Question
Asked 23rd May, 2018
• alex-korolev.jimdo.com

# Why an aircraft (rocket) does not shift relative to the land due to Earth rotation at low altitudes but does it at the high ones?

It is well-known that the ground velocity of a plane or helicopter does not depend on relatively fast Earth rotation (~2000 km/h). But a rocket's ground velocity at the high enough altitudes does it increasing if the flying direction is close to the direction of rotation and vice versa.
What is the dependence of such a shifting on altitude?

## Most recent answer

2nd Jun, 2018
Dear Guido,
Thank you for the comment. I think that the best way to convince people is when they can say yes and feel happy. For that I try to find out how they think. It is the others task to follow my thinking. I have to follow the others thinking and lead him from there some step further.
Regards,

## Popular answers (1)

25th May, 2018
Daniel Pfenniger
University of Geneva
Thierry,
You are wrong. Coriolis and centrifugal forces are named fictitious because they don't exist in a proper inertial frame. If you need to check, look what happens in a rotating vacuum chamber, you will see air is not necessary
for Coriolis force to act.
3 Recommendations

## All Answers (151)

23rd May, 2018
James Garry
Red Core Consulting ltd.
Alexander,
I am not sure that I understand your question.
Aircraft (airplanes, helicopters, balloons) are no different from rockets.
If I launch a modest rocket vertically, it will, after consuming its fuel, fall back to Earth at roughly the same location that it launched from.
Can you clarify your question?
And this is broadly true for brief vertical sojourns above most of the atmosphere too. The momentum that the craft has at launch is preserved as there is no sideways component of drag acting on it.
23rd May, 2018
Alexander I. Korolev
alex-korolev.jimdo.com
Dear James,
For a modest rocket- yes. But for big rockets Earth rotation is significant.
A vertically launched powerful rocket after achieving high enough altitude will not fall back into the startup point due to lesser entrainment of the rocket by rotating Earth. If this were not the case, geostationary satellites would be launched vertically from the equator, for example.
In contrast, some wonderful entrainment at low altitudes does not allow levitating aircraft (gravity force is compensated by the thrust force) to fly away from the Earth in the direction of linear velocity of the rotation obtained at the launching.
23rd May, 2018
James Garry
Red Core Consulting ltd.
Alexander,
I used to be a mission analyst for geosynchronous satellites.
I am still unsure as to what your point is.
Your final paragraph is quite perplexing. Please rephrase.
24th May, 2018
Leticia Peña Barrera
They are not questions of my area of experience
24th May, 2018
Alexander I. Korolev
alex-korolev.jimdo.com
Dear James,
Look please at the figure for better understanding the question. 0 is the startup point. An aircraft (rocket) is fully entrained at low altitudes (by unknown reason) but acquires lateral component of velocity V0 at high altitudes. This effect is successfully used for launching rockets but is negligible for planes and other low-altitude aircrafts.
24th May, 2018
Guido Colasurdo
Sapienza University of Rome
Just a basic concept, that perhaps helps you. Propulsion creates a velocity relative to the Earth; you pay the same propellant to create the same relative velocity either eastward or westward. The ballistic trajectory outside the atmosphere depends on the absolute ("inertial") velocity, that is greater launching eastward than westward (using the same amount of propellant).
24th May, 2018
Alexander I. Korolev
alex-korolev.jimdo.com
Dear Guido,
Could you tell us minimal altitude where such a dependence becomes measurable? Are there any investigations regarding influence of Earth rotation on flying bodies trajectories (ballistic or not) at different altitudes?
24th May, 2018
Daniel Pfenniger
University of Geneva
As seen in an inertial frame, the initial horizontal speed Vh of a rocket launched vertically is the local Earth rotation speed. This speed sets the initial specific angular momentum J =R*Vh of the rocket, where R is the distance to the Earth center of rotation. Angular momentum is conserved since no torque acts on the rocket as it elevates vertically.
As the rocket climbs, its position becomes R+h, where h is its height.
In the meanwhile the Earth surface rotates at the constant angular speed Omega = Vh/R, while the rocket keeps J constant, so the angular speed of the rocket at any height is omega = J/(R+h)^2.
The difference Omega-omega = J*(1/R^2-1/(R+h)^2) is growing as h grows,
the Earth appears to rotate with respect to the rocket.
2 Recommendations
24th May, 2018
Alexander I. Korolev
alex-korolev.jimdo.com
Dear Daniel,
1. What are the forces that bind a launched rocket and the Earth? Viscous friction forces from rotating atmosphere are too weak to do this and not even observed due to Earth rotation. It is well-known that small particles pulling away from a rotating ball fly tangentially to the surface.
2. Where is the evidence that a vertically launched rocket at all the altitudes bends its trajectory around the Earth? In the case of Fg=Ft it may drift along initial velocity V0 at the high enough altitude.
3. Why does the differential "omega-omega" effect not revealed at low altitudes? Your formula gives difference in corresponding linear velocities of about 0.5 km/h at 1 km altitude above equator. It seems pilots of balloons are able to observe such a lateral velocity component on a windless day.
24th May, 2018
Daniel Pfenniger
University of Geneva
1. Gravity is the dominant force. Viscous forces are parallel to the velocity vector so would not explain a lateral shift unless the rocket shape is asymmetric.
2. In practice real rockets themselves bend their initial vertical orbits in order to reach a circular orbit (by pushing a little more on one side for example). The reason to start vertically is to get out of the thick atmosphere as quickly as possible. But ignoring practical constraints a vertical launch would bend anyway due to Coriolis force, which is the fictitious force seen in a rotating (non-inertial) frame. (In another context snipers know bullets are deflected on the side due to Coriolis force). In an inertial frame the angular momentum conservation argument leads to the same conclusion. The only places on Earth where a vertical launch would not bend are the poles.
(My previous formula is only valid at the equator, for other latitudes one would need to replace the distances R and h by their projection on the equatorial plane, so multiplied by a cosine of the latitude.)
3. In practice at low altitude wind may first dominate a lateral deflection, but without atmosphere the effect would be immediately measurable provided the instruments are good enough.
1 Recommendation
24th May, 2018
Alexander I. Korolev
alex-korolev.jimdo.com
1. Gravity is the reason of attraction to the Earth center. It does not explain effective transfer of the angular momentum through atmosphere, as well as absence of vertical velocity gradient inside it.
2. Thus, the evidence is absent. Coriolis force needs binding of a rocket with Earth also. It might becomes zero at some high enough altitude. The interesting experiment may be in stoppings (in the solar rotating frame) of a spacecraft at different altitudes above the Earth surface and registering after that its trajectory- toward the Earth center only or with some entrainment around it.
3. This looks like hypothesis. Observed entrainment and its intensity might be described not only with your theory.
24th May, 2018
James Garry
Red Core Consulting ltd.
Alexander,
There's no magic involved. If I ascend (slowly) on a magic carpet, I start at rest with respect to the rotating atmosphere, and that continues to be true in a 'toy' Earth where the atmosphere rotates at the same angular speed as the Earth.
There's nothing 'unknown' about the 'entrainement'. We sit at the bottom of an ocean of air that co-rotates with the Earth. How *could* we move with respect to that air, without overcoming drag?
But that is not the case.
Now imagine a *rapidly* ascending dense rocket, with near-zero cross-section transverse to its length. It starts at rest with respect to the ground, and rises vertically. From the view of an observer it *of course* does not take a straight-line path. In a non-rotating frame the rocket takes a straight line path - it is *narrow* and the rotating atmosphere applies no drag.
I do not see that this is cause for great debate. I might be wrong.
Daniel, launches from the equator, due East or West would be immune to the Coriolis 'force'.
