1. Can anyone explain why addition of vectors does not apply to a satellite in orbit?

When a satellite accelerates towards the Earth at ‘g’ at an exact right-angle to its orbital velocity of ‘v’, why don’t you get a simple addition of vectors?

The vector addition I am talking about would involve a right-angled triangle with a long side of ‘v’, a small side at a right-angle to this representing the increase in velocity per interval of time resulting from ‘g’, and the hypotenuse the vector addition itself.

The centripetal acceleration required for a satellite to perform a circular orbit is g = v²/r, where ‘g’ is the acceleration due to gravity. But why does this ‘centripetal acceleration’ not add during each interval of time to the orbital velocity using vector addition and Pythagoras’ Theorem?

2.

3. Originally Posted by galexander
But why does this ‘centripetal acceleration’ not add during each interval of time to the orbital velocity using vector addition and Pythagoras’ Theorem?
It does.

Think of the satellite doing one half of the orbital circle. For convenience, and so I don't have to make drawings, let's imagine a huge clock face filling this circle, and suppose the satellite goes from the twelve o'clock mark to six (in the, erm, clockwise direction).

Now take all those tiny arrows, say a million of them, pointing inwards from the orbit. The one at 12 points downwards, the one at 1 points down and slightly left and so on, the one at 3 points left, the one at 4 points left and somewhat up.... If you move them around (by pure translation, no turning!) so they are arranged nose to tail, they will form a half-circle shaped like the bottom half of our clock face.

So the sum total of those tiny arrows is one diameter of the circle, pointing left.

Remember that these are velocity vectors, not positions of the satellite. Their dimensionality is in m/s, and the radius of the half-circle they form is equal to the orbital speed. The diameter is, of cource, twice that much.

Now while the speed of the satellite is constant (as a scalar), its velocity (vector) changes dramatically. At the 12 o'clock mark, the satellite was moving right. At 6, it's going left. The change in velocity over this time is a vector 2v long, pointing left - precisely what we found as the sum of the little arrows.

After a full circle, the total change in velocity is zero.

Does this help?

(note: summing a million tiny arrows is an approximation. The proper way to do this is integration - that is, summing infinitely many infinitely small arrows)

4. Originally Posted by Leszek Luchowski
Originally Posted by galexander
But why does this ‘centripetal acceleration’ not add during each interval of time to the orbital velocity using vector addition and Pythagoras’ Theorem?
It does.

Think of the satellite doing one half of the orbital circle. For convenience, and so I don't have to make drawings, let's imagine a huge clock face filling this circle, and suppose the satellite goes from the twelve o'clock mark to six (in the, erm, clockwise direction).

Now take all those tiny arrows, say a million of them, pointing inwards from the orbit. The one at 12 points downwards, the one at 1 points down and slightly left and so on, the one at 3 points left, the one at 4 points left and somewhat up.... If you move them around (by pure translation, no turning!) so they are arranged nose to tail, they will form a half-circle shaped like the bottom half of our clock face.

So the sum total of those tiny arrows is one diameter of the circle, pointing left.

Remember that these are velocity vectors, not positions of the satellite. Their dimensionality is in m/s, and the radius of the half-circle they form is equal to the orbital speed. The diameter is, of cource, twice that much.

Now while the speed of the satellite is constant (as a scalar), its velocity (vector) changes dramatically. At the 12 o'clock mark, the satellite was moving right. At 6, it's going left. The change in velocity over this time is a vector 2v long, pointing left - precisely what we found as the sum of the little arrows.

After a full circle, the total change in velocity is zero.

Does this help?

(note: summing a million tiny arrows is an approximation. The proper way to do this is integration - that is, summing infinitely many infinitely small arrows)
The problem that I have Leszek is that the vector addition appears to be a continuous process. It is happening each and every second.

When you do vector addition it is very much taken for granted that two vectors at a right-angle will both add and change the direction of the object with the second "new" vector applied. So why are satellites an exception to this rule?

5. Originally Posted by galexander
When you do vector addition it is very much taken for granted that two vectors at a right-angle will both add and change the direction of the object with the second "new" vector applied. So why are satellites an exception to this rule?
Satellites are no exception.

You seem to be struggling with the binary nature of vector addition. But that is not what is going on.

Satellites are governed by orbital mechanics, a branch of classical Newtonian mechanics. To understand it you need to understand and apply calculus in a multi-variable (vector) setting.

