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Orbital Mechanics

Posted by: campbelp2002 - Thu Dec 23, 2004 7:49 pm
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Orbital Mechanics 

Could an object spiral into the Sun?
Of course! 64%  64%  [ 16 ]
No way! 28%  28%  [ 7 ]
I used to think so, but now I don't. 0%  0%  [ 0 ]
I didn't think so before, but now I do. 8%  8%  [ 2 ]
Total votes : 25

Orbital Mechanics 
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Post    Posted on: Thu Jan 06, 2005 3:46 pm
slycker wrote:
Despite the wording of the poll question, my vote is in.

Would this be a better wording:

Is there any initial velocity that can be given to an object for it to spiral into the Sun purely under the influence of the Sun's gravity?

Or this:

If the Earth were slowed in it's orbit slightly, could it then spiral into the Sun purely under the influence of the Sun's gravity?


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Post    Posted on: Thu Jan 06, 2005 9:48 pm
Much better. It does seem to preclude all other possible answers, though, doesn't it? (ie, very self evident...)


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Post    Posted on: Fri Jan 07, 2005 3:27 am
Well, the "any velocity" wording encourages people to point out that some trajectories do hit the sun. Of course they will not be impressed by the fact that none of these trajectories are spiral.

And the "slowed slightly" wording encourages people to say that I am limiting the choices and that velocity changes that are not small could still result in a spiral. That is not true of course. And I don't want to limit choices.

On the whole I just wanted a simple question because it all seems so obvious to me.


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Post    Posted on: Fri Jan 07, 2005 11:45 am
Peter,

please take Garnetstar's issue exactly - he doesn't say that a slight change of velocity necessaryly will cause a spiral, he only says that "the object will eventually crash into the sun".

His second issue is that the shape of the course may be similar to what popularly is called a spiral but it won't be "a true exponential spiral".

He didn't say that the object would go a spiral in mathematical sense. I myself too didn't have in mind a spiral in mathematical sense.

And I never did consider a "spiral" to be the only possible result of a slight change in velocity. There are special conditions that have to be fulfilled to get a "spiral" - if they are not fulfilled there will be an elliptical orbit.

Besides - the mathematician Henri Poincare at the beginning of the twentieth century calculated planetary courses by considering all the planets of our solar system: He got spirals. He worked on the Three-Bodies-Problem which is unsolvable using Newton's mathematics and equations. So Poincare tried another approach. His results are considered to be one of the roots of the Theory of Chaos sometimes.

But we shouldn't discuss that here and now.



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Post    Posted on: Fri Jan 07, 2005 12:37 pm
I am willing to agree that real orbits under the gravity of many bodies will be chaotic. But I don't see anything in your mathematics that takes extra bodies into account. You appear to be only calculating the Sun's gravity based on an assumed circular orbit around it.

(EDIT) You also say "special conditions" are needed for a spiral. I agree. You need atmospheric drag or relativistic effects or constant thrust or some other mechanism to continuously remove energy from the orbiting body. Your calculations do not take any of these, or any other conditions, in to account. You are trying to take an orbit that does not hit the Sun and calculate your way into the Sun purely due to the Sun’s gravity, and that can’t work.

I think maybe one problem is that we do not agree on what is a tangential velocity.
In this diagram of an elliptical orbit I consider all three black vectors as tangential. Am I right is believing that you do not?
Image
In particular do you consider the middle black vector (number 2) as not tangential? Are you decomposing it into the green vector pointing at the Sun and the red vector tangent to the medium size yellow circle? And is that red vector the one you are calling the tangential velocity in all your posts?


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Post    Posted on: Fri Jan 07, 2005 9:07 pm
Ekkehard Augustin on Jan 4 wrote:
That comet has been detected first when it fell into the sun - so nothing has been calculated, no results are available. May be someone at this forum knows calculations and results. Then I would be wrong and I would be interested in the calculations and results.

http://umbra.nascom.nasa.gov/comets/SOH ... azers.html
says “the SOHO spacecraft viewed two sungrazing comets following similar but not identical orbits”.

So the SOHO observations DO contain enough information to determine the orbits of comets that hit the Sun. However the detailed results are not usually made available to the general public because the general public is not interested.

http://cdsaas.u-strasbg.fr:2001/ApJ/jou ... tract.html
says “More than 300 sungrazing comets, most of them discovered with the Solar and Heliospheric Observatory (SOHO) coronagraphs since the beginning of 1996, are known to belong to the Kreutz group or system. Moving about the Sun in similar orbits, they are of indisputably common parentage”

This makes it clear that all comets that are seen by SOHO hitting the sun have well known orbits. You will need to subscribe to the Astrophysical Journal to get the full text. Or maybe you could find it in a library somewhere.


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Post    Posted on: Fri Jan 07, 2005 9:31 pm
Your point, campbelp, is well taken. I will cover my hindquarters by pointing out that I never specified a CIRCULAR orbit. At any given distance from the sun, it is possible to give the projectile a tangential velocity small enough that the perihelion will be within the radius of the sun (or close enough thereto to be vaporized anyway). The path, of course, will be a segment of an ellipse of high eccentricity, not a true spiral.[/i]


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Post    Posted on: Sat Jan 08, 2005 12:41 am
...