24th May, 2018
Alexander I. Korolev
alex-korolev.jimdo.com
Dear James,
The first Newton law tells us that "In an inertial frame of reference, an object either remains at rest or continues to move at a constant velocity, unless acted upon by a force". But the reference frame with respect to the rotating atmosphere is non-inertial.
24th May, 2018
James Garry
Red Core Consulting ltd.
Alexander,
<cautiously>
Yes.
But we can treat it as an inertial frame for the few minutes a launch might require.
If I launch a wafer-thin rocket (or launch from an airless world - the outcome is the same), with some stupendous acceleration, and a magical fuel source, it might find itself, hovering, engines a-flame, at 400km, with the Earth rotating beneath it. It has all of the horizontal motion that it started with, and is *not* in an orbit - but slowly circles the globe (which rotates beneath it)
24th May, 2018
Guido Colasurdo
Sapienza University of Rome
It is late in the night and I do not read James' and Daniel's comments. Dealing with a launch vehicle and its payload, the spacecraft trajectory only depends on "inertial" velocity V. The rocket pays the relative-to-atmosphere velocity W. Being U the Earth (or atmosphere) velocity (due to East)
V=W+U (bold means vector)
So the same circular orbit (same V) require less propellant (less W) if direct (all 3 vectors eastward) than a retrograde orbit (V and W westward).
This applies to spacecraft (they fly higher than 200 km).
P.S. I cannot understand what you are exactly asking for
1 Recommendation
25th May, 2018
Alexander I. Korolev
alex-korolev.jimdo.com
Dear James,
What if your rocket with "magical fuel source" is levitating for a long time at 4 km, 400 m above the land? Even during a few minutes at sea level it would take off up to a few hundred meters on equator if there is no some binding of it with rotating solid land. This is because it would fly in a straight line along the rotation velocity at the startup (V0).
Dear Guido,
Is the direction of velocity component U in your formula changing during the flight or it is the same as it was at the startup moment?
Let us suppose one stops a spacecraft at the altitude of 200 km in "inertial" (Solar) frame, switches on thrust for fully compensation of Earth gravity and observes its further trajectory. Will the spacecraft accelerate in the direction of local U component? If one creates U artifically will the spacecraft move along it or bend its trajectory around the Earth?
The general question is about differency in flying dynamics at different altitudes related to Earth rotation.
25th May, 2018
Daniel Pfenniger
University of Geneva
James,
A vertical launch at equator is fully subject to the Coriolis force. Only horizontal motion is not.
25th May, 2018
Daniel Pfenniger
University of Geneva
Thierry: please elaborate, as fictitious forces depend only on the non-inertial frame of reference.
25th May, 2018
Alexander I. Korolev
alex-korolev.jimdo.com
Dear Thierry,
Coriolis force in atmosphere is impossible without mechanical contact, but is not caused by the air friction forces. Read about Foucault pendulum experiments. The difference between rotational velocities of Earth in the lower and upper points of the oscillated body trajectory is a few cm/min. Convectional flows in a laboratory are significantly faster.
One of the problems for modern popular physical theory is in avoiding the old and fundamental term "aether". The theory is written in such a way that around atoms and elemental particles there is empty space (vacuum). "Quantum fluctuations", "WIMP" and a number of modern classical ("alternative") theories save the situation.
25th May, 2018
Daniel Pfenniger
University of Geneva
Thierry,
You are wrong. Coriolis and centrifugal forces are named fictitious because they don't exist in a proper inertial frame. If you need to check, look what happens in a rotating vacuum chamber, you will see air is not necessary
for Coriolis force to act.
3 Recommendations
25th May, 2018
Alexander I. Korolev
alex-korolev.jimdo.com
Dear Daniel,
Do you suppose that a flying body in rotating vacuum chamber should deviate its trajectory due to Coriolis force? This should not be the case. To rotate "vacuum" an attracted force is needed- probably such like a gravitation.
1 Recommendation
25th May, 2018
James Garry
Red Core Consulting ltd.
Alexander,
> What if your rocket with "magical fuel source" is levitating for a long time >at 4 km, 400 m above the land?
Then it is stationary with respect to the air surrounding it, and also with respect to the ground.
<sincerely, I am at a loss as to why you think it should not be>
The fluid it moves through, continually constrains its motion.
Imagine, if you will, the _slow_ vertical launch of a rocket from an airless world's equator. It differs from the trajectory taken by a balloon being launched from the Earth.
Where is the mystery?
Daniel.
An equatorial launch, directly vertically, stays in the plane of the equator - it experiences no 'force' perturbing it north or southward. Please describe what you think happens in that case.
25th May, 2018
Alexander I. Korolev
alex-korolev.jimdo.com
Dear James,
The atmosphere is composed by gases not a liquid. Weak friction forces do not prevent your rocket from drifting in the direction of V0 away from the Earth at low altitudes. In a big vacuum chamber trajectory of the rocket should be the same at the same thrust.
The similar question may be about launching of a rocket from a rotating massive celestial body without atmosphere. When Apollo spacecrafts were taking off slowly from the Moon surface there were not observed the surface shiftings caused by Moon rotation- from what I remember.
25th May, 2018
Daniel Pfenniger
University of Geneva
Alexander,
At equator a vertical launch deviates westward, because the Coriolis acceleration formula is 2 times the velocity vector V cross product the angular velocity vector Omega, which for the Earth is directed towards the north pole. V is perpendicular to Omega and in the equatorial plane. If you know how to calculate a cross product you will find the acceleration is perpendicular to both V and Omega and directed to the West.
26th May, 2018
Alexander I. Korolev
alex-korolev.jimdo.com
Dear Thierry,
The Coriolis force, acting on the connection point, is negligible due to small deviations of the point from initial position- even if the roof are elastic.
Kepler's laws do not deal with rotation of gravitating bodies. And the mass in Foucault experiment is not in a Kepler's orbit. This is not the case.
Dear Daniel,
Thank you for explanation. The direction of Coriolis force at the equator was discussed by you with James.
Centrifugal force, as well as Coriolis force, is required some kind of mechanical contact with the rotating body. To check this one can moves a spacecraft along a circulal trajectory using its reactive engines and measures the value and direction of the thrust during the experiment. The component of thrust compensating hypothetical centrifugal force should not be revealed.
Both forces are observed during a flight in the atmosphere. The contact can not be explained by low viscous friction forces.
The experimental data on flying dynamics of a body near rotating center of gravitation (Earth) at different altitudes are needed for a clearer understanding the questions under discussion.
Maybe Guido will tell us something...
27th May, 2018
Alexander I. Korolev
alex-korolev.jimdo.com
Dear Thierry,
Your hypothesis regarding tidal forces is interesting. Can you prove that the rotation of Foucault pendulum is caused only by variations in tidal forces, not in combination of these forces with Coriolis forces? The reason of tidal forces in this case is the difference of centrifugial forces due to Earth rotation. For centrifugial forces, as well as for Coriolis forces, some mechanical contact with rotating body is required...
27th May, 2018
Guido Colasurdo
Sapienza University of Rome
Dear Alexander,
unfortunately I have no time in these days to read all comments. So I just answer to your questions directed to me.
