Thread: Curious about effect of earth rotation on air travel times?

Does it take less time when traveling in a commercial airliner in the opposite direction of the rotation of the earth? For instance, if you fly non-stop between Honolulu Hawaii to New York City, and since the earth is rotating toward the east, it seems when you take off from New York, Honolulu would get closer as a destination since the earth itself is moving while you are in the air. On the other hand, if taking off from Honolulu, it seems New York would be moving farther away and you'd have to 'catch up' to it. The distance between those two points is 4968 miles If there is a time difference, how much difference does it make and how do you calculate that? I'm not in school, but always wondered about this. Thank you.

Jean Tessmer Hawaii

No it doesn't take less time in the flight itself. Remember that the time difference already exists when you take off, you could change the time on your watch as soon as you're airborne if you wanted to.

Very important when flying across the Pacific because you can 'lose' or 'gain' a day because of the international date line.

Though when you're flying exactly west to east and vice versa, say between Sydney and Perth, the flights take longer/shorter times because the plane is either fighting or being aided by strong upper air flows.

Originally Posted by Jean Tessmer

Does it take less time when traveling in a commercial airliner in the opposite direction of the rotation of the earth? For instance, if you fly non-stop between Honolulu Hawaii to New York City, and since the earth is rotating toward the east, it seems when you take off from New York, Honolulu would get closer as a destination since the earth itself is moving while you are in the air. On the other hand, if taking off from Honolulu, it seems New York would be moving farther away and you'd have to 'catch up' to it. The distance between those two points is 4968 miles If there is a time difference, how much difference does it make and how do you calculate that? I'm not in school, but always wondered about this. Thank you.

Jean Tessmer Hawaii

Try this experiment. Jump straight up into the air. Now even a pretty weak jump will keep you off the Ground for half a sec or so. Now Hawaii is about 20 degrees from the equator which means that the ground beneath your feet is moving at 218 m/sec due to the rotation of the Earth. So, do you land over 100 meters away from were your feet hit the ground. (in other words did the Earth rotate out from under you while you were in the Air?). The answer is, of course, No. You go straight up and come straight down relative to Where you jumped. The reason is that while standing on the ground, you are also moving with the Earth when you jump into the Air you keep that motion and still travel right along with the Earth.

Now ask yourself another question: Is the air around you blowing at over 900 miles an hour? This would be the case if the atmosphere around the Earth didn't rotate with the Earth for the most part. Since the answer to the question is no, then we can assume that the atmosphere does share the Earth's rotation.

Now, what does a airliner travel relative to? The air. A plane that can fly 300 mph, measures this speed with respect to the air, and since the air is carried around the with the rotating Earth, then the plane is carried along with the rotating Earth when it is flying. IOW, as along as there is no wind, a plane will travel at the same speed relative to the ground no matter which direction it is headed. (again, this is disregarding the effects of the high altitude winds mentioned in adelady's post.)

There are many ways to answer this question, depending on how far you want to take it, and in which direction!

The simplest answer is no, an airliner does not exploit the rotation of the Earth, when travelling in the opposite direction to the rotation. If it travels through the air at a given speed, it will take the same time to reach the opposite side of the planet, regardless of the rotation of the Earth.

To make an extreme example, at the equator the rotation speed of the Earth is around 1600 km/h, which is something around 400 metres per second! This has no discernible effect on anyone standing on the Earth, or if they throw a ball in either direction, or do the long jump in either direction, or get in an aeroplane and fly in either direction.

One reason for this is that the Earth's atmosphere moves with its rotation, so there is no advantage to be gained from that rotation with regards to your speed through the air or on the ground.

But interestingly, even if there was no atmosphere the same would be true! You still wouldn't be able to exploit the rotation of the planet you are on to jump further in one direction than another, or to fly your rocket plane any quicker to the other side of the planet. Even if you flew your rocket fast enough to get into orbit, your orbital speed relative to the ground below you isn't helped by the direction you took off in.

But if you go fast enough to leave orbit, you can use the rotation speed of the Earth to boost your speed relative to the Sun and the other planets, in order to get somewhere else in the Solar System a little quicker, for instance. If you are heading for Mars, it makes sense to use the Earths rotation to give you over 1600 km/h more speed towards Mars than you would have if you took off against the rotation.

Originally Posted by SpeedFreek

But interestingly, even if there was no atmosphere the same would be true! You still wouldn't be able to exploit the rotation of the planet you are on to jump further in one direction than another, or to fly your rocket plane any quicker to the other side of the planet. Even if you flew your rocket fast enough to get into orbit, your orbital speed relative to the ground below you isn't helped by the direction you took off in.