Try reading an introductory book on Orbital Mechanics. The text by Prussing is quite good.

6. I think Galexander's difficulties would apply to any constant-speed circular motion, and explaining them does not - IMHO - require the use of such special case as orbital mechanics.

Calculus, on the other hand, would be very useful. How familiar are you with that, Galexander?

7. Originally Posted by Leszek Luchowski
I think Galexander's difficulties would apply to any constant-speed circular motion, and explaining them does not - IMHO - require the use of such special case as orbital mechanics.

Calculus, on the other hand, would be very useful. How familiar are you with that, Galexander?
In general I am slightly suspicious of Calculus and tend to avoid it if I can. There are usually other methods of working a problem out but often Calculus does present a good short-cut.

Returning to your 2v solution, i.e. the observation that there is an apparent vector change of 2v for a satellite of orbital velocity v in a circular orbit. If you actually calculate the vector addition using the triangle I specified (a right-angled triangle with a long side of ‘v’, a small side at a right-angle to this representing the increase in velocity per interval of time resulting from ‘g’, and the hypotenuse the vector addition itself), then the actual increase in v you get each orbit is just a fraction of 1.0% each orbit and does not in any way account for the 2v vector change you suggest.

In other words the vector addition I have suggested would cause a small but steady increase in orbital velocity which in turn would cause the satellite to shift to slightly higher orbits as time went on.

8. Originally Posted by galexander
In general I am slightly suspicious of Calculus and tend to avoid it if I can. There are usually other methods of working a problem out but often Calculus does present a good short-cut.
You have just said it all -- wrongly.

Calculus was invented to describe motion. It is not a shortcut but in fact a necessity.

9. Originally Posted by galexander
In general I am slightly suspicious of Calculus and tend to avoid it if I can.
Okay, if you don't like adding infinitely many infinitely small things (which is called integration, a part of that Calculus thing you loathe), then let's talk about a few big vectors instead.

Think of a point following a square "orbit". Never mind how it's implemented mechanically, it could be a ball in a flipper game (AKA pinball). It goes from A to B to C to D to A again, ad nauseam. Always at the same speed v.

What is the vector of change in velocity at every corner of the square? What is the angle between that vector and the velocity of arrival at that corner? What do those change vectors add up to, when the point has turned two corners? Four corners?

If you want some mathematical workout, do the same for a hexagonal orbit (regular hexagon), then octagonal etc, always looking at:
• the change of velocity at one corner,
the angle between the change and the velocity before the change,
the sum of such changes for one half of the "orbit", and finally
the sum of such changes for the whole "orbit".

If you don't mind doing some algebra (or is it another thing you are suspicious of?), you may want to find a general formulas for those vectors in a regular polygonal orbit with n corners, where n is any integer greater than 1.

10. Originally Posted by Leszek Luchowski
Originally Posted by galexander
In general I am slightly suspicious of Calculus and tend to avoid it if I can.
Okay, if you don't like adding infinitely many infinitely small things (which is called integration, a part of that Calculus thing you loathe), then let's talk about a few big vectors instead.

Think of a point following a square "orbit". Never mind how it's implemented mechanically, it could be a ball in a flipper game (AKA pinball). It goes from A to B to C to D to A again, ad nauseam. Always at the same speed v.

What is the vector of change in velocity at every corner of the square? What is the angle between that vector and the velocity of arrival at that corner? What do those change vectors add up to, when the point has turned two corners? Four corners?

If you want some mathematical workout, do the same for a hexagonal orbit (regular hexagon), then octagonal etc, always looking at:
• the change of velocity at one corner,
the angle between the change and the velocity before the change,
the sum of such changes for one half of the "orbit", and finally
the sum of such changes for the whole "orbit".

If you don't mind doing some algebra (or is it another thing you are suspicious of?), you may want to find a general formulas for those vectors in a regular polygonal orbit with n corners, where n is any integer greater than 1.
How I tend to do the calculation Leszek is by approximation using one second intervals. Over an orbit of say 100 minutes this is 6,000 points.

Again if you do the vector addition using the right-angled triangle for each second interval, then sum the effect for the entire orbit, the increase in orbital velocity is again just a fraction of 1% per orbit.

This is because 'g' at 9.8 m/sec2 is a comparatively small figure when compared to an orbital velocity of say 7,700 m/sec. When you apply Pythagoras' Theorem it makes that difference in ratio seem even larger.