Last edited by whonos on Thu Jun 07, 2007 7:05 pm, edited 1 time in total.



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Post    Posted on: Sat Jan 08, 2005 2:08 pm
Peter,

now you seem be misunderstanding me completely.

First - the nearly circular orbit I assume for Earth only which is jsutified by the extreme low excentricity of 0.017% only and difference of 2 million km only between the perihelion distance and the aphelion distance. I don't make any assumptions about the course of the vehicle because there a external forces causing and effecting this course: impulse by space elevator and/or engines and gravitational acceleration.

Second - I don't have any problem to consider your middle black vector to be tangential to the ellipse but please explain which is the orbit in your graphics - the ellipse or the yellow circles? If the yellow circles don't be orbits - what is their meaning?

Thrid - the observations of the comets by SOHO have been made short before they fell into the sun. These data are not as rich as the data concerning other comets observed by men. When a comet is seen first by an earthian telescope it's hundreds of millions of km away and date about their courses are gathered over weeks´or months. This means that a much larger portion of their course is known and based on this a whole ephemeride can be calculated. In this case the results are much closer to reality and they are much more stabel than in the case of the comets observed by SOHO and fallen into the sun - SOHO is observing the sun and not searching for comets and can see a small fraction of the courses of their orbits only.

Fourth - to repeat it once more: I never used the term "spiral" in a mathematical sense.

Fifth - the special condistions I have been speaking of don't have anything to do with friction and the like but they are existing. But wait for the moment when they will be of meaning.

Sixth - a mathematical orbit going through the body of a physical object doesn't be an orbit an object can go: the object will crash into the body and it will be destroeyd - this and only this is under discussion here. So there will be a mathematical orbit - elliptical perhaps - but it would be barred. The object wouldn't orbit. That's the reason why i said earlier that you should look at it from the vehcle and not from the course or the orbit.

Seventh - I really don't include additional bodies and beacuse of this the Three-Bodies-Problem isn't involved. I did mention it only to say that in reality the planets are going spirlas beacuse of physics. But in another sense additional bodies are involved: elevator, Earth, sun, vehicle, particles of propellant to be used later are more than two bodies and they are all effecting the course oof the vehicle.

In one of your posts far above you said that the momentum of Earth has to be removed to throw something into the sun. That was the moment when I said that it can be done another way. I understood your post in the sense of removing the angular velocity directly and then to let the object move to towards the sun - it would fall there. I consder this to mean to remove the tangential component. From my point of view it is possible to increase the vertical component instead - the tangential component would be kept. This way the object will go closer and closer to the sun vertical while it would be moving non-vertical too. Seen by a man's eye the course would be looking very similar to a spiral or a portion of a spiral. This doesn't mean a spiral in mathematical sense.

IWe are discussing a way to throw something into the sun.



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Post    Posted on: Sat Jan 08, 2005 4:17 pm
Hi Ekkehard,

We surely are not understanding each other.

First – The only explanation you have ever given about your calculations was when you divided the Earth's orbit into a large number of seconds and calculated the angle at the sun due to one second of motion along an assumed circle of 1 UA radius.

Second – The reason I think you don't understand what is tangential is due to statements like this:

(From your Dec 10 post - Now lets separate the movement of the object into two components - tangential to the orbit and vertical to the orbit.)
There is never any component vertical to the orbit if you are using tangential to mean the black vectors in my diagram. You can't define the tangential velocity as being along the orbit and then say there is another component that is not along the orbit. You can break the tangential velocity into 2 vectors, neither of which is equal to the tangential velocity, but you can't just add another vector to the motion. Now it is true that ACCELERATION is acting at an angle, but it would be physically wrong to say that the motion has a component that is not tangential.

(and also from your Dec. 10 post - There will be no tangential acceleration if no Fly-By at Venus or mercury occurs and no tangential acceleration rockets are fired and the launch gave vertical acceleration only - there will be vertical acceleration by gravitation of the sun only.)
This makes it sound like you don't believe there is a gravity acceleration component acting on the tangential vector. Since the angle between the black vectors and the Sun is not 90 degrees (except at perihelion and aphelion) then there IS a tangential acceleration due only to gravity at all points of the orbit (except perihelion and aphelion).

The yellow circles are possible circular orbits that I thought you were considering. If I am wrong, just ignore them.

Third – The SOHO scientists themselves have stated in the links I gave you that the orbits are well known based on the SOHO observations. If you refuse to believe them, there is nothing I can do about that.

Fourth – You continue to say that your mathematical calculation will show a spiral but you do not claim your math takes anything other than the Sun's gravity into account. That means you get a spiral using only two body orbital math, which is not possible.

Fifth – You have not said what special conditions you are considering or how your math will account for them.