You actually can use the Earth rotation to your advantage in an orbital situation. First off, imagine an orbit with a perigee at the surface of the Equator and a apogee at an altitude of 150 km. It would take an orbital speed of 8019 m/sec at the surface of the Earth to attain this orbit. Sitting at the Equator, you are already moving at 464m/sec Eastward. Therefore you would only have to launch at 7555 m/s from the Surface to attain such an orbit. Now imagine that you are launching Westward. You are starting off moving 464m/sec Eastward, so if you try to launch at 7555 m/s, you end up with a perigee orbital velocity of only 7091 m/s. This not only won't get you into the desired orbit, it won't get you into any orbit at all.

I think that an air "mile" is longer than a land mile, a land mile longer than a nautical mile. If, given that information, and accounting for a time variable, a person may be able to plugs in a set of given variables and likely the end result will produce a fairly justified answer. Only a thought.

Janus, you are correct, of course. Most launches into orbit are sent eastwards because of this and are thus prograde orbits, although there have been a few launches into retrograde orbits. What you are referring to is the difference in energy required to actually reach orbital velocity.

What I was referring to, however, is your speed relative to the ground below you, or how long it takes to reach the other side of the planet, which I think is the same whichever direction you choose to launch in. The way I understand it (and do please correct me if I am wrong) if you want to land at a point on the opposite side of the planet, you wont reach it any quicker when launching eastwards than when launching westwards. Assuming we want to consider the same amount of acceleration and thus the same velocity in either direction, then the velocity required in order to achieve orbit in the retrograde direction will produce a higher orbit when used in the prograde direction, so the trajectory is longer. Or if you use the slower velocity required for a prograde orbit, in order to trace the same trajectory as the retrograde orbit but in reverse, it will take longer to trace that trajectory due to the slower velocity, and it all cancels out.

ampwitch,
There's no such thing as an "air" mile. A nautical mile is longer than a statute (land) mile.

I thank you all for your input. Janus and Speedfreek, I really enjoyed your correspondence with vivid and enlightening explanations. I learned a lot and had have wanted to know about this for a long time. Great discussion.

Jean Tessmer Hawaii

Originally Posted by Janus

Originally Posted by SpeedFreek

But interestingly, even if there was no atmosphere the same would be true! You still wouldn't be able to exploit the rotation of the planet you are on to jump further in one direction than another, or to fly your rocket plane any quicker to the other side of the planet. Even if you flew your rocket fast enough to get into orbit, your orbital speed relative to the ground below you isn't helped by the direction you took off in.

You actually can use the Earth rotation to your advantage in an orbital situation. First off, imagine an orbit with a perigee at the surface of the Equator and a apogee at an altitude of 150 km. It would take an orbital speed of 8019 m/sec at the surface of the Earth to attain this orbit. Sitting at the Equator, you are already moving at 464m/sec Eastward. Therefore you would only have to launch at 7555 m/s from the Surface to attain such an orbit. Now imagine that you are launching Westward. You are starting off moving 464m/sec Eastward, so if you try to launch at 7555 m/s, you end up with a perigee orbital velocity of only 7091 m/s. This not only won't get you into the desired orbit, it won't get you into any orbit at all.

On that note, a plane flying Eastward should find itself just barely lighter, due to centripetal (pseudo) force. That should increase its fuel economy slightly, and allow the engines to focus more on forward motion, and just slightly less on lift.

How much? I'm guess probably not very much.

Originally Posted by SpeedFreek

Janus, you are correct, of course. Most launches into orbit are sent eastwards because of this and are thus prograde orbits, although there have been a few launches into retrograde orbits. What you are referring to is the difference in energy required to actually reach orbital velocity.

What I was referring to, however, is your speed relative to the ground below you, or how long it takes to reach the other side of the planet, which I think is the same whichever direction you choose to launch in. The way I understand it (and do please correct me if I am wrong) if you want to land at a point on the opposite side of the planet, you wont reach it any quicker when launching eastwards than when launching westwards. Assuming we want to consider the same amount of acceleration and thus the same velocity in either direction, then the velocity required in order to achieve orbit in the retrograde direction will produce a higher orbit when used in the prograde direction, so the trajectory is longer. Or if you use the slower velocity required for a prograde orbit, in order to trace the same trajectory as the retrograde orbit but in reverse, it will take longer to trace that trajectory due to the slower velocity, and it all cancels out.