But there is also a danger that things can become a bit confusing if you let them. For example if you know the orbital period in seconds and multiply this by the the acceleration due to gravity at the altitude in question and assume that this figure represents the total increase in velocity or vector change, then you end up with more than a 7v increase! But I would hold that this figure is meaningless anyway as the actual effect I am referring to is far more subtle than this and would represent an absolute increase in orbital velocity after each completed orbit.

11. Originally Posted by galexander
This is because 'g' at 9.8 m/sec2 is a comparatively small figure when compared to an orbital velocity of say 7,700 m/sec. When you apply Pythagoras' Theorem it makes that difference in ratio seem even larger.
g is only 9.8 m/s^2 at the surface of Earth.

It sounfs like you are doing a bunch of inappropriate approximations. You really do need to read a book on orbital mechanics.

What in the world do you think a ratio of speed to acceleration means ? They don't even have the same units. The Pythagorean theorem has nothing to do with it.

12. It sounds like he's using Euler's method or something similar. It's known to be a fairly inaccurate approximation of a proper integral, especially for timesteps as large as one second. (You can prove to yourself that this is what's happening because as you take shorter and shorter time steps, the error gets smaller and smaller.)

As DrRocket said, calculus is the appropriate tool for this job.

13. Originally Posted by galexander
How I tend to do the calculation Leszek is by approximation using one second intervals. Over an orbit of say 100 minutes this is 6,000 points.

Again if you do the vector addition using the right-angled triangle for each second interval, then sum the effect for the entire orbit, the increase in orbital velocity is again just a fraction of 1% per orbit.
As long as you are doing finite intervals (and not infinitely small ones), the triangle is not right-angled. I tried to emphasize this by using very big intervals in the "square orbit" example. The vectors would be tilted 45 degrees "inwards and back".

With your one-second intervals, the angle is much closer to 90 degrees, but still only close to it. The vectors will still have a small backward component, just so the new vector has the same magnitude and only a changed direction.

The simplest way to see this is to subtract , assuming the satellite is in uniform circular motion. But as that assumption is exactly what you are trying to question, you need to do some math the hard way. In any case, please try to do some precise math from beginning to end, and not use an approximation and then claim its error - however small - as a discovery.

14. Think of the vectors the other way around. Instead of getting hung up on the resultant vector, think of it in terms of what happens each second of orbit with the velocity and gravitation. For each second the satellite will move a distance x at right angles to the radius of orbit. Then during travelling the distance x, the satellite will fall distance y onto the surface of the orbital circle at another right angle. It will repeat this same motion each time, as g and speed stays constant. The resultant vector you get is the line you get between the point of origin at the beginning of the second and the point where the satellite touches the orbital circle again. Both start and end points fall on the orbital circle if g and v are in equilibrium. The shorter intervals you use, the closer you will get to the true orbital distance travelled in one second and the closer your total distance travelled to make a complete orbit will get to the circumference of the orbital circle, just like Leszek was talking about using larger and larger polygons.

You can imagine shortening the time bit by bit until you see the motion of the orbit as it happens in reality.

This is closely related to how pi is calculated if I am not mistaken.

15. Originally Posted by MagiMaster
It sounds like he's using Euler's method or something similar. It's known to be a fairly inaccurate approximation of a proper integral, especially for timesteps as large as one second. (You can prove to yourself that this is what's happening because as you take shorter and shorter time steps, the error gets smaller and smaller.)

As DrRocket said, calculus is the appropriate tool for this job.
If you could show me the calculus I would be delighted.

16. Originally Posted by KALSTER
Think of the vectors the other way around. Instead of getting hung up on the resultant vector, think of it in terms of what happens each second of orbit with the velocity and gravitation. For each second the satellite will move a distance x at right angles to the radius of orbit. Then during travelling the distance x, the satellite will fall distance y onto the surface of the orbital circle at another right angle. It will repeat this same motion each time, as g and speed stays constant. The resultant vector you get is the line you get between the point of origin at the beginning of the second and the point where the satellite touches the orbital circle again. Both start and end points fall on the orbital circle if g and v are in equilibrium. The shorter intervals you use, the closer you will get to the true orbital distance travelled in one second and the closer your total distance travelled to make a complete orbit will get to the circumference of the orbital circle, just like Leszek was talking about using larger and larger polygons.

You can imagine shortening the time bit by bit until you see the motion of the orbit as it happens in reality.