1) the "inertial" reference frame is geocentric and not rotating. ("inertial" is not completely correct)
2) U is the velocity of a rotating frame fixed to the Earth surface. So it is always eastward. It changes only with altitude if trajectory is equatorial (more correctly in general is proportional to the distance from the Earth's rotation axis)
3) the problem of rocket ascent trajectory is very complex (and it is very difficult to find a book dealing with it) because the rocket takes off from a moving point, flies inside the atmosphere and eventually will operate outside
4) when people study a flight all inside atmosphere, this is assumed rotating with the Earth. Aircraft velocities are low and centrifigal and Coriolis' forces can be neglected in a first instance
5) according to my main work on obits and interplanetary trajectory, I PREFER to integrate rocket ascent equations the non rotating frame, where neither centrifugal nor Coriolis' force exist
6) In the following I will not consider accademic problems of this discussion, but I will describe the unique solution whis is possible with the present technology of launch vehicles
7) Looking at the trajectory in the rotating frame, the rocket takes off vertically with a very low initial acceleration (0.2-0.3 g) that rapidly increases. At 100m of altitude the roket rotates and enters a zero-lift or zero-incidence trajectory (to sustain only longitudinal loads) wich is progressively made horizontal by gravity
8) (here the presentation becomes approximative) In order to improve performance, the final velocity acquired using propellant (W=7400m/s), should be direct eastward and horizontal in order to sum it algebrically to the frame local velocity U (approximately the launch pad velocity=400m/s), thus providing the orbital velocity V=7800m/s required for a 200km altidude circular orbit
I suggest you to fix these practical concepts, and then present me your precise question
Guido
PS sorry not reading Thierry, Daniel, James who have provided their own considerations
2 Recommendations
27th May, 2018
Alexander I. Korolev
alex-korolev.jimdo.com
Dear Guido,
Thank you for information. There are two practical questions for you below.
1. Let us suppose a rocket was launched from the equator in a meridianal plane. Is the variable velocity component U (maximal near equator and zero near poles) distorting the orbit? Or does only initial U near the startup point has an impact on the orbit?
2. Let me repeat my previous question: "Let us suppose one stops a spacecraft at the altitude of 200 km in "inertial" (Solar) frame, switches on thrust for fully compensation of Earth gravity and observes its further trajectory. Will the spacecraft accelerate in the direction of local U component? If one creates U artifically will the spacecraft move along it or bend its trajectory around the Earth?"
Dear Thierry,
For the tidal forces a physical contact is also needed.. It is simply because they appear due to the difference in corresponding centrifugal forces. Significance of these forces for Foucault pendulum is the interesting theme.
27th May, 2018
Guido Colasurdo
Sapienza University of Rome
I have just time to answer the first question. If one launches from an equatorial launch pad eastward or westward, the trajectory is plane. If the target orbit is polar (along a meridian) the trajectory is not planar (three-dimensional). Without discussing its (complicate) shape, and just give a "role of thumb" answer, one loses all the U contribute (400 m/s) which has to be paid burning propellant.
Second question to-morrow (hopefully)
28th May, 2018
Janusz Pudykiewicz
Dear Alexander,
The answer to your question can be obtained after solving the Equations of motion for flight over a spherical rotating Earth, for the explicit form of these equations please see section 1.3.4 in:
It is much simpler indeed to consider the equations in equatorial plane, they are presented in section 1.4.2. After a preliminary evaluation of these equations I can conjecture that the large horizontal component of velocity attained by the craft is the effect of a trust force and the change of the $\gamma$ angle. To obtain analytical or numerical solution we can try to use Mathematica.
Uncertain factors are associated with the upper winds; they are very often faster than the speed of rotation measured in the inertial Cartesian frame.
With my best regards,
Janusz Pudykiewicz
2 Recommendations
28th May, 2018
Alexander I. Korolev
alex-korolev.jimdo.com
Dear Guido,
Is this trajectory three-dimensional due to variable U component during the flight (in accordance with your formula V=W+U) or by the other reasons?
Dear Thierry,
The reason of the tidal forces is the difference in acceleration of free fall. The latest one can vary depending on attracting mass, distance from the center of mass or on centrifugal forces. To avoid bare assertions the forces acting on the pendulum should be drawn. Look at the fig. above. If bodies near the Earth surface could orbit independently, the aircraft would fly in a straight line along V0 away from the Earth.
Dear Janusz ,
Thank you for reference, but complicated theory, correctness of which is practically unproven, are unlikely to give us more understanding on the question. The official theoretical answer (at first approximation) was given by Daniel above. But the cases may exist where this theory is not working (see my questions to Guido). And nobody before answered on the basical question regarding absence of drift of an aircraft away from the land in fig. above. Maybe you can do this?
28th May, 2018
Alexander I. Korolev
alex-korolev.jimdo.com
Dear Thierry,
You have agreed that the tidal forces is determined by the difference in g. Look at the expression for g- there is no any velocity, but only mass and distance. The mass of oscillating body in Foucault experiment is the same, the variations of the distance lead to negligible variations of the gravity force in vertical direction. Which forces can rotate the pendulum if not Coriolis and centrifugal ones? Your explanation regarding "changing Kepler geometry for circular orbits" is not convincing because the pendulum is not on any Kepler orbit (circular or elliptical) and dynamics of the task becomes unclear.
For the levitating aircraft it is unusable because gravity force is compensated by thrust force and there is no situation of free falling.
P.S. I've mistaken by mixing centrifugal forces with tidal forces. Sorry. For Foucault pendulum only the first ones can have significance.
28th May, 2018
Daniel Pfenniger
University of Geneva
Thanks Janusz for a definitive and most complete answer. Those still believing that friction of some kind is necessary for Coriolis force to act should study the link given by Janusz.
1 Recommendation
28th May, 2018
Alexander I. Korolev
alex-korolev.jimdo.com
Dear Daniel,
Thank you for your first short theoretical answer. I hope it is in accordance with the theory in textbook given by Janusz.
Dear Thierry,
The centrifugal force as well as Coriolis force are real. About centrifugal force you can ask the pilots in a training center or children riding a carousel. Let us stop the discussion on Foucault pendulum here, it may be a separate topic and needs accurate drawings of the pendulum with acting forces.
28th May, 2018
Janusz Pudykiewicz
@Alexander I. Korolev
Dear Alexander,
Thank you for your comments. The theory is still limited, because the aerodynamic effects are not included realistically, but I’m convinced that it can be used as the first approximation. I will definitely try to go further with the analysis of equations (using Mathematica). In the meantime, I wonder if the effect mentioned in your question is the result of a pitch maneuver during the initial stage of launching?
For the entertaining and much simpler version of the problem you can consult the paper published by Franco Bagnoli: “We shoot a bullet vertically. Where will it land?”
With my best regards,
Janusz
@Daniel Pfenniger
Thank you Daniel for the positive opinion. I hope that my solutions will provide the answer that is acceptable to all participants of this discussion.
With my best regards,
Janusz
28th May, 2018
Alexander I. Korolev
alex-korolev.jimdo.com
Dear Janusz,
Thank you for the demonstrable theoretical paper. There are two problems for such an explanation:
1. At low altitudes. Coriolis forces can not explain why does a levitating aircraft not shift in the direction of its initial velocity obtained at the launching (see fig. and question above)? It is fully entrained by the Earth.
2. At high altitudes. Equation (3) for lateral velocity has too simple dependence on altitude above the land z. It is unphysical because of at the high enough altitude Earth gravitational field becomes negligible, and Earth rotation should not affect the trajectory of a spacecraft (see questions to Guido).
Introduction of atmosphere viscous friction and other side effects to the description will not allow to solve these basical problems.
28th May, 2018
Alexander I. Korolev
alex-korolev.jimdo.com
Dear Thierry,
This looks like a question: Which force does a standing man feel- gravitation or support reaction? Let us end this school discussion here.
28th May, 2018
Guido Colasurdo
Sapienza University of Rome
You (all) write too much and too fast for an old man.
Alexander, if the trajectory is not equatorial, it cannot be planar beacause atmospheric forces are not in the trajectory plane (i.e., have an out-of-plane component).
