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Paradoxes and Dilemmas
- tvanflandern
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21 years 11 months ago #4030
by tvanflandern
Replied by tvanflandern on topic Reply from Tom Van Flandern
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>Imagine the rotation of the earth is divided into 4 discreet increments. The impulse on the sun at increment 1 will cause it to accelerate and move towards earth position 1. Then, the gravity force on earth will also increase and the centripetal acceleration. That will cause an increase in orbital speed, according to Kepler's laws. Then, at increment 2, the sun is already closer and get another impulse. It moves again closer to earth and the force increases again, causing the orbital speed to change once more. WE NOW HAVE ANOTHER NEW ORBIT. At position 3, another yet impulse, decreases the orbital radious and speed increases. Now we have ANOTHER ORBIT. At the fourth and last increment, yet another impulse and another reduction in orbital radious with a proportionate increase in orbital speed. I DO NOT SEE ANY CANCELLING MECHANISM.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
If the Sun had no motion relative to the Earth, it would do as you describe, and the two would soon be in contact (igboring that the Earth would vaporize <img src=icon_smile_shock.gif border=0 align=middle>.
But that is not reality. Instead, the Sun has a transverse motion. If it were going fast enough, it would just keep on going forever, and we would be left to freeze. If the Sun's transverse motion were too slow, it would fall toward Earth. But if it is a reasonable amount, we have the normal orbiting case. Let's look at the simplest case where the orbit will be circular.
In December, the Earth is toward Gemini and the Sun is toward Sagittarius (opposite Gemini). The Earth is moving forward toward Virgo, and the Sun is moving very slowly in the opposite direction toward Pisces. This corresponds to your this position 1.
If Earth had no pull on the Sun, the Sun's motion would be a straight line in the sideways direction toward Pisces. But in fact the Sun is forced to fall (accelerate) toward Earth. So the Sun's actual path is not a straight line, but an arc that deviates more and faster from the original straight line at every moment. (That is what an acceleration does.) The result is that, instead of the Sun increasing its distance from Earth faster and faster, as it would on that straight line, it falls into the curved path that keeps it about the same distance from the Earth.
This might be easier to visualize with a cannonball (ignoring the atmosphere). Shoot it too fast and it escapes. Shoot it too slowly and it drops back to the ground. Give it just the right speed and it continually falls along a path that circles the globe forever.
In technical terms, you considered only the linear acceleration. But the force between Sun and Earth affects mainly angular acceleration, and has very little linear acceleration. If Earth's orbit were a perfect circle, there would be no linear acceleration.
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>Can you stand the heat? (Of the sun getting closer?)<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
Given the present temperature outside right now in Washington, DC, I'd welcome more linear acceleration in the Sun's direction. Ironically, we are getting closer to the Sun until January 5 (the date of perihelion). But that's a whole other paradox. <img src=icon_smile_evil.gif border=0 align=middle> -|Tom|-
If the Sun had no motion relative to the Earth, it would do as you describe, and the two would soon be in contact (igboring that the Earth would vaporize <img src=icon_smile_shock.gif border=0 align=middle>.
But that is not reality. Instead, the Sun has a transverse motion. If it were going fast enough, it would just keep on going forever, and we would be left to freeze. If the Sun's transverse motion were too slow, it would fall toward Earth. But if it is a reasonable amount, we have the normal orbiting case. Let's look at the simplest case where the orbit will be circular.
In December, the Earth is toward Gemini and the Sun is toward Sagittarius (opposite Gemini). The Earth is moving forward toward Virgo, and the Sun is moving very slowly in the opposite direction toward Pisces. This corresponds to your this position 1.
If Earth had no pull on the Sun, the Sun's motion would be a straight line in the sideways direction toward Pisces. But in fact the Sun is forced to fall (accelerate) toward Earth. So the Sun's actual path is not a straight line, but an arc that deviates more and faster from the original straight line at every moment. (That is what an acceleration does.) The result is that, instead of the Sun increasing its distance from Earth faster and faster, as it would on that straight line, it falls into the curved path that keeps it about the same distance from the Earth.
This might be easier to visualize with a cannonball (ignoring the atmosphere). Shoot it too fast and it escapes. Shoot it too slowly and it drops back to the ground. Give it just the right speed and it continually falls along a path that circles the globe forever.