Let's consider a Newton's cannon situation. We'll assume our Cannon has a muzzle speed of 8464 m/s. This way, The retrograde firing cannon will still give the projectile enough velocity to complete an orbit. Using the Vis Viva equation:



We find that the semi-major axis of this orbit is 6,504,920 m

From this, we find that the Orbital period is 1.45 hrs. (This is when the projectile will return to ground level.) The Earth will have rotated in that time by some 2400 km, so your projectile hits 2400 km from the cannon. Since the Earth's rotation is counter to that of the projectile's orbit, it will seem to the cannon that the projectile overshoots by this distance.

Now firing pro-grade:

The muzzle velocity remains the same, so the initial orbital velocity is 9828 m/s. Using the same vis viva equation, we get an semi major axis of 8,734,047 m with a period of 2.25 hr, during which time, the Earth will have rotated by 3758.4 km. Since the Earth's rotation and orbit are in the same direction, it will appear that the Projectile falls short of getting back to the cannon as seen by the cannon.

Thus when firing retrograde, as measured relative to the ground, the projectile flies 40076 km in 1.45 hrs, or at an average speed of 27639 km/hr
and when firing pro-grade, it travels 36316 km in 2.25 hrs, or at an average speed of 16140 km/hr.

Very interesting!

I thought there was no advantage to launching in either direction in this scenario, when in fact it seems there is an advantage to launching in the retrograde direction. So here we need to do the opposite of what we usually do when we launch satellites, which are mostly launched prograde. Rather than getting a boost from the rotational speed of the Earth adding to the speed of the projectile, we get a boost from the Earth moving under the projectile, in the opposite direction to the projectile, which is similar to the question originally asked by Jean, but here we are talking ballistics rather than powered flight through the air.

So, does a cannonball fired retrograde always land further from the cannon than a cannonball fired prograde (given that they have the same muzzle velocity)? I know gunners shooting over long distances have to adjust their aim to account for Coriolis force because of the Earths rotation, but I thought that was just when shooting in a direction at a tangent to that rotation.

Originally Posted by SpeedFreek

Very interesting!



So, does a cannonball fired retrograde always land further from the cannon than a cannonball fired prograde (given that they have the same muzzle velocity)? I know gunners shooting

over long distances have to adjust their aim to account for Coriolis force because of the Earths rotation, but I thought that was just when shooting in a direction at a tangent to that

rotation.

I realize that it been almost 2 weeks since I last posted to this thread, but I've decided that this deserves a more thorough answer than a quick reply would have given so I want ed to take the time to work up a good one:

The scenario I gave is somewhat of a special case in that one are completing a complete orbit and going from a perigee to perigee path. Not your typical ballistic trajectory.

So instead, let's look at a more "normal" situation, with a cannon with an elevation of 45°, with a given muzzle velocity on the Equator. I'll show how you can get the ground distance when firing East or West. It is a bit more involved than my last example.

First you have to do a little vector addition to get the orbital speed and radial velocity (The velocity component measured perpendicular the line joining the projectile to the center of the Earth.) For instance, for a muzzle velocity of 1000 m/s, you get a pro-grade orbital speed of 1368 m/s and a radial velocity of 1171m/s, and for retrograde, you get 748 and 243 m/s respectively.

As above, you use the orbital speed and Earth's radius in the vis viva equation to get the semi-major axis(a) of the trajectory. And then we find the period(P) of this "orbit".

Then we can find the "Areal velocity" with the Earth's radius and horizontal velocity.



Areal velocity is also related by:



Where e is the eccentricity of the orbit.

This and the information arrived at above allows us to solve for e.

Now we use the equation



To find the "true anomaly" (The angle from perigee) at the point where this orbit intersects the surface of the Earth.

This subtracted from 180° gives you the angle from firing point to apex (apogee) of the trajectory. Twice this gives the full angle traveled from firing to impact. The distance traveled is

determined from this and the circumference of the Earth.

This does not take into account how much the Earth rotated while the projectile was in the air. For that, you have to determine the time of flight.

For this first you need to find the "eccentric anomaly" (E)

which is related to the true anomaly by:



Then we need the "mean anomaly" (M), which is found by




This can then be used to find the time taken to go from perigee to the surface intersection point:



This subtracted from 1/2 the period (P/2) gives half the projectile trajectory, and doubling this answer give the full projectile flight time. Multiplying this by the Earth's rotation speed

gives you how much the Earth has turned, which, when combined with the trajectory distance, give you the Distance traveled with respect to the ground. (Be sure to keep your orbital

and Earth rotation directions straight.)

Now with our 1000 m/s muzzle velocity, we get a ground distance of 103.5 km firing East (with the Earth's Rotation) and 102.1 km firing West, so it turns out that the pro-grade trajectory travels a greater ground distance by almost 1 1/2 km.

