Jupiter's Orbital Velocity & Equatorial Velocity cancel?

Discussion in 'Pseudoscience' started by nebel, Jul 31, 2017.

1. billvonValued Senior Member

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Complete nonsense.

Do the math. Compare the gravitational attraction that Jupiter exerts on a particle in the upper atmosphere, then compare that to the gravitational attraction that the Sun exerts. See if one is larger than the other.

Tides are not caused by a "lack of energy." They are caused by gravity.
Just because you imagine something does not mean it must be true.
Solar gravitation acts no matter how fast a planet is rotating. And in all cases in our solar system, it is minuscule compared to the planet's gravity.
Incorrect again. "Zero orbital velocity" means nothing to particles in the atmosphere. The Sun's gravity acts on the entire planet, generating a small tidal force in two places - the closest point to the Sun and the farthest point from the Sun. This is true whether or not the planet is rotating. Rotation makes the planet a little more oblate, but again compared to Jupiter's gravity it is insignificant.

The comparison to a tire on a snowy road is just silly.

The above is why this should not be in amateur astronomy. It should be in pseudoscience, since it is not based on science.

3. nebel

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EC: the point,--or vertical floater -- at noon does not move at orbital velocity. It did 2. hours 30 minutes ago at sunset. from head to toe, now,
At noon, it is at standstill in orbit. orbital velocity cancelled out by rotation. 2 1/2 hour later, at sundown, it will be cruising at 10 km/sec again toe to head. look at the sketch please.
At midnight it will go at double orbital speed. Think of the tire moving at twice the speed forward in the fender.
Bsw: I support your notion to rename the "Cesspool" as "bilge", more benign.

Last edited: May 15, 2018

5. nebel

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True, as I corrected in post # 57. By comparison, the solar gravity at Mercury's level would be overwhelming. But as stated, with a strength of only 4% at 5.2 AU, it is less. but I am not reasoning about ratios here. If you are stuck with no orbital speed supporting you, anywhere out from the Sun, inward acceleration will be present. Give us the figures please.
Even if its small, it is there. remember the Sun gravitationally reaches out to infinity.

underlined for emphasis.
Yes, the normal solar/Moon tides are standing waves, in which the planets rotate. But
Gravities' action are modified, cancelled by the orbital velocities (thankfully). By stopping the gas particles in their tracks at noon on the Jupiter and Saturn equators, they are stationary in space (sidereal), deprived of the lift from "centrifugal" forces. Not that they have stopped in their rotational movement, but in space. That is why, this tidal effect (small as it might be, only 4% of Earth's), is energy generated, not purely gravitationally, which would be the case, if the planet would not rotate, as you proposed. and
Jupiter is a very oblate planet with a large girth and a 10 hour day, gravity notwithstanding.
science of the minuscule effects. mouse milk, but not pseudo, nor bilge. imho.

Last edited: May 15, 2018

7. nebel

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These additional, particular predicted energy tides, peaking every ~5 hours on both planets, are occurring, because a given equatorial surface particle goes from normal balanced zero solar gravity effect at sunup (the leading point in the orbit) to zero sidereal velocity,
upward lift toward the sun at noon, (high tide)* and back to normal again at sunset. (the trailing point in orbit).
This is not the normal tidal effect based on the diameter of the planet, giving differing distances from the Sun's centre. (as you correctly observed for non-rotating bodies). no,
This is a proposed additional Energy- caused tide, it happens because of lack of energy of motion in particles with zero sidereal motion at noon.
*It might actually have little to do with the local gravity, which is after all ~ constant from sunup to sundown.
Solar gravity strength from full to zero, cause by acting magnitude of motion energy. Additionally,
It has also to do with the material tided, Earth tides are smaller than ocean tides, by a magnitude, so imagine having the atmosphere extra stirred every 150 minutes.

Last edited: May 15, 2018
8. billvonValued Senior Member

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21,338
Again, nope. An object on Mercury's surface will experience an acceleration of 3.7 meters per second squared towards the planet due to the mass of mercury. The acceleration due to the sun is .04 m/s^2 - about 1% of the gravity of Mercury.
See above for the acceleration for Mercury. I will let you calculate the acceleration for Jupiter due to solar gravity. (It will be much, much smaller.)
No, they are not.
No, they are not. This is the fallacy that the Dean Drive was based on (which didn't work.) Centrifugal forces are exactly the same, even if you choose a frame where the rotation seems to stop.