In technical terms, you considered only the linear acceleration. But the force between Sun and Earth affects mainly angular acceleration, and has very little linear acceleration. If Earth's orbit were a perfect circle, there would be no linear acceleration.
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>Can you stand the heat? (Of the sun getting closer?)<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
Given the present temperature outside right now in Washington, DC, I'd welcome more linear acceleration in the Sun's direction. Ironically, we are getting closer to the Sun until January 5 (the date of perihelion). But that's a whole other paradox. <img src=icon_smile_evil.gif border=0 align=middle> -|Tom|-
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21 years 11 months ago #4031
by Jim
Replied by Jim on topic Reply from
The issue of the sun being moved is the same as the Earth being moved by the moon and this does not happen in the real world. Only in models do dumb things like occur and Makis is correct in that if the model was anywhere near correct what he says would happen. In the real world without a model the tide is raised on Earth by the moon and by extention the Earth/moon raises a tide on the sun as does all the planets and no doubt is why sun spots have a cycle. The tidal effect is a part of gravity effects.
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21 years 11 months ago #4363
by Jim
Replied by Jim on topic Reply from
For my part the orbit of Earth is not as you describe and the proof fo this is the MEASURED solar constant that can be seen on NASA. The solar constant is only a few watts/m2 different all year. And this FACT can only lead anyone to the conclusion the Earth is always about the same distance from the sun. Not an eccentric orbit but almost round.
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21 years 11 months ago #3951
by tvanflandern
Replied by tvanflandern on topic Reply from Tom Van Flandern
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>You realize that angular acceleration is NOT and were not considered by Newton in his formulations.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
Yes, but I mis-spoke. I did not mean omega-dot. I meant angular momentum, or acceleration of angular motion. The radial acceleration of Earth toward Sun (and Sun toward Earth) would be in distance only if the two bodies moved only along the line between them. But when the bodies have mainly or only transverse motion, that same radial acceleration produces little or no r-dot, instead changing the direction of motion rather than the distance.
With that wording problem corrected, please read my previous response again.
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>if a force law is inverse square in the form of Newton's law, will the trajectory be an ellipse AND STABLE. ... If you have read [Feynman's] proof, what do you think of it?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
Assuming we are talking only about the two-body problem, almost any celestial mechanics book today can prove that the trajectory in an inverse square force field must be a conic section (e.g., an ellipse if the relative velocities are right). My preferred text is Danby's <i>Fundamentals of Celestial Mechanics</i>, but it is heavy on vectors, which not everyone is comfortable with.
I've never read Feynman on celestial mechanics. But I do realize he dabbled in many fields far from his own expertise. He was that kind of genius.
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>And the $42 question is? How is pushing gravity account for this in the INVERSE manner possed by Halley? If you have pushing gravity, will the orbits be STABLE ellipses? Feynman thought no, but after reading his proof, I realised that he lacked an understanding of the problem, God forgive me. Because it is the fact of the elliptical orbit (even with 0.02 e for earth) that saves the day. Do you agree?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
There is no difference at all in the math, orbits, or stability of the two-body problem between pushing and pulling gravity (ignoring the small, special effects in the former).
In the two-body problem, perfect circles are just as stable as other orbits. So I cannot agree with you here.
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>It is not a paradox, it just happens that although the sun is closer the winter, the angle of inclination of earth's orbit is such that sun rays come inclined and reflect away most of their intensity. At the same time, the souther hemosphere get them straight down. In the summer, the sun is further but rays come vertical so intensity absorption is maximum. In the south, the opposite happens. Do you agree with this explanation?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
Yes, I agree with this. You might have added that the length of daylight in summer is longer, whereas the length of night is longer in winter. But the angle of the radiation is the main factor.
BTW, that <i>is</i> a paradox. A paradox is an apparent contradiction that has a resolution. A paradox without a resoilution becomes a logical contradiction. -|Tom|-
Yes, but I mis-spoke. I did not mean omega-dot. I meant angular momentum, or acceleration of angular motion. The radial acceleration of Earth toward Sun (and Sun toward Earth) would be in distance only if the two bodies moved only along the line between them. But when the bodies have mainly or only transverse motion, that same radial acceleration produces little or no r-dot, instead changing the direction of motion rather than the distance.