This is closely related to how pi is calculated if I am not mistaken.
Vector diagrams apart, everyone knows that falling bodies accelerate i.e. their overall velocity vector increases. It's just that the angle of the acceleration due to gravity for a satellite changes during that fall.

A satellite is in free fall after all and falls at exactly the same rate as the Earth's surface curves away. Newton used this analogy himself.

17. Originally Posted by galexander
Vector diagrams apart, everyone knows that falling bodies accelerate i.e. their overall velocity vector increases. It's just that the angle of the acceleration due to gravity for a satellite changes during that fall.
wrong

And satellites are a clear counterexample. In a circular orbit speed is constant. Further the acceleration vector is radial, of constant magnitude. It is always at a right angle to the velocity vector.

Originally Posted by galexander
A satellite is in free fall after all and falls at exactly the same rate as the Earth's surface curves away. Newton used this analogy himself.
And here you contradict your statement above. but here you are basically correct.

18. @galexander, it's a common misconception that acceleration means an increase in the absolute value of velocity. It doesn't. It only means a change in velocity.

Also all the calculus for Euler's method is in the Wiki page. (Though if you have more specific questions, I can answer them.)

19. Originally Posted by MagiMaster
@galexander, it's a common misconception that acceleration means an increase in the absolute value of velocity. It doesn't. It only means a change in velocity.

Also all the calculus for Euler's method is in the Wiki page. (Though if you have more specific questions, I can answer them.)
Unfortunately I could not see any relevant equation in the Wiki page.

20. You're going to have to ask more specific questions. I'm not sure what it is you're confused about.

21. Originally Posted by MagiMaster
You're going to have to ask more specific questions. I'm not sure what it is you're confused about.
I don't know how I can be more specific.

I can't see the equation in the Wiki page. Which one is it, if any?

http://en.wikipedia.org/wiki/Leonhard_Euler

22. Originally Posted by galexander
Originally Posted by MagiMaster
You're going to have to ask more specific questions. I'm not sure what it is you're confused about.
I don't know how I can be more specific.

I can't see the equation in the Wiki page. Which one is it, if any?

http://en.wikipedia.org/wiki/Leonhard_Euler
chees

http://en.wikipedia.org/wiki/Euler_method

23. Originally Posted by MagiMaster
It sounds like he's using Euler's method or something similar. It's known to be a fairly inaccurate approximation of a proper integral, especially for timesteps as large as one second. (You can prove to yourself that this is what's happening because as you take shorter and shorter time steps, the error gets smaller and smaller.)

And if you can't be more specific than "what calculus," I don't think I'll be able to fit the answer in a forum post.

24. Originally Posted by DrRocket
Originally Posted by galexander
Originally Posted by MagiMaster
You're going to have to ask more specific questions. I'm not sure what it is you're confused about.
I don't know how I can be more specific.

I can't see the equation in the Wiki page. Which one is it, if any?

http://en.wikipedia.org/wiki/Leonhard_Euler
chees

http://en.wikipedia.org/wiki/Euler_method
Fascinating DrRocket.

But how does that relate to the addition of vectors in our satellites in orbit problem?

I see no mention of vectors on the webpage you cite.

25. Originally Posted by galexander
Originally Posted by DrRocket
Originally Posted by galexander
Originally Posted by MagiMaster
You're going to have to ask more specific questions. I'm not sure what it is you're confused about.
I don't know how I can be more specific.

I can't see the equation in the Wiki page. Which one is it, if any?

http://en.wikipedia.org/wiki/Leonhard_Euler
chees

http://en.wikipedia.org/wiki/Euler_method
Fascinating DrRocket.

But how does that relate to the addition of vectors in our satellites in orbit problem?

I see no mention of vectors on the webpage you cite.
I give up.

You need to take real physics class, one that uses calculus.

26. Originally Posted by galexander
Originally Posted by DrRocket
Fascinating DrRocket.

But how does that relate to the addition of vectors in our satellites in orbit problem?

I see no mention of vectors on the webpage you cite.
It may not be mentioned explicitly, but everything on that page can be applied to either scalar or vector differential equations.

27. Originally Posted by MagiMaster
Originally Posted by galexander
Originally Posted by DrRocket
Fascinating DrRocket.

But how does that relate to the addition of vectors in our satellites in orbit problem?

I see no mention of vectors on the webpage you cite.
It may not be mentioned explicitly, but everything on that page can be applied to either scalar or vector differential equations.

And have you fully worked the problem through yourself?