Sixth – I don't really understand your point here. I have said that a highly eccentric elliptical orbit with perihelion at or below the Sun's surface if the best way to send waste into the Sun, but you continue to disagree.

Seventh – If you don't want to use other bodies to calculate your spirals then don't claim that other bodies are the cause of your spirals.

Yes, I claim angular momentum must be removed. Not all of it, but most of it. However, with a high enough velocity component directly toward the Sun, you COULD hit the sun without removing any angular momentum. But you fail to understand how extremely large such a velocity would have to be. It would have to be many times greater than the Earth's orbital velocity. It would take many times more propellant to use such a path than it would to just remove the required angular momentum. And if you use too small a velocity the object would miss the sun and enter an earth crossing orbit, or even possibly escape the solar system on the far side of the Sun.

You continue to think in terms of just pointing your rocket at the desired target and firing the engine. Such an idea shows a complete lack of understanding of the physics of rotating dynamical systems.


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Post    Posted on: Sat Jan 08, 2005 7:22 pm
Peter,

it seems to me that several questions are required:

Do you agree that I exoplained my calculations too by saying that I will calculate vectors and that I did the initial calculations you are mentioning to get the first of these vectors?
Do you agree that a tangent is a straight line touchiing a circle, an ellipse or any other curved line at only one point?
Do you agree that a movemnt along a circle, an ellipse or any other curved line doesn't go along that tangent except that one point of touch?
Do you agree that to get the motion along the curved line, circle or ellipse a second vector is required?
Do you agree that the sum of the two vectors is avector the end of which is lying on the ellipse, cirved line or circle as well as the touching point where all these vectors are beginning?
Do you agree that if acceleration is distance per square-second and velocity is distance per second following is possible "NEW distance per second = OLD distance per second + distance per square-second * second"? Do you agree that this is an addition of the amount of acceleration to the amount of velocity for only one second? Do you agree that this is allowed beacuse all this is considering one second only?
Do you agree that sun's gravity is causing acceleration in the sense of distance per square-second and that this acceleration can be added to any velocity an object might already have towards the sun? Do you agree that this already given velocity of the object might be zero and that I'm considering it to be zero at the point of touch in the case of Earth? Do you agree that I can add to this velocity the amount of acceleration when considering only one second? Do you agree that for this one second I can consider this to be a second vector that is representing velocity? I myslef agree that I cannot do so if I were considering more than one secoind per step of calculation - but this I don't do so: Remeber - I refuse to calculate hours instead pof each second by each second.
Do you agree that my second vector pointing from the touching point of the tangent towards sun is for the one first second the acceleration by sun's gravity as well as the velocity component got by this acceleration during this one first second?

I will read your links concerning SOHO and I don't doubt the astronomers - there allways is a difference in knowledge about the compelte orbit or course of a comet if there is adifference in the portion of course observed. Joachim Hermman is explaining this in his book "Gesetze des Weltalls" ("Lwas of Space").

Do you agree that I saidrepeatedly now that I'm using the term "spiral" in a non-mathematical sense? Do you agree that I used the term "spiral" first when I not yet had decided to do mathematics? In one post I said that I got a spiral years earlier while calculating something quite different. I got the spiral when I was calculating vectors - I didn't know wether the shape I got were an exponential mathematical spiral. I only saw a spiral shape recognized that I was wrong and used another method than calculating vectors then. I used the term "spiral" in a non-mathematical sense.

The special conditions I will explain later a conditions valid at a special point the course of the vehicle will go to necessaryly. I will explain them later because I first want to move the package there. These special conditions don't have any effect on the course before that point is reached.

Please keep aside my disagreement - it has two reasons at least: 1. I want to see what I get doing my mathematics. 2. I am calculating a movement of a vehicle/package and look at the movement from the vehicle/package - because of this I'm not interested in looking on the movement from outside. I have a vehicle the course of which can be controlled and steered by using the pulsed fusion engine if I want. As long as the package isn't released things like your extremely elliptical orbit doesn't be of any interest. I'm not calculating to prove any advantages.

Do you agree that AFTER going under sun's surface the orbit BEFOR entering that surface is will be invalid?

The seventh point I didn't claim - I simply added an example showing that spirals cuase by gravity are possible. I did it simply beacsue I gave other examples much earlier - forget it if you don't like it. I t doesn't have to do with my calculations.

Concerning the extremely high velocity - the 650 km/s I got when I was experimenting first with step two was leading only 2° off the direct direction to sun - seen from Earth. This direct direction is the direction to the center of the solar disk which is going a qaurter-second of angle to the sides of the center. The pellet-sized microfusionbombs igneted by laser will not cost that much compared to the velocity got by them. There has been serious work on it to make use of it in practice.

Your last two issues are wrong or misundertandings - I'm experimenting simply. You were right if I would use chmeical rockets as is usual during NASA missions, But new engines explicitly provide impulse and thrust during the whole flight to another planet - DS1 was the first example. In this case things are different to chemical rockets beacuse the vehickle is steerable and controllable.

