; in which cases the ambiguity in the problem statement invalidates the problem.

Enough already.

Bret Huggins, Arroyo Grande, CA

But plane speed is measured in air speed not speed relative to the land. So two planes same speed same distance same time

Once A(Terry) and B(Adam) depart, there are no forces that accelerate or brake them as they don't receive any influence from the earth (the air influence is negligible) so the earth movement can be safely ignored for them. The point C(Tina), on the other side, continues moving around an earth parallel at a speed given by omega * r C = omega * R * cos(lat C) this makes the distance to point B to enlarge as time elapses, while distance from point C to point A maintains constant. As both, A and B are intelligent, it's supposed they will fly directly to the meeting point of the they routes and the trajectory the point C is following, this is maximum circles on a perfectly spheric earth. As the distance to point C' grows for B and doesn't for A, always it will take for B more path (and more time) to reach C', except if it happens that C' is in circle segment of parallel that connects C to it's symmetric C'' over the meridian that connects A to B. In that case, the distance from B to C' will be lesser than the distance from A to C' (and to C) and it will take less time for B to reach C' than for A.

@Terry Chadwick , please, the velocity of planes is the same or not?.... have you heard about galileo relativity. Are you considering any at all reference system? if you consider the equal velocity as the ground velocity (with respect to the moving earth) then the distance to Tina's point of arrival is the same for both all the time... as the problem states, the distance is the same for both, so despite any other consideration, T = V/D is the same for both (V constant and the same for both, and D is the same and constant for both, except allowing the earth to deform itself), and so, they should arrive at Tina's at the same time. Have you considered the fact that Tina is moving also, but always maintains the same distance to Terry's ground departure point?

I'm not so sure on the current given answer to this. Here is my solution, which gets Terry as the answer -

I think the addition of a few assumptions are necessary:

(Minor assumptions might include: 'North Pole' referring to the geographic one, the Earth being considered to be a sphere, and the Earth being modelled as a rotating ball with period 24 hrs in a fixed space (all of which are fairly obvious))

1. We can probably assume that the planes travel along a great circle of the fixed sphere the Earth lies in to reach their destination (as opposed to continuously 'chasing' their target which is constantly moving). To do this, they need to predict the position of their destination at time of arrival, and we'll need to assume they can do this perfectly.

2. How does the speed of the planes vary? Do we just assume a constant speed throughout the flight? If this isn't the case, and the planes initially build up to a maximum speed from take-off, then the rotation of the Earth will mean that Adam's initial speed (relative to the fixed space the Earth lies in) isn't $0$ , but Terry's is. This will give Adam a slight advantage or disadvantage (depending on the location of Tina) over Terry, although it's hard to quantify this advantage (it depends on what the maximum speed and acceleration of the planes are). So we probably will have to assume the planes move at a constant speed throughout the flight in order to avoid this problem.

3. We also need to make an assumption about Tina's initial position and the speed of the planes which I shall explain in the solution. (imo this is a pretty big assumption which we can perhaps deduce from the diagram, but needs the question to be edited, preferably).

Notes:

To find which of the distances between two pairs of points on a sphere is larger, or if they are the same, comparison of the straight line distance and the shortest distance over the surface of the sphere produces the same result. Therefore only the straight line distances from Tina to Terry and Adam need to be considered in this problem.

Given the first assumption, the plane which has the shortest of the distances between the planes' initial and landing positions will be the first to arrive.

Over time, relative to the fixed space the Earth lies in, Terry and Adam's positions move due to the motion of their planes (and are independent from the rotation of the Earth, given the assumptions in the question), while Tina's position moves due to the rotation of the Earth.

Solution:

Let the sphere $x^2+y^2+z^2=1$ represent the Earth with the equator being the intersection with $z=0$ .

Let the initial positions of Terry and Adam be $(0,0,1)$ and $(1,0,0)$ respectively.

The set of points which are equidistant from their initial positions is the plane $x=z$ .

Let Tina's initial position in cylindrical coordinates be $(r_0,\theta_0,z_0)$ where $r_0=\sqrt{1-z_0^2}$ and $z_0=r_0\cos\theta_0$ (satisfying the sphere and plane equations).

Then Tina's position over time is $(r_0,\theta_0+\frac{\pi}{12}t,z_0)$ where $t$ is the time since take off in hours.