If you have what are the results?

28. Those are still very general questions. I have not applied Euler's method specifically to circular orbits, but I've used it often enough to work through what would happen in my head. You'd gain a little altitude at each step. How much would depend on your step size. In the limit, which is an integral, you get a perfect circle.

Different methods might lose altitude instead (I'd expect backwards Euler to lose altitude), but again, it'd depend on the step size, and go away in the limit.

If you ask some specific questions, I'll try to give specific answers, since I'm actually very familiar with this topic, but these very general questions are hard to answer (without writing a book) and (without writing a book) probably don't teach you much.

29. Originally Posted by MagiMaster
Those are still very general questions. I have not applied Euler's method specifically to circular orbits, but I've used it often enough to work through what would happen in my head. You'd gain a little altitude at each step. How much would depend on your step size. In the limit, which is an integral, you get a perfect circle.

Different methods might lose altitude instead (I'd expect backwards Euler to lose altitude), but again, it'd depend on the step size, and go away in the limit.

If you ask some specific questions, I'll try to give specific answers, since I'm actually very familiar with this topic, but these very general questions are hard to answer (without writing a book) and (without writing a book) probably don't teach you much.
A book is precisely what is needed here.

My earlier recommendation of Prussing and Conway's Orbital Mechanics stands.

30. I don't actually know any good books on numerical integration methods. My professor mainly teaches from his own notes compiled from papers he's read.

31. V in fysics is a change in distance between two bodies (dS/t) ór a movement séén at a perpendicular angle but not necessarily with a change of distance (not a fysical speed then). These two are extremes ; physical or mentally (the classic two : fysical and spirit)

Example closer to the first type of velocity is when something is coming straight towards someone. M/s is the amount of meter the distance between the two decreases then.

For this case V=dS/t as the change of fysical distance related to time experienced.

Someone walking along a river can have the same speed to the river (for fysics) but without any change of distance (except for a point halfway the river if you want, the river is not endless).

When the river would be endless to walk by (for instance the ecquator instead of the river) it,s possible to move relative to it and have a constant distance that,s when the extreme is real. In fact there is no movement at all then relative to the river but only to the water, the trees along the river etc. For fysics and in reality also.

It,s also possible to accelerate or slow down running along a river but the distance is not affected by that it stays the same. Therefor the speed relative to the river (if it was endless) does not change and for fysics there is not an acceleration to "the river" but only relative to specific things and places ; a boat tied to the wall, a tree, a bridge or anything to really determin a dS/t).

In case of the satellite the distance seems to be constant each time it passes over his (or anyones) head except for some loss of speed (energy) and correspondingly altitude but the accelleration in the fysics sense is absent. Same time "fysics teaches" that there is an acceleration. Why does no-one see it then ?

When something would accelerates from the river perpendicular to it while I run alongside it that would result in a change of distance to the river and for me. But not for the orbit. The satellite would accelerate perpendicular to the line of the orbit and same time stay in it

That,s not a question that,s a problem.

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32. Velocity is the change in position (S) with respect to time, and it is a vector quantity, with both magnitude and direction.

Acceleration is the change in velocity with respect to time, and is also a vector quantity, with both magnitude and direction.

A satellite in orbit accelerates towards the body it is orbiting due to gravity, providing a large amount of the energy that keeps it in orbit. It's speed, however, would be roughly constant, which may be the concept you are thinking of. Velocity changes direction, and therefore sees an acceleration, even though it may not be changing magnitude.

33. But still for one coordinate in one direction (altitude) there is no change for position while the accelleration is in that direction. I think it is clear in my post that,s what I meant.
For a 3-d euclidean space on paper and someone carrying that euclidean space with him while moving east-west the ecquator will still be in rest because there is no dS relative to the ecquator.

Do you express time as dt ? It,s been a long time I studied these things and maybe we did this also but t would not stand for time as duration then but for position in time.
In daily life there are two meanings for the word time and they are used in fysics as well. One meaning comes forth in a question like "what time is it" ? Asking for a position in time as "how late is it" and you would not answer "one hour" but "one a clock". The answer could also be "one second past twelve" but not "it,s one second". That,s a strange answer to the question.

The other meaning comes forth with a question as : "how much time did it take to travel this distance" ? Here if it was one hour it doesn,t matter if this was from twelve a clock to one a clock or from one - to two a clock. One hour means duration here and one a clock can be on different days and does not express duration.