Try this thought experiment. Let's say you get on a merry-go-round and they turn it on. You will feel some centrifugal force pushing you towards the edge. Now let's say they speed it up until the tangential speed at the edge is 60mph. You will feel a LOT of centrifugal force.

Now let's say you have a car drive by the carousel at 60mph. For a fraction of a second they are stationary with respect to you. Does this mean that you no longer feel any centrifugal force since you are now "stationary in space" for that fraction of a second relative to that car?

Now let's say you put the entire carousel on the back of a truck and move it at 60mph. Then you start it up and spin it at the same speed. Will you feel 2x centrifugal force at one point, and zero at the other?

Now let's say you move the entire carousel to a point near the North pole where the Earth is spinning at 60mph under it. Then you start it up and spin it at the same speed. Will you feel 2x centrifugal force at one point, and zero at the other?

No, it's not. Nothing was generated. Kinetic energy was _maintained._ That's it.
Jupiter is 18% wider at the equator due to its rotation. That would be true whether it were orbiting the Sun or drifting in interstellar space.

9. nebel

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here is where your proposed analogy is skewed from reality. the car has to be driving retrograde ON the carousel, not on a nearby road. If the horsies on the carousel could gallop backward to keep mom watching the kids full time, there would be no centrifugal force. Noontime equatorial particles on Jupiter, Saturn move back to be stationary , sidereal,with the stars, even our own star, not close zipping by @ 60.

thanks for the number. first let me say that the local planetary gravity constant has little to do with the strength of the tides, except perhaps for viscosity due to pressure (Reynold numbers). I read the Solar gravity at IAU is . 0006 of 1 G and that is responsible for 1/3 of our tidal heights (rest is moon gs). So: the solar pull near Jupiter would 4 % of that. Average water tide is a ~ a meter on Earth, ground has less, gases more, but sticking with water, 2/3 caused by the moon here; so~ a foot derived from the sun, so, ~.5 of an inch in water on Jupiter. much higher in the less dense, Helium Hydrogen gas of Jupiter. and:
There is a thermal tide of an expanded atmosphere on the evening side of the earth all the time. locally every 24 hours. Caused by stored daytime insolation energy., not gravity.

10. nebel

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billvon, having slept over your proposal, I like to rescind this above misapplication, hasty response to your analogy, but
your extended trucked amusement park has merit, and I will get back to it. thank you.

11. originHeading towards oblivionValued Senior Member

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I am really sorry if I had anything to do with people addressing this crank. He just craves attention and is beyond redemption. I have no idea why he is permitted to post his junk in the science section.

12. billvonValued Senior Member

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Yes. If there were particles in Jupiter's atmosphere that were not rotating with the the rest of the atmosphere, they would not be affected by centrifugal force.

But they are all rotating with the atmosphere. Just because you choose a reference frame where, to you, it looks like they do not rotate does not change this. Again, from your perspective, that part of the atmosphere might not be rotating, just as the guy in the car doesn't see you moving. That doesn't change the fact that the atmosphere is spinning and thus experiences centrifugal force.

Again completely incorrect. A tide is a ratio of external gravitational influences (i.e. the sun) compared to local gravitational influences. At 1G there's very little tide - just a few meters on the ocean. If the Earth's gravity were .001G (everything else staying the same) the tides would be enormous.

13. nebel

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Of course, if everything stayed the same, but with little mass, commensurate with the weak G, the Moon would have great pull, outweighing the earth. I thought the dominant factor for enormous tides was the proximity of of an external mass. The Earth had enormous tides in the past, and the local gravity has not changed, but the distance to the Moon has, so figure.