For Tina's position over time, $x-z=r_0\cos(\theta_0+\frac{\pi}{12}t)-z_0=r_0(\cos(\theta_0+\frac{\pi}{12}t)-\cos\theta_0)$ .

Now we'll need to make the last assumption which is that $0<\theta_0+\frac{\pi}{12}t<\pi$ throughout the whole problem. This assumption imposes restrictions on $\theta_0$ and the speed of the planes. Note that this assumption is sufficient but not necessary for the rest of the solution to produce a unique answer to the problem.

$(0<\theta_0+\frac{\pi}{12}t<\pi) \land (r_0>0) \implies x-z=r_0(\cos(\theta_0+\frac{\pi}{12}t)-\cos\theta_0)<0$

The region $x \lt z$ is closer to Terry than Adam while $x \gt z$ is closer to Adam than Terry.

Therefore Tina's position lies closer to Terry than Adam throughout the problem.

Therefore Terry will arrive first.

All things considered, perhaps it'd make things much clearer in this problem and remove quite a number of ambiguities and assumptions which currently necessary if it asked which plane has to travel the shortest distance rather than for the shortest amount of time.

To say that the wind is negligible is to say that there is no wind. By "no wind" we mean that the air is moving at the same rate as the earth under it. If not, then people on the ground at the equator would experience a constant westward gale.

To say that two planes are moving "at the same speed" means that they are doing so relative to the air they are moving through.

The net result of these two points is that when you are considering speed across the ground in "non-moving" air, you can consider the Earth not to be moving at all. The movement of the Earth and the air on it cancels out of any equation. If it didn't, then if Adam were to travel due north (in dead air) he'd arrive somewhere northeast of his starting point. He would not.

Also, you never said that Tina was east of Adam, only indicated it on the picture, which we always know not to trust relative positioning in pictures unless they are stated.

@Terry Chadwick The rotational speed of the Earth is going to slow Terry/Adam down if Tina is where she is according to the diagram. You need to think if the rotation gives a positive or negative effect to the speed.

Got this right according to the officially posted "answer", but am feeling uneasy about it. Mr. Chadwick's idea seems to be that the plane sitting on the ground at the equator acquires the eastward vector of the ground and, when it takes off, maintains that eastward vector in the air, which as others have noted below, must be assumed not moving relative to the ground at any latitude. Since Terry's plane does not have that eastward vector, he will be at a disadvantage in getting to a point on the earth that is to the east of his initial position. Sounds good, but that seems to say that no one starting from the equator can fly directly north; would have to aim a little to the west to fly due north. That seems absurd. I guess I have to agree with what seems to be a general sense of confusion about this question as indicated in these postings.

What are you talking about? "velocity of the ground'? If that was a real term then explain to me this...how the hell does a helicopter hover? Once we ignore wind effects, there is no such thing as velocity of the ground. Everything hovering in the sky moves at the exact same speed as the ground absent wind effects. If there was a "velocity of the ground component" then a balloon launched at the ground would move at a fairly hefty speed eastward, ignoring wind effects.

Someon brought up the case where a west to east plane goes faster than an east to west. True. But that is due to wind direction which was ignored in this problem

Equal distance + identical speed = arrive at the same time.

If Terry just flies due south and 'leads' his trajectory (starting omega) such that, accounting for Earth's rotation, his path will intercept with Tina's position as he crosses her latitude, he will arrive at least as soon as Adam, whether Adam needs to fly due north or take some other route. The answer is incorrect; the question is also not specific enough as mentioned by others.

Disagree with answer. The planes follow the same air speed and the problem stated that the air [correction: WIND] speed was negligible. Therefore there is no Coreolis Effect. By the problem statement, we were to ignore the effect that the equator is moving faster than the pole. Air speed was all that mattered and the winds were zero. Therefore ground speeds of the planes were equal, Time of arrival was the same.

I don't agree. Tina is at a fixed ground point x equal distance from her two friends. While the Adam initially has a greater relative ground speed, as he approach Tina his relative ground speed decreases. Terry may have a slower relative ground speed but as he approaches the fixed spot x his relative speed increases. Since they are both flying east the same distance, air speed,altitude and same weather, the gain and loss in relative ground speed offset and they should arrive at the same time. Your explanation seems to ignore the gain in relative ground speed accrued by Terry and lost by Adam

Is it fair to change the question after you realize the error and people have answered? Don't ethics come into play at some point and you admit your error instead of changing the pretense to suit your answer?