It's fine you stated critics now that are more concrete but answer the questions please to find out the details of the reasons why we don't underatand each other.



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Post    Posted on: Sun Jan 09, 2005 2:30 am
Hi Ekkehard,

I will not use the usual quote boxes but will just paste my answers into your text. I will be P and you will be E

E: Do you agree that I exoplained my calculations too by saying that I will calculate vectors and that I did the initial calculations you are mentioning to get the first of these vectors?

P:Yes

E: Do you agree that a tangent is a straight line touchiing a circle, an ellipse or any other curved line at only one point?

P: Yes

E: Do you agree that a movemnt along a circle, an ellipse or any other curved line doesn't go along that tangent except that one point of touch?

P: No. The instantaneous velocity is exactly the straight line vector. You are thinking of an average velocity over a finite time. More in my answer to the next question.

E: Do you agree that to get the motion along the curved line, circle or ellipse a second vector is required?

P: No. I believe this is your central error. The instantaneous velocity is all complete with no other component. This is the central requirement in physics of motion problems. At one instant in time the speed and direction is whatever it is. Acceleration may change it, but that is not to be considered part of the instantaneous velocity. The instantaneous velocity is tangent to the curved path at only one instant in time, not over some small but finite interval of time. We are simulating the changing velocity with a series of short straight lines. We are not doing calculus where infinitely small changes are all summed up over an infinitely large number of infinitely small times. Our calculation is an approximation, but good enough for our purposes. If you try to add some assumed other velocity to the tangential (or instantaneous) velocity and then also add the acceleration, you have used one too many vectors.

E: Do you agree that the sum of the two vectors is avector the end of which is lying on the ellipse, cirved line or circle as well as the touching point where all these vectors are beginning?

P: Well, I don't agree on which vectors to use, but the velocity change that WOULD occur over 1 second due to the acceleration of gravity can be though of as an extra velocity component in a straight line and constant direction and to be added to the instantaneous velocity to get a new vector that touches the curve at another point. That is 2 vectors, not 3.

E: Do you agree that if acceleration is distance per square-second and velocity is distance per second following is possible "NEW distance per second = OLD distance per second + distance per square-second * second"?

P: Yes. This is exactly what orbit.xls does.

E: Do you agree that this is an addition of the amount of acceleration to the amount of velocity for only one second?

P: No, I think you are referring to your unnecessary 3rd vector here.

Do you agree that this is allowed beacuse all this is considering one second only?
P: Yes.

E: Do you agree that sun's gravity is causing acceleration in the sense of distance per square-second and that this acceleration can be added to any velocity an object might already have towards the sun?

P: Yes. Again, this is exactly what orbit.xls does.

E: Do you agree that this already given velocity of the object might be zero and that I'm considering it to be zero at the point of touch in the case of Earth?

P: If you mean that velocity toward the Sun in a circular orbit is 0, then yes, I agree.

E: Do you agree that I can add to this velocity the amount of acceleration when considering only one second?

P: Yes, if you are speaking of the acceleration of gravity and not some extra vector to stay on an assumed curve. Orbit.xls does this.

E: Do you agree that for this one second I can consider this to be a second vector that is representing velocity?

P: Yes. Again if it is the gravity acceleration and not the 3rd vector. Orbit.xls does this.

E: I myslef agree that I cannot do so if I were considering more than one secoind per step of calculation - but this I don't do so: Remeber - I refuse to calculate hours instead pof each second by each second.

P: I don't agree. One hour or one day is small enough for our needs. Orbit.xls is computing Earth's orbit in only 200 steps, so each step is over a day and a half. And the result is good to within a few percent.

E: Do you agree that my second vector pointing from the touching point of the tangent towards sun is for the one first second the acceleration by sun's gravity as well as the velocity component got by this acceleration during this one first second?

P: Yes. Orbit.xls does this.

E: I will read your links concerning SOHO and I don't doubt the astronomers - there allways is a difference in knowledge about the compelte orbit or course of a comet if there is adifference in the portion of course observed. Joachim Hermman is explaining this in his book "Gesetze des Weltalls" ("Lwas of Space").

P: OK.

E: Do you agree that I saidrepeatedly now that I'm using the term "spiral" in a non-mathematical sense? Do you agree that I used the term "spiral" first when I not yet had decided to do mathematics? In one post I said that I got a spiral years earlier while calculating something quite different. I got the spiral when I was calculating vectors - I didn't know wether the shape I got were an exponential mathematical spiral. I only saw a spiral shape recognized that I was wrong and used another method than calculating vectors then. I used the term "spiral" in a non-mathematical sense.

P: If you got a spiral and that made you realize that you calculation was wrong, then I am glad, because that is what I would think if I got a spiral. I would think I made a mistake somewhere. I do recall you saying that spirals were possible because you had gotten one mathematically before, which seems to be the opposite of what you are saying here.

E: The special conditions I will explain later a conditions valid at a special point the course of the vehicle will go to necessaryly. I will explain them later because I first want to move the package there. These special conditions don't have any effect on the course before that point is reached.