14. nebel

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Yes you would
, if it is a circular track like a planetary orbit. If the peripheral speed of the carousel matches the speed of the truck, any horse in radius line with the centre of the track would be at zero speed with respect to the ground for a moment . It would be the ideal point to pass refreshment to the kids, no apparent movement, ideal meshing of hands. Centrifugal forces from the centre of the carousel are constant and maintained, but the forces from t tracking the track are cancelled at "noon" and doubled at "midnight" . because
the free particle, the kid on the horse, is stationary to the ground at the inside perimeter and has double the speed at it flings along the outside.
Constant speed on the truck, but zero and added speed at the longer outside journey. .
It might churn your stomach, and churn the atmosphere of a planet. Besides, it is so easy to pick up energy while you are at a standstill. A win win situation. energy tides. thermal tides, gravity tides working for you.

15. billvonValued Senior Member

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Nope.

Gyroscopes in airplanes flying hundreds of miles an hour work just like they do on the ground. One side doesn't see more force than the other.
Helicopter blades see the same centrifugal forces on them whether the helicopter is moving or stationary. (Aerodynamic forces change of course.)
A person in a spinning habitat in Earth orbit will see the same centrifugal forces even if the habitat is orbiting at 18,000 miles per hour.

These are all facts.
Correct. There would be zero relative motion for an instant. (However, there would be constant acceleration.)
It doesn't matter what frame of reference he is stationary with respect to. He sees exactly the same acceleration towards the center that he always did.
It is no easier or harder to pick up energy when you are at a standstill with respect to any given reference frame.

You are engaging in a common fallacy - that there is a "preferred" reference frame that is "real" - and thus there's a "real" zero velocity that has some meaning. There is not. There is only relative motion.

16. nebel

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Yes! Using the carousel in post #42, at point #41 the horsy moves at truck speed, with respect to the mums waiting to pass energy drink,- bar to kid, it now decelerates from there to zero ground speed at point #7. It is moving slower and slower as it reaches the mum at location #7, which is fixed, sidereal.
From that easy, slow motion, handover, (not just an instant jerk), -- your mentioned acceleration takes slowly over to get the kid on the pony back up to truck speed at #42 during his rotation. That is a delayed, longer process, all along the the inner, sunny, mummy, energy - giving side.
If the truck were on the same turning circle, as the size of the carousel, there would be no centrifugal force at that moment on the recipient child. As it is, the centrifugal force in a 2 (or more) rotation body system sees a proportional reduction if not cancellation. That why counterrotating Gyros, Blades, favoured being so stable.
At the position #8, the kids better hang on, because, as an apparent acceleration (slow at first), in ground speed occurs from position #42 toward #8. At that point the seat on the carousel follows a double curvature, adding the carousel's diameter and the curved road of the truck. At the same time at #8 the speed of the truck and the speed of the Carousel diameter have been slowly but fully added.
I dare you to suggest there is no added precaution to be taken for the carousel passengers to hang on, as the ground races by, being pulled outward at twice the force as the truck driver.
From blurry point #8, the perimeter's groundspeed slowly decelerated back to forward truck speed, same as looking at the driver through his rear window. Kids experiencing a pull forward toward the driver from the carousel rotation, and backward out from the trucks tracks.
While the total time inside and outside between #41 and #42 must be equal from the points of the children, the mums would be out of luck to pass on the need energy. (there is none on the night side anyway).
Sorry, All this Tide - kind of orbital mechanics can be tiring to read. but not gravity related, but energy. energy tides.

17. billvonValued Senior Member

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No decelerating to zero ground speed. The carousel turns and the people on board see an acceleration towards the center, caused by the horse they are sitting on pushing them back to the center (since the horses are attached to the platform.) If they got off and stood on a frictionless surface they would keep going straight, quickly ending up off the carousel.
There is no "fixed, sidereal." There is no one privileged frame where speeds can be called "absolutely zero." This has been proven hundreds of times, over a span of time well over 100 years.
"Acceleration" does not slowly take over. The kid is feeling it all the time, constantly.
There is always centrifugal force on the rotating child. If the truck were turning along with the carousel so as to stay next to the child, the people in the truck would feel it as well.
[quote As it is, the centrifugal force in a 2 (or more) rotation body system sees a proportional reduction if not cancellation. That why counterrotating Gyros, Blades, favoured being so stable.[/quote]
Counterrotating objects are certainly not more stable. Indeed they are LESS stable than the equivalent spinning mass. Spinning masses have a gyroscopic moment which tends to stabilize them in at least one plane; counterrotating masses do not.