Tanks to Janusz who cites my colleague Giulio (and his thesis). In a difficult problem it is easier to write the complete equations and to integrate them. It is more difficult to give a simple picture of the problem
And now I will try to answer your question number 2). If I understood, you want to split the trajectory into two phases: an ascent to stay without motion at 200km altitude, and use of thrust to get the orbital velocity (7800 m/s).
Let me assume the absence of atmospheric drag and two impulsive maneuvers. The first is at ground to obtain a vertical ballistic ascent. Taking off with vertical velocity 2000 m/s the rocket climbs and stops at 200 km altitute. Than the engine provides an horizontal velocity (in any direction) of 7800 m/s (V). The total propulsive effort is 9800 m/s
In a previous comment I tried to explain the correct ascent trajectory (zero lift). The total propulsive effort is (launching from equatorial pad - U = 400m/s)
W=7800-400= 7400 m/s eastward
W=7800 m/s northward
W=7800+400= 8200 m/s westward
To each of these 3 values one has to add 800 m/s to get the potential energy relate to the higher altitude
Be aware that payload mass is a very non-linear function of W and one loses a big amount of payload by launching westward (retrograde orbit). The payload loss is huge, if Alexander gets the the final orbit by means of "his" strategy
Have a nice night
Guido
1 Recommendation
28th May, 2018
James Garry
Red Core Consulting ltd.
Alexander,
Your example is bogus.
The LEM did not ascend vertically - it was aiming for the CM after all - and so would follow the opposite of a 'gravity turn'.
Your failure to grasp the notion that the launcher keeps the horizontal component of its momentum at launch is troubling.
As the kids say these days, "I'm outta here"
1 Recommendation
29th May, 2018
Alexander I. Korolev
alex-korolev.jimdo.com
Dear James, Thierry
You don't understand that when the gravity force is fully compensated by the thrust force Kepler's laws are not applicable. In accordance with mechanics an aircraft should fly along the initial lateral velocity component. Small deviations in the gravity force direction during this flight could be simply compensated by corresponding small deviation in the thrust vector with fixed module (the difference in value of g at low altitudes is negligible). At least for aircrafts near the Earth surface such a movement is not observed. The interesting question- what it would be if a LEM was levitating near lunar equator for a long time..
Thank you for participation in discussion on my question.
Dear Guido,
Thus you cannot validate your formula V=W+U for each point of a polar orbit (with variable local U)- due to great effect of atmospheric forces on a rocket trajectory. Complication of the theory without understanding of basical aspects will not lead to progress on the question.
You have not understood my second question. I wrote about two situations:
1. A spacecraft (rocket) is stopped at ~ 200 km altitude in "inertial" solar frame- with compensation of rotational 400 m/s velocity (U-component) obtained at the launch and of the gravity force- by continuously working engines with controlled thrust parameters (thrust force is equal and oppositely directed to mg all the time). Will it conserve its position for a long time?
2. A spacecraft (rocket) after such a stopping acquires lateral U-component again by small lateral thrust pulse. Will it fly in a straight line along U or bend its trajectory around the Earth?
29th May, 2018
Alexander I. Korolev
alex-korolev.jimdo.com
Dear Thierry,
Kepler's laws describe the motion of a body (of planets in the solar system originally) under gravity and centrifugal forces only. One can read about it in a school textbook. Addition of a compensating thrust can let the body move independently from the gravity centers. Let me stop the discussion with you here to avoid flood in the topic.
29th May, 2018
Janusz Pudykiewicz
Dear Thierry,
Thank you for your comment. The Coriolis force is always included in the rotating reference frame. This fact results from the principles of covariance of physical theories. For this reason Giulio Avanzini's work is correct.
The Coriolis force problem is particularly important in atmospheric flow; for derivation using a 4d tensor formalism please see
(Please consult eq. (40) where the inertial forces are defined by the Christoffel symbols). The same techniques can be also used in the case of a material point in the gravitational field.
Sincerely,
Janusz
Dear Guido,
The paper by Giulio Avanzini is very useful in the context of this thread, in fact it is one of then best self-contained references.
With my best regards
Janusz
Dear Aleksander,
All the problems associated with modeling of the rocket in the Earth's atmosphere are quite difficult without formal mathematical methods. Until now, we mainly referred to Newtonian mechanics, perhaps the formulation using Lagrangian mechanics will be more intuitive. I would strongly recommend to read paragraph 14 in the book “Classical mechanics” by Landau and Lifshitz.
With best regards
Janusz
29th May, 2018
Guido Colasurdo
Sapienza University of Rome
I can validate everything, of course. In a problem ruled by complicate equations I need to simplify and present it in the best way to understand the relevant point
question 0. It is split into two parts
0.1) during the continuous-thrust ascent into a non-equatorial orbit, even though the thrust is in the trajectory plane, the atmospheric force will bend the trajectory that will not planar (compensation to keep it planar is possible but expensive and useless)
0.2) looking to the global propulsive effort, imagine a tower 200km high with a gun on the top (located on the equator). You pay the the velocity relative to the gun (W) wich must be summed to the tower velocity (U) to obtain the inertial velocity (V). The circular velocity for a 200km altitude is 7800m/s no matter the inclination of the orbit. So the propulsive effort W depends on the orbit inclination according to my previous post
1) First: the problem is ill-posed. The "inertial" frame must be geocentric. And I assume this in the following. You are hovering (zero-velocity) outside of atmosphere, no drag, only gravitation. (You may reach that condition by firing vertically a gun with exit velocity 2000m/s). Soon after you arrive, please compensate gravitation and you will stay there until you run out of propellant
2) you are free of the tower; my U velocity does not matter. Only inertial velocity vector V matters. If V is zero, you must continue to compensate gravitation until the spacecraft reaches the circular velocity (7800 m/s). You have an uncorrect point of view. Decide the trajectory you want and you can derive the necessary thrust law. For instance, to get a circular orbit, the simplest and most efficient solution is the one on point 0.2). Fire a very big engine and acquire instantaniously 7800m/s orthogonal to the radius and the spacecraft will enter a circular orbit.
I hope that my answer is clear enough
Guido
30th May, 2018
Janusz Pudykiewicz
Dear Thierry,
The Coriolis force is a fictitious force arising in a rotating system.
A good discussion of the meaning of "fictitious", without resorting to mathematics, is presented in the article "Space, time and gravitation” by E. B. Wilson
(para. 10, pages 225 and 226)
After checking many source materials, I must confirm my opinion from the previous post, namely: the Coriolis force is always included in the equations describing the mechanical systems in the rotating frame of reference. It is also observed during the reentry of the space vehicles (outside of the atmosphere).
For interesting historical notes about the Coriolis force, starting with the experiments of Giovanni Borelli, please see:
The problems with the interpretation of the Coriolis force are really not trivial when we consider the general theory of relativity, as discussed in the article:
“Coriolis effect in the Einstein universe”
With my best regards,
Janusz
30th May, 2018
Guido Colasurdo
Sapienza University of Rome
I am reading Janusz's last post (this discussion has no boundaries, unfortunatly). I will repeat what I usually teach to my students
1) equation F=ma is valid only if a is the second derivaty of vector position, r expressed by means of cartesian (the simplest case) components in an inertial frame
2) you can express vector a using the derivatives of the components of r in a rotating frame; in this frame the same acceleration vector has (at least) three terms (relative, centrifugal and Coriolis acceleration)
3) if one moves the last two terms on the left-side of Newton equation, close to the real force F, part of the acceleration is assumed to be (i.e., is tranformed into) two "fictitious" forces
4) at this point for any problem two approaches are available: write equations in the "inertial" frame; use instead a rotating frame
5) the first approach is perfect for orbital mechanics
6) the second approach may be used when motion in atmosphere is analyzed (I usually prefer the inertial frame also in this case)
Going back to ascent trajectory, the propulsive effort (using, for instance, a gun) creates a velocity W (parallel to the gun barrel) measured in the rotating frame. In order to use simple equations for the orbit, it is necessary/convenient to shift to the inertial frame where velocity V=W+U.