P: OK, but no special conditions are needed. Really.

E: Please keep aside my disagreement - it has two reasons at least: 1. I want to see what I get doing my mathematics. 2. I am calculating a movement of a vehicle/package and look at the movement from the vehicle/package - because of this I'm not interested in looking on the movement from outside. I have a vehicle the course of which can be controlled and steered by using the pulsed fusion engine if I want. As long as the package isn't released things like your extremely elliptical orbit doesn't be of any interest. I'm not calculating to prove any advantages.

P: I am only calculating the total velocity change from leaving Earth until releasing the package. I don't care what kind of engine is used. I have been assuming it is being released fairly near Earth since that seemed to be your original proposal. And I am looking for the lowest energy cost, or lowest velocity change.

E: Do you agree that AFTER going under sun's surface the orbit BEFOR entering that surface is will be invalid?

P: NO NO NO! The part of the orbit above the surface is still valid even another part is below the surface, like your statement that SS1 is really in an elliptical orbit before it enters the atmosphere. The portion of it's course outside the atmosphere could be easily calculated using Kepler's laws or orbit.xls. And we don't care about after it hits. Waste in Sun, job done.

E: The seventh point I didn't claim - I simply added an example showing that spirals cuase by gravity are possible. I did it simply beacsue I gave other examples much earlier - forget it if you don't like it. I t doesn't have to do with my calculations.

P: I will forget it if you do.

E: Concerning the extremely high velocity - the 650 km/s I got when I was experimenting first with step two was leading only 2° off the direct direction to sun - seen from Earth. This direct direction is the direction to the center of the solar disk which is going a qaurter-second of angle to the sides of the center. The pellet-sized microfusionbombs igneted by laser will not cost that much compared to the velocity got by them. There has been serious work on it to make use of it in practice.

P: I got 6,367 when I did that calculation. See below.

E: Your last two issues are wrong or misundertandings - I'm experimenting simply. You were right if I would use chmeical rockets as is usual during NASA missions, But new engines explicitly provide impulse and thrust during the whole flight to another planet - DS1 was the first example. In this case things are different to chemical rockets beacuse the vehickle is steerable and controllable.

P: I am only calculating the path of the package after you drop it. Your original proposal was to fly a short distance from Earth whit minimum velocity change, drop the package and return to Earth. I am only calculating the velocity needed, not how you will create it.

E: It's fine you stated critics now that are more concrete but answer the questions please to find out the details of the reasons why we don't underatand each other.

P: It is harder if we can't speak in person.

This problem is not really as complicated as you are making it seem.

Here is my “bullet at the Sun” calculation. It will neglect all gravity and just use straight line velocity calculations.

The radius of the Sun is 695,000 km. The Earth is orbiting the Sun at 29.5 km/s. 695,000 km / 29.5 km/s = 23,560 seconds. This means we will move sideways a distance equal to the radius of the Sun every 23,560 seconds. If we suddenly gain a velocity straight at the center of the Sun without canceling any of the sideways velocity, we need to get there in less than 23,560 seconds or the sideways velocity will cause us to miss the Sun (remember for this simple example we are neglecting the Sun's gravity). The distance to the Sun is 150,000,000 km and has to be covered in 23,560 seconds so the speed is 150,000,000 / 23,560 = 6,367 km/s. Even if the Sun's gravity could pull in an object missing by 100 solar radii, we would need 64 km/s or three times the Earth's orbital velocity to hit.

OK, so that was just too simple and unrealistic. We need a more precise solution.

Now I will use Kepler's laws.

The problem can be though of as an elliptical orbit with perihelion at 695,000 km and aphelion at 150,000,000 km. (This is the Hohmann transfer orbit. It has been shown mathematically be the German scientist Hohmann as the LOWEST energy path between two orbits. Our starting orbit is that of the Earth around the Sun and our ending orbit is a circular orbit at the Sun's surface. We will not complete the second rocket firing to actually enter the lower orbit, we will just hit the Sun.) 150,000,000 / 695,000 = 216. That means perihelion will be 216 times closer to the Sun than aphelion. Kepler's equal area law can be reduced to a simple distance ratio for the simple case of perihelion and aphelion because triangles drawn from the sun to perihelion and aphelion will have area ratios equal to the the distance ratios. In other words the object needs to be going 216 times slower at aphelion than at perihelion to be in this orbit. Now I don't know either speed, only the ratio between them. But I can assume that the perihelion speed must be slower than escape velocity for an elliptical orbit. So I will use escape velocity of the Sun as an upper limit for the perihelion speed and divide by 216 to get the aphelion speed. You have posted a solar escape velocity of 618 km/s which I will accept. 618 / 216 = 2.9. The elliptical orbit is tangent to Earth's orbit so the speeds can be subtracted to find the velocity change from Earth's orbit needed to enter this orbit. 29.5 – 2.9 = 26.6 km/s. This is much slower than the no gravity value above, but much faster than 7.74 km/s you are proposing. And if we decide that the perihelion could be slower, it just makes the aphelion speed slower and increases the difference with the Earth's orbital speed, increasing the velocity needed to get to the Sun.