With that wording problem corrected, please read my previous response again.
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>if a force law is inverse square in the form of Newton's law, will the trajectory be an ellipse AND STABLE. ... If you have read [Feynman's] proof, what do you think of it?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
Assuming we are talking only about the two-body problem, almost any celestial mechanics book today can prove that the trajectory in an inverse square force field must be a conic section (e.g., an ellipse if the relative velocities are right). My preferred text is Danby's <i>Fundamentals of Celestial Mechanics</i>, but it is heavy on vectors, which not everyone is comfortable with.
I've never read Feynman on celestial mechanics. But I do realize he dabbled in many fields far from his own expertise. He was that kind of genius.
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>And the $42 question is? How is pushing gravity account for this in the INVERSE manner possed by Halley? If you have pushing gravity, will the orbits be STABLE ellipses? Feynman thought no, but after reading his proof, I realised that he lacked an understanding of the problem, God forgive me. Because it is the fact of the elliptical orbit (even with 0.02 e for earth) that saves the day. Do you agree?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
There is no difference at all in the math, orbits, or stability of the two-body problem between pushing and pulling gravity (ignoring the small, special effects in the former).
In the two-body problem, perfect circles are just as stable as other orbits. So I cannot agree with you here.
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>It is not a paradox, it just happens that although the sun is closer the winter, the angle of inclination of earth's orbit is such that sun rays come inclined and reflect away most of their intensity. At the same time, the souther hemosphere get them straight down. In the summer, the sun is further but rays come vertical so intensity absorption is maximum. In the south, the opposite happens. Do you agree with this explanation?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
Yes, I agree with this. You might have added that the length of daylight in summer is longer, whereas the length of night is longer in winter. But the angle of the radiation is the main factor.
BTW, that <i>is</i> a paradox. A paradox is an apparent contradiction that has a resolution. A paradox without a resoilution becomes a logical contradiction. -|Tom|-
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21 years 11 months ago #3952
by tvanflandern
Replied by tvanflandern on topic Reply from Tom Van Flandern
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>[Jim]: The issue of the sun being moved is the same as the Earth being moved by the moon and this does not happen in the real world.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
Jim, let's try to stay connected to reality. Direct observations of the Sun and Moon daily with transit circles show that these motions exist and are real. It is wrong for you to claim otherwise.
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>For my part the orbit of Earth is not as you describe and the proof fo this is the MEASURED solar constant that can be seen on NASA. The solar constant is only a few watts/m2 different all year. And this FACT can only lead anyone to the conclusion the Earth is always about the same distance from the sun. Not an eccentric orbit but almost round.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
Again, the problem is with your understanding, not with reality. The "solar constant" is not the measured amount of solar radiation, but rather is the measured amount projected to a constant distance of one astronomical unit. Please avolid making dogmatic claims where questions are more appropriate. -|Tom|-
Jim, let's try to stay connected to reality. Direct observations of the Sun and Moon daily with transit circles show that these motions exist and are real. It is wrong for you to claim otherwise.
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>For my part the orbit of Earth is not as you describe and the proof fo this is the MEASURED solar constant that can be seen on NASA. The solar constant is only a few watts/m2 different all year. And this FACT can only lead anyone to the conclusion the Earth is always about the same distance from the sun. Not an eccentric orbit but almost round.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
Again, the problem is with your understanding, not with reality. The "solar constant" is not the measured amount of solar radiation, but rather is the measured amount projected to a constant distance of one astronomical unit. Please avolid making dogmatic claims where questions are more appropriate. -|Tom|-
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21 years 11 months ago #3989
by Jim
Replied by Jim on topic Reply from
About the solar constant being different than the modified constant posted by NASA- Why is the solar constant posted in a form that is totally faked by corrections that are made so it conforms to the designed orbit od Earth? This is how I see the process you seem to feel is quite OK. The real constant must not be constant at the true distance of Earth so why not post this data as well as the posting in service now? Two questions which I hope are less odious that the dogmatic statements TVF objected to in earlier posting here.
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