34. You can't just make altitude one of your coordinates. That's not Eulcidean, that's spherical. The equations of motion are different in different coordinate systems. (Also, physics almost never deals with "1 o'clock", it almost always deals with "1 hour".)

35. What Galexander means is mostly meant for satellites that are geo-stationary maybe.

There is -seemingly - no angle speed then. With a game of roulette a ball needs an angle speed to at least one given radius (possibly given by an observer at the table) drawn or imagined from the center of the table to have an orbit and stay at a distance. Only the radius of the ball to the center can,t have an anglespeed. without an anglespeed the ball accellerates straight to the center. The roullette table gives a primitive idea of gravity as a well and the ball is a satellite. In this case it would seem that a roullette ball could have no speed on the table and still keep it,s heigth. In case of something on the ecquator with the satellite directly above there is solid earth to compensate by weight. The sattellite is in rest to that point on the ecquator in the sense that there is no change of distance to that point if it keeps it,s altitude.

You can't just make altitude one of your coordinates. That's not Eulcidean, that's spherical. The equations of motion are different in different coordinate systems. (Also, physics almost never deals with "1 o'clock", it almost always deals with "1 hour".)
I suppose my way of formulating made it confusing. I was not taking altitude as a coordinate but as a dimension and coordinates for that dimension don,t change if the altitude is constant.
A 3-d euclidean space can have (and has in most practical cases like architecture) one of the dimensions vertical and two horizontal offcourse.

and one hour is ? M/h an hour is from x o'çlock to (x+ 1) o' clock isn,t it ?
Or from second zero when a stopwatch is started to second one which comes down to the same. That's a distance in time or duration of time timed with a clock. Arcane Mathematician uses the notation dt. So dt is obviously meant as distance between two moments of time. Hence t (without the d on itself) would stand for time as moment of time then or dt would mean difference of difference in time. Hence the question.
Difference of difference of time may seem absurd but it isn,t that absurd. (but to me this is dt then not d(dt) as it should be for arcane mathematician with his notation. These things can be just differences of lokal usage for notation, I live in Holland and studied here and even in Holland different teacher could use different notations in some cases (or same notation meaning different things.

difference in time as duration (d(dt) according to arcane mathematician) is quite common in classic relativity. Someone rapidly throwing balls from a train driving towards me I could count maybe ten balls a second pass by or hit me. I experience only 1/10 second between two balls pass by.

The person on the train could only throw one ball each second and the distance in time beteen two balls for him is only one second then. Related to the different speeds between me and the balls or between the train and the balls. For these two (me - balls and train - balls) there is a dV then but dV is also used for acceleration sometimes if I,m not mistaken.

Difference in speed dV between two moving things for instance two cars approaching each other with a dV=10 this is the same for everyone including the drivers and no matter what reference is taken. It can be an euclidean system connected to one of the cars or a tree along the road. As long as all observers would use that one euclidean system dV between the cars is constant for all observers no matter the motion of an observer as long as they refer to one and the same euclidean system to define space. Offcourse this changes when the observers accellerate or with different altitudes.

36. No, you can't make altitude a dimension (coordinate, same thing, in this case), because the direction of the altitude vector changes over time, even if its length doesn't. You can write the equations of motion in spherical coordinates, and then a circular orbit will have a constant altitude, but the equations for the other coordinates will be different than what they are in Euclidean coordinates.

37. Architects do this daily. Y-axis is mostly vertical where I live thus height. Besides euclidean spaces often have three dimensions so at least one dimension is vertical or it would not fit with the angles would it ?

Offcourse the direction can changes over time relative to the sun or the moon typically because they are in a orbitational relation with the earth. But in this case I thought these satellites are hanging more or less stationary above a fixed point on the ecquator.
That,s the typical problem (without being sure if this is the same as what Galexander meant starting he topic) ; A satellite hanging stationary above a fixed point there is no change of direction seen from that fixed point. Hence there seems to be no orbitational relation between the earth and the satellite.

38. That only works near the surface, where altitude is approximately Euclidean. For a satellite, this is no longer true.

For a geostationary satellite in particular, it's still moving, it just doesn't look like it from the surface, since the observer on the surface is moving at the same speed. You can't do the analysis of such a satellite without taking such things in to account.