Magnitude and direction of U depend on the time (or the rocket position) when/where you prefer to shift from a frame to the other.
I hope that my presentation can help Alexander, while, I am sure, it is useless for other people
Guido
30th May, 2018
Guido Colasurdo
Sapienza University of Rome
Dear Thierry, dear all,
Coriolis force does not exist (as a force); it is a trick to get a similar equation to apply Newton's 2nd law in a rotating reference frame.
The second reason of the non occurence outside the atmosphere, is that a rotating frame is useless. Therefore also the trick of Coriolis' force is never needed.
Guido
30th May, 2018
Alexander I. Korolev
alex-korolev.jimdo.com
Dear Guido,
Unfortunately you have misunderstood all my questions to you because of:
0) The question was about your formula V=W+U applied to a polar orbit. U here is the local Earth rotation velocity (of the land under the rocket for each time moment), isn't it? You cannot prove it due to a big "atmospheric force will bend the trajectory that will not planar".
1) A spacecraft is stopped in "inertial" heliocentric (solar) frame.
2) "A spacecraft (rocket) after such a stopping acquires lateral U-component again..." U component in your formula is the Earth rotation velocity which equals to about 400 m/s at equator. And if the thrust is continiously working with compensation of the gravity force (Ft= mg) the spacecraft is not on a Kepler's orbit.
Dear Janusz,
Thank you for recommendation. I studied Landau-Lifshitz "Theoretical Physics. Mechanics" a long time ago. Formal mathematical methods are not able to solve any physical problem. This is because mathematics for physics is only language. Physics is the natural science which deal with natural phenomena. Theoretical descriptions, which are pretended on comprehensiveness, impede scientific progress.
30th May, 2018
Alexander I. Korolev
alex-korolev.jimdo.com
Dear all,
Please, let us stop flooding here. For discussion on the nature of inertial forces a separate question may be asked.
30th May, 2018
Guido Colasurdo
Sapienza University of Rome
Dear Alexander, if you do not fix any point after receiving suggestions, all we others waste our time.
If you are obstinate and want to use a heliocentric frame for orbital problems, please give up this discussion.
In order to study Earth orbits, the frame MUST be geocentric (either rotating or non rotating).
Before playing cards we need to fix play rule
Guido
1 Recommendation
30th May, 2018
Guido Colasurdo
Sapienza University of Rome
Thierry believe me, Coriolis force does not exist.
I try to take a differnt approach
Let we select any mechanical problem. The motion of a car or of a rocket (in the following).
Let me be an observer who looks from the sky and you are sit on the Earth.
I say: "Yes Newton's law is correct; acceleration is according to engine thrust and gravity"
You say: "Newton was a mad; the rocket does not move according to the engine thrust (and gravity). I am discovering two more forces: Coriolis and centrifugal force"
And I say: "Thierry, you are the mad, because Newton introceded the inertial reference frame and you are not an inertial observer. You see an acceleration (the relative one) that is different from the one that I see. If you correct your acceleration by adding two more terms, we get the same acceleration. If you like, change the sign and call these acceleration "forces". I allow you this, but "force" is just a nickname for acceleration terms".
Believe me, I am old and wise
Guido
30th May, 2018
Guido Colasurdo
Sapienza University of Rome
Dear Thierry,
I read just your last post and did not consider your previous ones and your example.
You did not understand that I is not Guido but the inertial observer and you is not Thierry but any man sitting on the Earth.
I do not deny any experiment, because I integrate rocket equations and routinely compare results with actual launches.
I am an engineer and can work using or not using Coriolis force (in an inertial or in a rotating reference frame) getting the same trajectory for the rocket.
I tried to explain you that the presence of "Coriolis force" depends only on the reference frame. If a scientist makes an experiment in his lab, he needs to account for "Coriolis force" as mesures, pictures and so on are all in the rotating room of his lab. Nevertheless this force is just "appearance".
True forces are the same vectors independently of the reference frame. Apparent forces (Centrifugal and Coriolis) appear and disappear depending on the frame.
By the way, I say that I am old and wise only because is more polite than telling the opposite to the worse among my students.
Guido
1 Recommendation
30th May, 2018
Dear Alexander I. Korolev
Your question is of cause a hypothetical question so we can assume a rocket with a special drive that is extreme efficient with nearly no propellant consumption. Now you want to take this rocket and lift this rocket of the surface of the earth in a vertical direction. Because you have such amazing rocket engine you don't need to lift at high speed and, at 200km hight, you even can make that your hight increase becomes completely zero. You hover. Now the point of the question is, you lifted off in a vertical direction with your trust vector always perpendicular to the surface but when you are at 200km hight the earth rotates underneath.
Is this a correct interpretation of the question?
My answer would be that you have to remember that the rotation of the earth is an absolute motion. The earth rotates and space doesn't rotate. At the surface of the earth you were moving with the speed of the earth surface around the rotation axis. The radius of the circle for an orbit around the centre at an altitude of 200 km is 200*2π = 1366km longer. So if you would have intended to hover above the place of liftoff you should have added the extra speed in rotational direction to get an about 24 hour rotation at a longer orbit length. Because of that you now will see that you fall behind with the rotational speed of the earth. In a case where there would be no atmosphere like with the LEM on the moon the same effect would already be there for the first meter above the surface. Only the moon rotates once every month. So the effect would be even smaller.
Regards,
2 Recommendations
30th May, 2018
Alexander I. Korolev
alex-korolev.jimdo.com
Dear Paul,
Yes, you correctly understood my general question. Such a spacecraft is not hypothetical. Look at Dragon produced by SpaceX, for example. It can be lifted up to a high altitude and hover there within a few minutes.
Your answer is in accordance with the first theoretical answer given by Daniel in the beginning of discussion. There are two problems for such an explanation:
1. At low altitudes. Why does a hovering aircraft not shift in the direction of its initial velocity obtained at the launching (see fig. above)? Thrust force compensates gravity force but the aircraft trajectory bends around the Earth.
2. At high altitudes. Bending of a hovering spacecraft trajectory around the Earth due to small lateral velocity is still unproven (see answers given by Guido). In the case of explanation due to Coriolis forces (look at the paper given by Janusz) it is unphysical because of at the high enough altitude Earth gravitational field becomes negligible, and Earth rotation should not affect a trajectory of the spacecraft.
Lunar rotation is too slow for good observation of the effect, thank you for remark. Let me suggest another example on the question. Let us consider fast rotating asteroid with weak gravitation near its surface. Near its equator there is fixed a spacecraft. At some moment the spacecraft switches on thrust- only to compensate the gravity force (the same situation as in fig. above) and detaches from the surface. What will be its trajectory?
30th May, 2018
Guido Colasurdo
Sapienza University of Rome
Dear Paul and Alexander,
any problem with an engine that does not need propellant and has sufficient thrust magnitude can be solved. The user decides the trajectory and someone will get the necessary thrust law (magnitude and direction). The real problem is to define a target orbit (no propellant consumption) and inject a spacecraft into it minimizing, for instance, the propellant/payload mass ratio. I can discuss this kind of problems.
On the other hand, I just guess that any problem with a magic engine has its own solution. Use thrust and overcame gravity, drag, earth rotation and so on, without any cost.