This is not a misunderstanding of astronomy. This is a basic misunderstanding of kinetics.

18. nebel

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No, Not on as illustrated by a chair swing carousel, carried on a truck following a circular pass at matching speeds. because
Forces act in a sidereal, combining, cancelling way.
In this case, ignoring time delays and dampening motions,
The truck, carrying an initially at rest,- ring of chained chairs,-- driving around it's quasi- orbit, would have all the chairs swing out to the night, right side, and some inside ones bumping against the central mast. Now,
As the carousel starts to rotate prograde as the truck drives around the ring road, these inner chairs, which were, while not revolving had bumped outward against the central support, now swing free , to the left, inward toward the centre by centrifugal force**, while the outer chairs start to reach out more and more over the outside, the right side of the truck bed.
Once the speeds are synchronized, the inner chairs should linger, one by one, at zero groundspeed over the inner road curb, allowing for a clean, non-shear- panic energy pass, --while the outer chairs at double the sidereal, or groundspeed are fully lifted out, and up, bulging tide- like from the vehicle - swing chair carousel - combo.
Energy tides. cancelled and doubled by prograde rotation, orbit velocity matches, in amusement trucks and planets.
of course the effects will be diminished with vast orbits, 10 hour days. but the effects are there. then add gravity, the other tidal cause.
** not an oxymoron here, in such a system, centrifugal force will drive free particles, chained chairs toward the centre.

Last edited: May 16, 2018
19. billvonValued Senior Member

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You are completely and utterly incorrect. Kinetics does not predict this, nor does any real world example work in this way.

20. Neddy BateValued Senior Member

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Let's say that there is a carousel ride which imparts 1 "gee" of apparent (centrifugal) force on its riders when it is rotating at its full speed. All carousel riders say the direction of this apparent force is always directed away from the center point of the carousel.

Let's also say that there is a truck driving on a circular racetrack which imparts 1 gee of apparent (centrifugal) force on its riders as it drives the circle at its full speed. All truck riders say the direction of this apparent force is always directed away from the center point of the racetrack.

Now place the carousel on the back of the truck, and let them go at their respective full speeds. It seems to me that there would be moments in time when the riders of the carousel would feel a maximum of 2 gees, and there would also be moments in time when the riders of the carousel would feel a minimum of 0 gees. I think this is the cancellation effect that nebel is trying to discuss.

However, it should be noted that the speed of the truck and the speed of the carousel need not be equal in order for both to impart 1 gee in the first place. That would only be for the special case where both circles have the same radius, which would not hold true in general. So in the general case, the moment when the carousel riders feel 0 gees does not have anything to do with them being momentarily at rest with the road.

Similarly, Jupiter's rotational speed at the equator being approximately equal to its orbital speed would not provide the type of cancellation that nebel is hoping for. The different masses of the two celestial bodies and the different distances from the center of each 'circle' would have to be considered from a gravitational standpoint first, and then compared to the apparent (centrifugal) forces involved. My guess would be that the centrifugal forces would be negligible compared to the gravity issues. Consider how much larger the orbital radius is than the radius of Jupiter.

Last edited: May 17, 2018
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21. nebel

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Correct, and that would happen always at a given point during the rotation, but only at that series of points, and a force gradient would exist from there up and down in latitudes towards the poles, and also toward the leading, sunup and trailing, sundown areas. Energy gradients are always interesting.

22. Neddy BateValued Senior Member

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To summarize my example, the radius of the carousel is analogous to the radius of the planet Jupiter, and the radius of the racetrack is analogous to the radius of Jupiter's orbit around the sun, which is many times larger.

In that case, Jupiter's orbital speed (truck speed) would have to be many times larger than Jupiter's rotational speed (carousel speed) in order for them to cancel out periodically. And that simple scenario also neglects the gravity due to the masses of both Jupiter and the Sun.

23. nebel

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Are not both velocities, orbital and rotational nearly equal , within 3 % of 13 km/sec?
The same with Saturn at ~ 10 km/sec?