39. Originally Posted by MagiMaster
No, you can't make altitude a dimension (coordinate, same thing, in this case), because the direction of the altitude vector changes over time, even if its length doesn't. You can write the equations of motion in spherical coordinates, and then a circular orbit will have a constant altitude, but the equations for the other coordinates will be different than what they are in Euclidean coordinates.
You ca also use a rotating reference frame, but then you get "pseudo forces" because your frame is not inertial. That is why long range ballistics calculations must consider the "coriolis force" -- a frame attached to the surface of the Earth is not inertial.

This whole thread is getting weird. There are perfectly good conventions used in orbital mechanics that can be conveniently adopted. Orbital calculations are made all the time, and are extremely accurate. Why fight it ?

Ghrasp and the OP deserve each other.

40. I,m not interested in what is true for the satellite but for myself.
When something orbits (or better orbitational relation between two) there is time. The sun comes up, goes down and up again... Moon samewise but with it,s own time. These distances are even further. Here there is nothing like that the satellites don,t seem to orbit if they hang above a fixed point on the ecquator. That sort a justifies the question Galaexander came up with.

Because in normal situation there would be sort of a balance between an anglespeed (and kinetic energy) and gravitational energy. Here I can see one line that is the line between satellite and the fixed point it hangs above but what,s the other line ?
The idea that there is a possible speed between two points without a change of distance and without any anglespeed is just weird.

Put two dots on a piece of paper and rotate the paper do the dots move relative to eah other ?

Not asking for a specific point of view here but about the movement between the two dots. There is none or the whole paper would deform or the dots getting out of place on the paper.

41. I didn't say that the satellite moves relative to a ground observer. What I said was that to correctly analyse why the satellite can just 'sit' there overhead, you have to take in to account the fact that the Earth is moving, even if you don't see or feel it. If you do that, and write the equations out properly, all the weirdness goes away and it all makes perfect sense.

BTW, this has very little to do with the OPs question, which was basically mistaking a discretization error for something real.

42. I see the earth moving everyday when the sun comes up east and goes down in the west. If I would look at the satellite I would not see this motion. So why is it referred to as an orbitational relation then ?

I have no idea if the topic starter still feels this as related.

43. Ghrasp, you're failing to realize that you are moving on the earth, as the satellite moves in orbit. The satellite moves at approximately the same rotational speed as the earth, but it is still making one full revolution a day, moving more that 25000 miles.

44. It's orbital because the satellite is orbiting, even if you can't see that. The Earth is spinning, and that means that the reference frame of a person standing on Earth is not an inertial reference frame, which means that we can't simply apply basic physics to the problem. There are more complicated ways of taking such things in to account, but the easy way is to choose an inertial reference frame to do the calculations in.

For most architects, the ground frame is close enough since the observer and any part of the building are at any instant moving in very close to the same direction. This doesn't apply to tall skyscrapers or especially satellites.

45. It feels like a merry go round and you try to explain that the horses move to each other. Between the horses on the merry go round there is no movement. The horses move to the environment and different as they are different horses with different lines to the rotational center and to the environment. But these lines have no anglespeed between each other even if they are different length.

I know the circumference of a cirkle gets bigger if the radius increases that,s not the point I can do these calculations just as well as anyone here. Done such things numerous times. But for the "merry go round" (or a rotating piece of paper) different horses / dots can be at different distance from the rotational center and they really don,t move relative to each other or the wood would not surfive it.

If the merry go round had a hole in it it would be different maybe. Someone can stand in the middle and see the whole thing rotate. That,s the owner of the merry go round then or an assistent. But the earth has no owner and there is no hole. So the rotating paper is a better comparison maybe.

46. It feels like a merry go round and you try to explain that the horses move to each other. Between the horses on the merry go round there is no movement. The horses move to the environment and different as they are different horses with different lines to the rotational center and to the environment. But these lines have no anglespeed between each other even if they are different length.

I know the circumference of a cirkle gets bigger if the radius increases that,s not the point I can do these calculations just as well as anyone here. Done such things numerous times. But for the "merry go round" (or a rotating piece of paper) different horses / dots can be at different distance from the rotational center and they really don,t move relative to each other or the wood would not surfive it.

If the merry go round had a hole in it it would be different maybe. Someone can stand in the middle and see the whole thing rotate. That,s the owner of the merry go round then or an assistent. But the earth has no owner and there is no hole. So the rotating paper is a better comparison maybe.

The Earth is spinning
Relative to what ? To the moon the earth has different spin then to the sun. The sun is not really particular for this I think (that,s why it is intriguing).