Before solving a problem, one needs a meaningful definition of it
Guido
PS Looking at last Alexander's post, it seems that Alexander is persuaded that a rocket can hover in space. Without a better definition of "hovering" I cannot agree
31st May, 2018
Guido Colasurdo
Sapienza University of Rome
Dear Thierry,
I was forced by your words to read the previous posts (in a very fast way).
You trust in experiments. So the fact that Daniel, James, Janusz and Guido agree (using different words and points of view) that Coriolis' force is only apparent results in an interisting text of your vision of that force.
Regards
Guido
31st May, 2018
Dear Alexander I. Korolev,
Your problem seems to be that at low level above the earth this effect is non existent. I think you are wrong. For every 15.9 cm above the earth the circumference of the earth is 1 meter longer. So after one earth rotation the hovering rocket would have shifted by one meter. In other words it will shift with 4.167 cm per hour. This of cause under the condition of no atmospheric drag and perfect vertical takeoff. Also the rocket would have tumbled a bit because its orientation has rotated a complete 360° but his orbit around the earth is not yet complete. It has still 1 meter to go.
Regards,
31st May, 2018
Dear Guido Colasurdo,
Guido>: The real problem is to define a target orbit (no propellant consumption) and inject a spacecraft into it minimising, for instance, the propellant/payload mass ratio.
The real problem is the problem of this question. The problem you refer to is another interesting problem. The problem of this question is highlighted with this hypothetical environment but infinitesimal existent in any space flight. When the distance between the centre of the earth and the rocket doesn't change while the velocity tangential to the surface of the earth only depends on the initial conditions of the departure then I think we can call that situation a special case of hovering.
Of cause in reality this situation in isolation is difficult to observe at low altitude. Rocket operators don't need to take it into their trajectory considerations because the speed differences between the rocket and this effect are too large.
Regards,
31st May, 2018
Alexander I. Korolev
alex-korolev.jimdo.com
Dear Guido,
The same hovering is also possible at high altitudes above the land. The value of lateral shift of a spacecraft (rocket) due to Earth rotation is the point of interest.
Dear Paul,
You haven't given the answer- why does the trajectory bend? In the case of straight movement along initial velocity V0 obtained at the startup an aircraft would lift up to a few hundred meters above the land after 1 min. It is the simple task on dynamics at first sight, see fig.
And you exact explanation of the question is less physical than Daniel's theory. Conservation of the rotational speed of an aircraft at lifting up leads to nonconservation of summary angular momentum in the system Earth-aircraft.
31st May, 2018
Guido Colasurdo
Sapienza University of Rome
No Paul, "the problem" of the real world is the one exposed by me. People work routinely to solve it.
Alexander poses an academic problem. It can be solved but one needs to pose it in a correct and clear way.
Alexander mixes continuosly questions before anyone is fixed. If you both agree, I will pose a simple problem close to what Alexander is probably asking. I will provide the correct (in my opinion) solution. We discuss it and all can move to a more complex problem
Guido
PS Alexander refuses to specify what "hovering" is. Constant-H ? Zero-velocity? In a rotating or non-rotating frame?
PS2 Paul consider hovering equivalent to "constant-H or -r". If we neglect the atmosphere (and thrust), just one value of velocity is permitted (it is named "circular velocity")
2 Recommendations
31st May, 2018
Janusz Pudykiewicz
Dear Alexander,
The answer to your question depends on the tangential velocity of the spacecraft. Starting from the surface and going through the permissible orbits, we see that this key parameter changes significantly. A brief summary can be found in the table "Tangent speeds at altitude" in the Wikipedia article below
The above data (in a geocentric, non-rotating system) should explain the relative movement of the spacecraft with respect to the surface.
Please also take into account the fact that the transfer trajectory from the launching pad to the stationary target orbit is always inclined. This also applies to the spacecraft with the ability to hover, as shown in the following footage:
Best regards,
Janusz
1 Recommendation
31st May, 2018
Guido Colasurdo
Sapienza University of Rome
Dear Janusz,
the first step is correct; usually in astrodynamics the velocity is named "circular velocity" and it is inversely proportional to the sqroot of the radius. It is easy to compute the ground track of the spacecraft on the Earth's surface.
As far as the second question is concerned, your sentence must be mitigated. If "stationary target orbit" means "geostationary orbit", it must be equatorial and the transfer trajectory has the same inclination as the latitude of the launch pad. You should agree that if one launches from a ship or from the italian (rusting) San Marco platform, the transfer trajectory is not inclined.
Bye
Guido
PS I am afraid that Alexander is far away from the content of your post
1 Recommendation
31st May, 2018
Dear All,
Guido, I completely agree with your PS2 text.
The original question was:
Alexander>: Why an aircraft (rocket) does not shift relative to the land due to Earth rotation at low altitudes but does it at the high ones?
It is well-known that a plane or helicopter do not response to relatively fast Earth rotation (~2000 km/h). But a rocket at the high enough altitudes does it by increasing its own velocity- if the flying direction is close to the direction of rotation and vice versa.What is the dependence of such an entrainment ability on altitude?
I have some problems with the statements. Planes and Helicopters respond to the earth rotation. They rotate with the same speed as the earth and on top of that they change their position with respect to the earth. The difference is that planes and helicopters counteract the effects of gravity through the movement of their wings or rotors through the air. With this effect the gravity is no more a parameter in their path. For the rocket the path it follows is completely defined by the gravity field the speed and the direction. So here is the principle of flying through space that was described in the hitchhikers guide through the galaxy as: falling to the earth and miss. To miss the earth while falling you need enough speed. with only the orbital speed of the earths surface you have not enough speed to mis. When a rocket takes of from a planet surface then it has an initial orbital speed. That is part of his motion. On top of that motion he needs to add more speed before he can fall around the earth and miss. So, as Guido correctly explained, the first part the take_off has to counteract simple gravity, the rocket moves straight up. But soon it tries to get in a trajectory that allows to stay above the atmosphere. For that speed in the direction of the orbit is needed.
So why does the trajectory bend? It is much cheaper in propellant cost to do like this and not just move straight out of the gravity well of the earth. As soon as you have enough speed that your orbit, without rocket engine running, increases the distance to the earth, then you can wait till you are at the most distant point of that orbit before you start the engine again. At that distance the gravity of the earth is much less and thus the same impulse of your engine gives you a larger speed change. How to do that in the most efficient way is something a specialist as Guido Colasurdo can tell you much better.
What is the difference in altitude? You want to have a speed so that you fall to the earth and miss. That speed is so high that you would burn up in the atmosphere. So you better climb out of the atmosphere with a speed that keeps your rocket in one piece and accelerate to orbital speeds above the atmosphere. In footage of rocket launches they talk about max Q. The thinner the air the lower the drag, the higher the speed the higher the drag. somewhere these two effects have a maximum. Accelerate to fast and the Q is too high. Airplanes are always with their speed below this. They fly with wings and not with inertia.
Regards,
2 Recommendations
31st May, 2018
Guido Colasurdo
Sapienza University of Rome
Paul your description is very good. I would just change some word, not because inccorrect but because I use a language that is a bit different.
Just a question. Why did you go back to "the problem"?
Guido
31st May, 2018
Alexander I. Korolev
alex-korolev.jimdo.com
Dear Janusz,
Thank you for references. I agree with the answer on your message to me given by Guido. The problem is that an aircraft/rocket in my example is not on any Kepler's orbit due to compensation of gravity force. So, description of an orbital motion is not applicable.
Dear Paul,
Thank you for remark on the word "response" in my question, I'll correct this. To stay above the atmosphere a speed in the direction of the orbit is not required if there is powerful vertical thrust. You, as well as Guido, haven't understood my example about the hovering aircraft. And regarding the example with asteroid you said nothing.