The following chart gives the relative ground distances for different muzzle velocities:



At lower muzzle velocities, the distance doesn't vary much between retrograde and pro-grade (Though retrograde does fall a little shorter)

Then you start to see some visible separation, but then it started to close up again with the Retrograde distance starting to exceed the pro-grade distance. Then the prograde distance flattens out and begins to decrease.

Here's what happening, as your muzzle velocity increases in the prograde direction, you are launching your projectile into an orbit with a higher and higher apogee. Eventually the apogee becomes so high that the projectiles avergage speed between firing and impact decreases so much that the Earth's surface rotating beneath it is allowed to "catch up". The "in the Air: time increases quite a lot while the true anomaly angle between firing and impact does not increase by much.

In the Retrograde direction the higher apogee increases the "in the air time", but this works to the projectile's advantage. Since the Earth rotates in the opposite direction to the orbital trajectory a longer time in the air allows for more rotation and a larger net ground distance.

Now a couple of more considerations to consider.

The above example was for when you were sitting on the Equator. What if you are not on the Equator? Let's say that you are on the 45th parallel? Well for one thing, you have to reduce the effect of the Earth's rotation speed. But that's simple.

A more complicated adjustment is what I'll call the "great circle problem".

If I launch a projectile from the surface of the Earth, its trajectory will follow a great circle or geodesic. When we were on the Equator firing due East or West this was not a problem because the Equator is a great circle itself.

The 45th parallel however isn't. If I fire my projectile directly West, it will follow the great circle and tend to "curve South", and this is without considering any effect due to the Earth's rotation. So, even before I consider the Earth's rotation, in order to hit a target due East or West, I have to aim a little to the North of due West.

But this means that I am no longer firing in the same direction as the ground is moving due to rotation. So now I have to compensate for this and it gets pretty messy.

And then there is another issue I haven't mentioned yet: Firing elevation. In basic ballistics (flat surface, constant g), a 45° elevation gives you the longest flight distance regardless of muzzle velocity. With orbital ballistics, this isn't the case, as the muzzle velocity changes, the "best range" elevation changes.

So as you can see, long range ballistics is a fairly complex subject.

This all reminds me of something else I've thought about, hovering with a jetpack or somesuch other device.
I realize that because of conservation of momentum an aircraft takes an equal amount of time to travel between equidistant points regardless of direction.
To keep this simple I will assume no atmosphere.
Because a line maps to a sphere on a great circle, if a person or object were to hover over a rotating sphere at a point between the axis and the equator wouldn't they then pass back and forth over the equator of the sphere they were hovering over?

Originally Posted by Janus

So as you can see, long range ballistics is a fairly complex subject.

Indeed it is! I thank you for taking the time to work up that treatment of the situation, it was very informative and just goes to show how complicated a seemingly simple scenario can turn out to be.

Janus - Thank you for the detail analysis of the muzzel velocity on the equator. I really appreciate the clarification on all the complexities of ballistics. So if I understand correctly, going pro-grade (West) is going in the opposite direction of the earths rotation, and we actually gained some distance due to more air time about 1.5 km?
Would that apply to air liner flight or is most of the gain lost due to lower speed of the air craft, and friction of the atmosphere?

Originally Posted by Jean Tessmer

Janus - Thank you for the detail analysis of the muzzel velocity on the equator. I really appreciate the clarification on all the complexities of ballistics. So if I understand correctly, going pro-grade (West) is going in the opposite direction of the earths rotation, and we actually gained some distance due to more air time about 1.5 km?
Would that apply to air liner flight or is most of the gain lost due to lower speed of the air craft, and friction of the atmosphere?

There is a difference between ballistics and air flight. With ballistics, the projectile is given an initial velocity and it follows a trajectory determined by gravity. An air liner pushes itself through the air like a car pushes itself along the ground. You wouldn't expect a car driving along the ground to gain distance with respect to its destination due to the rotating Earth, and neither should you expect the same from an airplane. Planes fly at a specific speed with respect to the air. The only advantage or disadvantage is when the air itself is moving with respect to the ground. A tailwind will get you there faster and a headwind will slow your progress. (and a crosswind forces you to adjust your heading to compensate)

Originally Posted by Janus

The only advantage or disadvantage is when the air itself is moving with respect to the ground. A tailwind will get you there faster and a headwind will slow your progress. (and a crosswind forces you to adjust your heading to compensate)

This is a slight deviation. On one west-east transatlantic crossing (Houston-Amsterdam) with a 220 mph tail wind, our velocity relative to the ground was, for a time, supersonic. Quite neat and much cheaper than Concorde. Even better, in those days KLM served Drambuie.