SO... if you don't cancel the Earth's orbital velocity at all, you have some variation of the no gravity example. Basically you need to go REALLY fast to get to the sun before the sideways speed makes you miss. And if you do cancel the Earth's orbital velocity, then the Kepler's laws example tells how much must be removed.

It is so simple!


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Post    Posted on: Sun Jan 09, 2005 4:39 pm
Hello, Peter,

it seems that I was right to post the list of questions because I got "No"s from you that are marking differences in our concepts of thinking and in our definitions.

1. "No":

"E: Do you agree that a movemnt along a circle, an ellipse or any other curved line doesn't go along that tangent except that one point of touch?

P: No. The instantaneous velocity is exactly the straight line vector. You are thinking of an average velocity over a finite time. More in my answer to the next question. "

I don't know why you are thinking this were an average velocity. We have several informations about the movement of Earth: 1. It's moving by 29.8 km/s along its nearly circular orbit; 2. It's accelerated by a number of km/s^2 I don't have in mind this moment - I will have to look into one of my books. The number are not of meaning this moment.

What's of meaning is the fact that we do know as velocity a distance per second only. One second isn't an infinite short moment and this means that a distance per second isn't an instantaneously velocity. It has been measured - it's not the result of differential calculation (infinitesimal calculation). As well the known acceleration isn't an instantaneous acceleration but the average acceleration during one second. For this reason both the known velocity of Earth and its known acceleration by sun's gravity should be considered to be averages.

But one second is an extremely short moment compared to all the seconds of one year - one second is a close approximation to an infinite short moment. Consequently the known velocity and acceleration are close approximations to instantaneous velocty and acceleration.

Because of the only approximative but really average nature of the known numbers I can take the known velocity of Earth as the result of two vectors. One vector is the gravitational vector pointing to the sun. I don't want to repeat what we discussed earlier in this thread but to get the known velocity by using this gravitational vector there necessaryly must be another vector too. This vector is a tangent to the orbit - according to Joachim Herrmann who in turn calls Newton as his witness in explaining the orbit of the moon around Earth. Joachim Herrmann too is talking about the centrifugal force as the counterpart to the gravitational pull towards the sun - centripetal force? - but he says explicitly that the centrifugal force only is preventing us from falling to the sun and too explicitly he is using the tangent by calling it the line of insertion into a course passing the sun. He says that if we would insert the Earth into a course rectangular to the direction to the sun we would get an exact circular orbit of Earth around the sun.

This is one of the concepts of thinking I am using here. One second, distance gone during that second and acceleration got during that second are averages allowing me to do as I'm doing but extremely close approximations to infinite numbers preventing me from getting to large deviations from reality.

2. "No"

"E: Do you agree that to get the motion along the curved line, circle or ellipse a second vector is required?

P: No. I believe this is your central error. The instantaneous velocity is all complete with no other component. This is the central requirement in physics of motion problems. At one instant in time the speed and direction is whatever it is. Acceleration may change it, but that is not to be considered part of the instantaneous velocity. The instantaneous velocity is tangent to the curved path at only one instant in time, not over some small but finite interval of time. We are simulating the changing velocity with a series of short straight lines. We are not doing calculus where infinitely small changes are all summed up over an infinitely large number of infinitely small times. Our calculation is an approximation, but good enough for our purposes. If you try to add some assumed other velocity to the tangential (or instantaneous) velocity and then also add the acceleration, you have used one too many vectors. "

This already has been answered under the first "No" above but it seems that I should say something here. Up to now it's not clear yet that I'm in error absolutely - I may be in error if you use your concept of thinking as the context - but necessaryly your concept isn't the context of my thoughts in fact.

I already said that I don't consider the known velocity and acceleration of Earth as instantaneous ones. "Instantaneous" and infinitesimal do exist only in theory and in mathematics but not in reality - this has been proven by experiment during the last few years. As long as you use the formulars of Newtonian physics you are right - but I myself don't use these formulars here. Perhaps you argue now that NASA is using them and the engineers are using them and you are right in this - but the success they get by doing so simply means that these formulars and "infinitesimals" are extremely excellent approximations to nature. And they do save much time to use them - but it only means that theory and its formulars are extremely good approximations to reality... Please remember - these are explanations of my concept of thinking in this thread.

Your "instantaneous" or infinitesimal numbers or vectors still can be got as a result of vectors similar to my vectors - my vectors can be made subjects of infinitesimal calculations. They would be reduced more and more then down to infitesimal small lenghts. It would be possible then to calculate your "instantaneous" vector out of them.

3. "No"

"E: Do you agree that the sum of the two vectors is avector the end of which is lying on the ellipse, cirved line or circle as well as the touching point where all these vectors are beginning?

P: Well, I don't agree on which vectors to use, but the velocity change that WOULD occur over 1 second due to the acceleration of gravity can be though of as an extra velocity component in a straight line and constant direction and to be added to the instantaneous velocity to get a new vector that touches the curve at another point. That is 2 vectors, not 3."