Dear all,
Let me describe the example with a hovering aircraft/rocket better. Look at the fig. Fg is complitely compensated by FtThere is no orbital motion, Kepler's laws are not applicable. Only rotation. What is the reason of this rotation together with the land downstairs? Why does an aircraft not fly away from the Earth in the direction of its initial velocity obtained at the startup (v0)? What would be in the similar case near fast rotating asteroid with weak gravity?
P. S. Please, all who see no physical problem behind my question- leave this discussion. Let's let a people who see it speak.
31st May, 2018
Guido Colasurdo
Sapienza University of Rome
Dear Alexander,
sorry, the message starting with "Paul" was to both you.
Let me distinguish between spacecraft motion in a vacuum (1) and aircraft motion inside atmosphere (2). The vehicle has already taken off. Let me assume that velocity Vo is "as seen by a non rotating ("inertial") observer".
1) earth rotation does not matter. The spacecraft moves with constant velocity (magnitude and direction) in a stright line. It is only necessary, according to the motion, to change the thrust direction (always radial) and magnitude (decreasing as the distance from the Earth center increases).
2) the earth and the atmosphere rotate with angular velocity omega; it is important to compare the initial velocity with the local atmosphere velocity, in order to get the velocity relative to atmosphere
W = Vo - omega*R
Positive values means eastward flight, negatve westward. Aircraft need a sufficient W magnitude.
An aircraft can flight in a straight line (the same as the spacecraft) but only for a short time. So in a real world it moves along a circle, keeping the altitude constant. The straight line cannot be maintaned with constant velocity because
- air density decreases and should be compensated by increasing velocity (with the same incidence)
- incidence increases until the aircraft stalls
- engine performance decrease with altitude
I hope that your question was understood by me and my opinion understood by you.
Bye
Guido
1st Jun, 2018
Dear Alexander I. Korolev,
Alexandr>: Let me describe the example with a hovering aircraft/rocket better. Look at the fig. Fg is complitely compensated by Ft
There is no orbital motion, Kepler's laws are not applicable. Only rotation. What is the reason of this rotation together with the land downstairs? Why does an aircraft not fly away from the Earth in the direction of its initial velocity obtained at the startup (v0)? What would be in the similar case near fast rotating asteroid with weak gravity?
Why would the rocket not flay away in direct straight line. In your picture you have Fg and Ft. These two compensate each other. The force Fg is a function of the distance to the centre of gravity of the earth/planet/astroid. If you move to the side, then the direction will change. Now, If I assume that your Ft also will maintain to compensate the momentary Fg, then the only speed involved is the tangential speed of the surface. If we observe this situation from a reference frame where the origin is moving with the planet but where the planet is seen rotating, then we see from that reference frame your spaceship move in a straight line away in horizontal direction equal to the surface speed exactly in the direction of the V0 in your picture. If you would have a massless wire wrapped around the planet then this wire could be connected to the spacecraft and it would look as if you unwrap that wire.
Now if you observe the same from the viewpoint of an observer at the surface, then this observer stands on the rotating planet. The spacecraft maintains is straight line and the observer rotates. What is straight ahead of the observer is a rotating point. In that view the spacecraft will move backwards.
Why does an airplane or helicopter not follow this path? Because they don't have this special rocket engine that can counterbalance the gravity. In your example you ave something that can counterbalance a changing gravity field. So your problem reduces to the question what is the path of an object when gravity doesn't exist. Then the object will follow a straight line. This straight line will look like some kind of spiral as seen from an observer who is rotating. Why does an airplane or helicopter not do like that? Because gravity exists. Why does in real world at some distance of a planet this effect becomes visible even when there is gravity? Because that spiral path is related to the rotation of the observer. At large enough distance even Kepler orbits will look like spirals.
Can this answer the question?
PS Guido, why did I go back to the problem? Because I had the impression that Alexander was not happy with the answers and I did my best to reinterpret his question and find an answer to it that is satisfactory for him.
1st Jun, 2018
Guido Colasurdo
Sapienza University of Rome
Dear Paul (and Alexander, who is the user),
I was joking. I had distinguished between "the problem" (of the real world) and academic problems. In a previous post you appeared to give them the same importance. And then, you magically gave a very good description of "the problem".
Now you provide a good description of Alexander's academic problem. For the sake of clarity, let me add that initial velocity in your description is
Vo = omega*Ro
corresponding to W=0. (In my post Vo is free)
In this case you correctly state that a rotating observer sees the spacecraft moving backward (westward is more precise).
I prefer my point 2) to your third block. And while I am writing, I note that you, to be coherent, should consider in this third block only a "hovering" helicopter (W = 0) that will remain above the head of the rotating observer.
Regards
Guido
1st Jun, 2018
Dear Guido,
You are correct with your comments. A hovering helicopter manages to get in a equilibrium. Increasing the height will result in lower air pressure and lower gravitational attraction. By setting the lift of the rotors correct they can stay at one height level. But to stay at one place above the observers head is an unstable situation. It needs constant correction but again with the right technique also that can be stable and then the helicopter will stay above one point and co-rotate with the earth surface.
Regards,
1st Jun, 2018
Guido Colasurdo
Sapienza University of Rome
Paul, you cannot mix the academic world to the real one! In academia a hovering helicopter hovers, and does not fall on observer's head. Leaving any joke, Alexander's problem (spacecraft or aircraft) requires to use thrust or lift in order to follow the path.
I hope that Alexander wil consider this discussion "complete" and I will answer your (complicate) e-mail.
Have a nice day
Guido
1st Jun, 2018
Alexander I. Korolev
alex-korolev.jimdo.com
Dear Guido, Paul
Thank you for attention to my question. I have no possibility to repeat the same things for many times. In short:
It is physically that Earth rotation does not matter at the high enough altitude. But this altitude should be experimentally determined. The aircraft does not move in a straight line near the land even for a short time. Rotating atmosphere is not a solid body and cannot transfer angular momentum of the Earth up to the aircraft effectively.
The effect described in my question does not depend on a reference frame. Look at the sand particles flying away from a rotating ball. They fly in a straight line in contrast to the aircraft/rocket hovering near the land in my example.
The discussion is not complete. Let other people provide their view on the question.
1st Jun, 2018
Guido Colasurdo
Sapienza University of Rome
Dear Alexander
you posed a correct problem and you got a similar answer from me and Paul. Now you do not like the answer (why ?) and change the problem that becomes ill-defined and cannot be understood.
"It is physical that Earth rotation does not matter at the high enough altitude."
The Earth rotation matters when taking off. Soon after the rotation of atmosphere (a bit different thing) becomes important. It is impossible to define the upper limit of the atmosphere, which depends on the problem. For a spacecraft one can neglect atmosphere from an altitude between 200 km (very few orbits) and 800 km (entire life).
"But this altitude should be experimentally defined."
Every one knows this altitude for his own problem. Fix a problem.
"An aircraft does not move in a straight line near the land even for a short time."
False. One has just to control the flight.
"Rotating atmosphere is not a solid body and cannot transfer angular momentum of the Earth up to the aircraft effectively."
One is interested to the flight. No one asks the atmosphere for trasferring the angular momentum of the Earth. May be or may not. Who cares?
"The effect described in my question does not depend on a reference frame."
Which effect? Any "effect" can be achieved in your problem by changing the control law. You posed a problem and you got the correct answer, no matter which frame is used.
"Look at the sand particles flying away from a rotating ball. They fly in a straight line in contrast to the aircraft/rocket hovering near the land in my example."
A more complicate problem cannot be a valid example for a simple problem.
You are acting as Thierry who got 4 answers by people who "consider" Coriolis' force as apparent and not real: therefore he decided to discard experts' opinion instead of cleaning his mind.