47. Why does it feel like I'm talking to a recording on the other side of a brick wall?

Rotating means rotating. While you can't say "X is moving" without saying what it's moving relative to, you can say "X is rotating." The whole point of this second discussion is that rotation is significantly different than linear motion.

BTW, the difference between the satellite and a building isn't radius or circumference, but speed. Also, paper's no better than a merry-go-round at describing an Earth-satellite system, since in both cases there's a rotating medium (the base of the merry-go-round or the paper) that's also rotating. There is no such thing in the Earth-satellite example.

48. Originally Posted by MagiMaster
Why does it feel like I'm talking to a recording on the other side of a brick wall?
Because you are talking to Ghrasp. You would have better luck talking to the wall.

49. Originally Posted by DrRocket
Originally Posted by MagiMaster
Why does it feel like I'm talking to a recording on the other side of a brick wall?
Because you are talking to Ghrasp. You would have better luck talking to the wall.
At least the wall knows more then Ghrasp

50. Funny, but maybe the wall you see is inside you're heads and you see no differences for distances. Then I have no other choices then either point that wall out for you or give up.

For a rotating frisbee the rotational axis is constantly perpendicular to the plane of rotation and the frisbee. It,s a mathematical line with no thickness hence no other lines given as a radius for the axis. All lines connecting points on the frisbee have a constant straight angle to the rotational axis hence no anglespeed for these lines to the axis possible as anglespeed implies change of angle.

So how can different points on the frisbee have different anglespeed to the axis if they have constant angle.

draw a line on the floor with an angle to the rotational axis or noangle but on a distance from it.....Then there is an anglespeed for the lines on the frisbee to that point as it rotates. But the rotational axis is not an axis as a wheel's axis. A wheel is a wheel because it has different parts with anglespeed to each other. All these parts are included in the term "wheel".

It would be great if the rims of my bikes rotated to the hubs also and not only to the axis. With given rotational speed for the hubs I would have a higher rotational speed for the rims...

51. All the points on the frisbee have the same angular velocity, but points farther from the center have greater linear velocity. All the points have the same small angular acceleration (from friction slowing the spin down), but each point has a different linear acceleration, all pointing towards the center of the disk.

I think I've said enough that any lurkers will understand what's going on, even if you don't. I'll probably continue to argue for a while though since I find it mildly entertaining.

52. I prefer keeping it as radialchange between lines, frecquency's and radius. 2 pi * r *f for this as a term (just use other terms thus) is radials *m/s with m a radial. without another line there is no radius. You can,t use the same line where it was to orientate a radius change to or one line would have a change of radius to itself.

a cycle as a change of angle to time has one continuous direction while speed can change during the cycle from positive to negative depending what is determined as positive or negative direction.

Example as the merry go round : A speed for one of the horses will be opposite to a horse on opposit side. But also opposite for the same horse for two parents atopposite sides of the mill. So what,s the direction of the speed then ? If it,s positive for one parent it,s negative for the other. You have two opposite horses with same speed for opposite parents and same horses with different speeds. That,s highly confusing. Once a given line on the floor is determined each horse it at a 180 degree different angle and each horse becomes distinguishable. That can be one line from one or more from each parent to the rotational axis (if each parent knows the position of the other and understands each has it,s own line of reference this is no problem).

rotational speed derived as 2pi rad * r * f is only for specific position in phase in the cycle : 2 ór 4 ór 6 ór 8 , aso. because 2 pi stands for radials not just a number therefor it is rad*m/s as unity's not just m/s and the meters for radius are opposed 90 degree in langle to the circumpherence of the swingmill. This radiuslength does not change during a cycle or phase the radials do.

If the satellite hangs above one spot on the ecquator f=0 hence there seems to be no orbital relation. The layer in space at that height can have a orbital relation but not the satellite.

For the frisbee the center has a speed towards me and a point on the side can move faster towards me but next moment (half cycle later) just as much slower.

Hence for a longer trajektory it seems to move just as fast as the center of the frisbee.

But this is not true either. Simply because the frisbees speed is not zero as the merry go round.

The center travels a straight line towards me befor I catch it and a marking on the edge not. Hence travels a longer trajektory as a wave.

For V=10m/s
cyclefrecquency is 2phi / s the wavelength is 10 m. No idea what the trajektory is then but suppose this can be calculated and also difference to this 10m/s.

But still for the frisbee itself there is no movement within it between parts of it. It,s a solid frisbee not a gyroscope.

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