I don't know wether you didn't read my earlier explanations, don't remember htem right now or something else but I repeatedly said that I identified TWO INDEPENDENT vectors and ONE RESULTING DEPENDENT vector. In sum these are three vectors really. It seems this is a repetition of another post of mine whih I wrote while thinking about orbit.xls - the concept of orbit.xls is different from my concept of thinking. For this reason it's NOT POSSIBLE to use the concept of orbit.xls to understand my thinking. One of the two independent vectors is the vector of the velocity Earth can be thought to be inserted rectangular to the direction to sun by an "Gedanken-Experiment" -the tangential velocity component - and the other the velocity caused by the gravitational acceleration to sun - the verticla velocity component.

4. "No"

"E: Do you agree that this is an addition of the amount of acceleration to the amount of velocity for only one second?

P: No, I think you are referring to your unnecessary 3rd vector here. "

Not that good answer - you just say what you are supposing about my doing but you don't explain your "No" by arguments.

You agreed that for one second the acceleration per second multiplied by second could be added to the old distance per second to get the new distance per second. This means to calculate a new velocity by adding the acceleration to the old velocity - valid in that one second. May be there is a misunderstanding because I said "for only one second" - I wrote these words to clarify that I'm doing that second by second and not for two or more seconds together.

It doesn't have anything to do with a third vector - I only get the result of acceleration this way. If I choose one point of a course and consider it as an insertion into course rectangular to the direction of sun at the beginning of the first second there is no velocity towards sun. But sun is accelerating Earth towards sun during that second. At the end of that first second Earth has velocity towards sun - necessaryly because there has been acceleration during that second. The fact that Earth isn't moving to the sun but going along an orbit is due to the component got by insertion. Please don't argue that Earth hasn't been inserted - the dust and gases Earth is made of have been inserted when our system began to form and evolve. All movements in our system are steming from that time 4.6 billion years ago.

5. "No"

"E: Do you agree that AFTER going under sun's surface the orbit BEFOR entering that surface is will be invalid?

P: NO NO NO! The part of the orbit above the surface is still valid even another part is below the surface, like your statement that SS1 is really in an elliptical orbit before it enters the atmosphere. The portion of it's course outside the atmosphere could be easily calculated using Kepler's laws or orbit.xls. And we don't care about after it hits. Waste in Sun, job done. "

As long as an object is ABOVE or just AT sun's surface the total mass of sun is concentrated within the gravitational center and your elliptical orbit is valid. But just that moment the object has gone UNDER the surface the TOTAL mass of sun isn't concentrated in that center any longer - it now begins to be around the obejct more and more. A growing amount of gravitation behind the object will start to decelerate the object now. Other portions of the total mass wil be at the sides of the object having gravitational forces on it. The gravitationa force in front of the object will be reduced mnore and more causing gravitational acceleration to be reduced. Your orbit isn't valid inside the sun - it's valid above and at the surface only.

It seems you did misunderstand me - I said all this to show that the object never will have the chance to terminate your orbit only once even if will be kept intact and move through the whole body of sun.

1. Remark

"E: I myslef agree that I cannot do so if I were considering more than one secoind per step of calculation - but this I don't do so: Remeber - I refuse to calculate hours instead pof each second by each second.

P: I don't agree. One hour or one day is small enough for our needs. Orbit.xls is computing Earth's orbit in only 200 steps, so each step is over a day and a half. And the result is good to within a few percent."

What I'm doing here for one second only isn't allowed for one hour - look at the gravitational acceleration by sun during one second. I accnot multiply it simply by 3600 because it's changing during these 3600 seconds permanently and by changing numbers. I use all these seconds not to calculate Earth's orbit but the course of the vehicle I launched. It's approaching sun and close to sun the gravitational acceleration is significantly larger which have to take into account. To use hours or days instead of seconds would cause a relatively large error. The approximation I explained earlier in this post will be lost - with awful results. Remeber - the gravitational acceleration is growing by square.

2. Remark

"E: Do you agree that I saidrepeatedly now that I'm using the term "spiral" in a non-mathematical sense? Do you agree that I used the term "spiral" first when I not yet had decided to do mathematics? In one post I said that I got a spiral years earlier while calculating something quite different. I got the spiral when I was calculating vectors - I didn't know wether the shape I got were an exponential mathematical spiral. I only saw a spiral shape recognized that I was wrong and used another method than calculating vectors then. I used the term "spiral" in a non-mathematical sense.

P: If you got a spiral and that made you realize that you calculation was wrong, then I am glad, because that is what I would think if I got a spiral. I would think I made a mistake somewhere. I do recall you saying that spirals were possible because you had gotten one mathematically before, which seems to be the opposite of what you are saying here. "

When I got the spiral I had been calculating vectors too - it wa not an error in calculation and not an error in physics - the error was the interpretation of the vectors. They were valid per second but I did the calculation graphical by drawing them on a piece of paper. To do so forced me got overscale their lenghts. I knew what the result had to be but I got a different result. So I recognized that there was an error but didn't clarify it - I changed form vector calculation to another method.