Please clean your mind instead of waiting for a miracle.
Guido
1st Jun, 2018
Alexander I. Korolev
alex-korolev.jimdo.com
Dear Guido,
Due to continuous misunderstanding let me stop discussion with you. Thank you for some useful information given above. Good bye.
1st Jun, 2018
Janusz Pudykiewicz
Dear Guido,
Thank you for your comments; I certainly agree with your remarks and I hope that one day the Italian San Marco Platform will be rejuvenated.
After reading your letter, I have realized that there is an interesting aspect of this thread related to brachistochrone and the general optimization of launching a space vehicle.
With my best regards,
Janusz
1st Jun, 2018
Janusz Pudykiewicz
Dear Alexander,
1. The atmosphere sup-rotates on average 10 m/s relative to the solid Earth. If we assume that the craft is suspended in air it is natural that it will translate with respect to the surface with the speed equal to the speed of super-rotation. The parameter U0 in your drawing is approximately 10 m/s.
2. For the aircraft hovering at (very) low elevation there is no translation with respect to surface, because the whole system, including the atmosphere, the earth and craft, rotates as a solid body (there is no observable super-rotation).
3. For the hovering space capsule located outside the Earth’s atmosphere the craft is stationary with respect to the geocentric Cartesian frame and effectively moves relative to the fixed point on the surface. The orbit in this case is indeed non-Keplerian; based on my search of literature, we can describe it as “special orbit”.
The subject is covered in the book
You can get more additional information about this interesting subject by simply typing: "special orbits" in the Google search.
I hope that my simple answer is satisfactory.
With my best regards,
Janusz
1st Jun, 2018
Dear Alexander I. Korolev,
I wonder how I could answer you more. The effect that you seem to have clear in your mind is something that with my best imagination I can't recreate in my mind without ignoring large parts of real life physics. When you describe a behaviour that needs the absence from gravity and then wonder why real airplanes don't follow that effect then what do you expect from real life? You can describe a special effect in theory, but then you have to look to the real world and see if that effect is overshadowed by a much larger effect. It seems that gravity is quite important near a planet. Its angular momentum together with the heat it receives from the sun is capable to generate a weather system where air can transfer a lot of angular momentum to the an airplane. If you ever studied why the airplanes move west over a much more south route over the Atlantic then when they move east, then you would know that at the 500millibar level there is a strong yet stream moving from west to east. This approximately 300 km/h wind is the effect of this angular momentum. It saves a lot of fuel if you can fly with tail wind and it saves a lot of fuel if you can fly the other way round outside that strong headwind.
So any effect that might be what you try to describe for sure has no minimum radius to start to work. It might have a minimum radius where it is larger than other effects. But all the different effects we described seem not to fit with your thinking.
Sand particles that fly away from a ball seem to go in a straight line. The gravity between the ball and the sand particles is too small as that you can see it. So all you see is the overwhelming larger gravity effect of the earth. But even in outer space with slow enough speed you would see that such sand particles would follow ballistic orbits around that ball. Gravity does not stop to exist. Only the force is on such small scale very small.
Regards,
1st Jun, 2018
Alexander I. Korolev
alex-korolev.jimdo.com
Dear Janusz,
1. The velocity V0 in my drawing is the velocity of Earth rotation at the startup point as it was defined above (about 400 m/s at equator). The "super-rotation" of the atmosphere does not explain- why does a craft participate in this super-rotation also? Imagine a wet rotating propeller with scattering drops. The rotation of atmosphere near it does not prevent the scattering.
2. This is physically. We can go further and discuss the interesting physical problem behind this: how can the angular momentum be effectively transfered in a gaseous (not a solid) planet atmosphere? What is the nature of the binding forces required for such a transfer?
3. It is the good physical point of view (unproven experimentally). But what to do with explanation of the lateral shift with Coriolis forces given by you above? As it was pointed above these forces should appear above the atmosphere also. This appearance shows Earth rotation influence on the trajectory of a moving spacecraft. But why then there is no the influence in the case of motionless spacecraft? In mechanics such a situation is impossible.
Regarding an orbital motion: the definition of the word "orbit" is:
"In physics, an orbit is the gravitationally curved trajectory of an object,[1] such as the trajectory of a planet around a star or a natural satellite around a planet." (see wikipedia). No gravitation influence- no orbit. And historically this word was introduced for the movement under gravity forces.
Dear Paul,
1. I read about similar W-E winds at altitudes ~ 10 km above Eurasia. Entrainment of gases in a rotating atmosphere is well-known from aerodynamics. But place into a whirl some solid particle with velocity equal to the local gas velocity and the particle will fly away from the whirl due to absence of any binding forces.
2. Yes, the effect of the earth rotation on an aircraft/rocket trajectory should be hardly observable due to atmospherical effects (winds and other). But it does not mean that it cannot be observable- even for low altitudes. And a hovering aircraft in my drawing at equator, in the case of its movement along V0, would be lifted up to about 300 meters in a minute.
3. Every force can be compensated by another force. The gravity is not an exception. In the case of summary force acting on a body equals zero the body moves independently on the forces and their sources (gravitational field). It is well-known from mechanics.
2nd Jun, 2018
Dear Alexander I. Korolev,
You have an unconventional view of reality.
Alexander>: 1. I read about similar W-E winds at altitudes ~ 10 km above Eurasia. Entrainment of gases in a rotating atmosphere is well-known from aerodynamics. But place into a whirl some solid particle with velocity equal to the local gas velocity and the particle will fly away from the whirl due to absence of any binding forces.
A hurricane is a whirl of gas. It caries with it clouds that have a lot of solid ice crystals in it. They follow the path the wind forces them to go. Your assumption that there are no binding forces makes me think you never have experienced wind blowing you away. I was in a hurricane. There raindrops fall (nearly) horizontally.
Alexander>: 2. ... And a hovering aircraft in my drawing at equator, in the case of its movement along V0, would be lifted up to about 300 meters in a minute.
Only if the aircraft would adapt his lifting trust to the decreasing support of the air. Because air density decreases with height a helicopter can hover and an hot air balloon can float at constant height. The balance of gravity air density and lift is used to keep constant height. When a ship is in a rough sea and his deck goes up and down you can see that a helicopter is hovering at constant height above this rising and sinking surface. Is that a miracle because there is no stick to keep that distance? Of cause not! it is the incredible skill of the helicopter pilot to alter the trust of the engine to stay at constant distance. It is the intention of a helicopter/airplane pilot to fly at constant height. Why do you ignore that?
Alexander>: 3. Every force can be compensated by another force. The gravity is not an exception. In the case of summary force acting on a body equals zero the body moves independently on the forces and their sources (gravitational field). It is well-known from mechanics.
Completely correct. You only problem is that you assume that these airplanes and helicopters completely compensate the forces. They compensate exactly so much that, with that tiny rest force, the object remains at rest in the moving frame of the earth and does not fly away.
If you ask why I make a curve while driving a curvy road because I was initially driving straight, then my answer is that I turned the wheel. My intention is to follow the road. The intention of the helicopter and airplane pilots is to follow the curved earth. You will not see an effect that is not intended to be followed.
Regards,
1 Recommendation
2nd Jun, 2018
Guido Colasurdo
Sapienza University of Rome
Dear Paul,
It is a pleasure to read you. There is a bit of poet in you. And kindness and delicacy at a level I will never reach.
Guido
2nd Jun, 2018
Dear Guido,
Thank you for the comment. I think that the best way to convince people is when they can say yes and feel happy. For that I try to find out how they think. It is the others task to follow my thinking. I have to follow the others thinking and lead him from there some step further.
Regards,