To repeat it - I do see spiral now too but not in the sense you are understanding the term by. The spiral I got by the detected error mathematically may have been something different instead of a spiral. But beacuse I#m calculating vectors I never can say wether the result is fulfilling the equation of an ellipse, of a circle, of a parabola, of a hyperbola or of a spiral. I just get something reminding me to the shape of a spiral. The term isn't worth that discussion - I'm wondering that you are so urgently discussing this point.

In your calculation you seem to have overseen what I have done - or I missed to explain it. When I experimented first in my Step Two I ignited the pulsed fusion engine which accelerated the vehicle by 1 g per second. I kept it working for several hours calculated the distance after each second got by the permanently increasing velocity and calculated the gravitational acceleration seperately - second by second. The acceleration of 1 g per second by the pulsed fusion engine I'm missing in your calculation. That's the reason why you get quite another result than me.

I will return to the release of the package after termination of my calculations -I don't remember to have said that the package will be released near Earth this moment but I have to read my post again. The special conditions I mentioned will occur by nature - I don't introduce them artificially.



Dipl.-Volkswirt (bdvb) Augustin (Political Economist)


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Post    Posted on: Mon Jan 10, 2005 12:00 pm
I tried to read the bothe links now you listed earlier but only the first one has been successful.

And now I'm but to recognize that two objects have different courses or orbits doesn't mean that the courses or orbits themselves are recognized and known completely. The reason is that to know one orbit or course completely requires data about much more elements of the course than to identify that two objects are going different orbits.

I have to wait until ´the second link is successfull to say more about it.

Some additional words concerning your last issues. I know very good what Kepler's laws say - but they don't forbid my concept of thinking. What I'm experimenting on isn't a proposal - it is a way to simplify the experiment. In my concept of thinking to cancel Earth's orbital velocity means to reduce the tangential component - but I want to see the results of increasing the vertical component as an alternative. I can artificially increase ( or decrease) that vertical component whenever I want by activating the pulsed fusion engine or other engines that might be available at the vehicle. Such engines are not considered by Kepler's laws. While they are working the resulting course cannot be explained by Kepler's laws - this only is possible while no engines are working.

Another reason not to use Kepler's laws is to get the numbers of velocity, acceleration etc. for each second directly.



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Post    Posted on: Mon Jan 10, 2005 1:36 pm
Hi Ekkehard,

You have posted many points I would like to discuss, but the one that is most important is the instantaneous velocity question. The most important thing you need to understand is that instantaneous velocities are real physical quantities, not just mathematically convenient tools.

Here is a link I found that may help:
http://library.thinkquest.org/10796/ch2/ch2.htm

When I say a car is traveling at 60 km/h, that is it’s real velocity, even if it only travels at that speed for 10 seconds. If you get in your car, start the engine, accelerate to 60 km/h, drive to a place 1 km away and then stop, you are only going 60 km/h for one minute of the 60 minute hour, but during that one minute your real velocity is 60 km/h. Your average speed is only 1 km/h because you only covered a distance of 1 km in the one hour.

Suppose I start a clock at the same moment I fire a bullet at 1000 m/s at a target 10 meters away. It travels for 1/100 of a second at 1000 m/s then stops for the other 99/100 second. After 1 second I stop my clock and compute the average speed as 10 meters in one second. The instantaneous velocity was 1000 m/s while the bullet was moving, but the average speed over the time interval being considered was only 10 m/s.

If fact both these examples show the instantaneous velocity is actually the REAL PHYSICAL velocity and the average speed is only a mathematical idea used in some types of calculations. So if you want to know how long it takes to drive to work, you need to know the average speed, including stops at traffic lights and slow driving in traffic. But if you want to tell your friends how fast your car can go, you tell them it’s top speed.

Now there is nothing physically special about any given choice of units for representing speed. It can be meters per second or miles per hour or furlongs per fortnight. All these units of measure can use different numbers to represent the same speed.

(EDIT) So there is nothing physically significant about 1 second. A speed of 10 meters per second only means that IF the object moved at constant velocity for one second then it would move 10 meters. It is NOT meant to imply the motion will continue for the full second.


In the case of curving motion, the example whirling of a rock on a string around your body is often used. The string constantly changes the direction of the rock, but it’s instantaneous velocity is always a straight line tangent to the circle. If you suddenly let go of the string, the rock flies away in a straight line in exactly the direction of the instantaneous velocity at the moment you released it. And the direction is a line tangent to the circle formerly described by the length of the string.

Our package of waste, or even our space craft, is not intelligent. It only tries to move in a straight line at a constant velocity represented by the current instantaneous velocity. It also reacts to forces (either gravity or rocket thrust) according to Newton’s F=MA law. It is only you, the intelligent observer, that sees an ellipse and then tries to impose extra rules on the motion to make it stay on the ellipse. You are thinking too much about the problem and not letting the physical laws dictate the path.


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