THE WASP-18b & Space Probes Mysteries

[The Man]

Yes it is – speed is the same factorm - like´you want...



The speed of the bull is the same.
I am not saying that the angular speed is the same……
This mean each time the bull in orbit radius 4 moves one time round the orbit, it will move 4 times round the orbit, at radius 1.

No one said you did say “angular speed is the same..” but you certainly do seem to be ignoring it when you guess about the forces involved in circular motion.

So how much more force does it require to keep to bull in orbit in orbit / radius 1?

See F₂ in pervious post.

( What is ”radian” ? – 1 orbit)..

A radian is the ratio of the length of a circular arc segment over the radius of that circle. That ratio represents the angle subtending that arc segment in radians. One rotation (or orbit as you might say) equals 2*p radians.

Please see.

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

An angle of 1 radian results in an arc with an equal length to the radius of the circle.

 

[ArcturusA]

So how much more force does it require to keep to bull in orbit in orbit / radius 1?

This has been answered at least twice now.

The equation for centripetal force is:

F = mv^2/r

So, assuming the mass of the bull is the same, and that it keep trying to walk at the same speed perpendicular to the rope (which is pretty unphysical, but this is just a thought experiment), then m and v are constant and the force is proportional to 1/r.

So for a radius 4x smaller, you'd need a force 4x larger.

But before you start getting ideas, that doesn't apply to gravity at all.

 

[jsfisher]

As I thought
Still a poison question ?

Still? It is not now nor was it ever a poison question. It is, however, a poorly constructed bogus question.

Come on man this is bull, cow and cowboy logic, isn’t it ?

You are correct. It is bull. Why do you refuse to provide the missing information?

 

[Giggywig]

I do not undertand that "radian" english word...

Do you understand Danish word radian?

 

[Myriad]

This has been answered at least twice now.

The equation for centripetal force is:

F = mv^2/r

So, assuming the mass of the bull is the same, and that it keep trying to walk at the same speed perpendicular to the rope (which is pretty unphysical, but this is just a thought experiment), then m and v are constant and the force is proportional to 1/r.

So for a radius 4x smaller, you'd need a force 4x larger.

Right. At the same speed, with a radius 4x smaller, there would be 4x as much force (that is, tension in the rope).

Now consider two bulls that are circling with the SAME radius, but one of them is going at twice the SPEED of the other.

Does the faster one need twice as much tension (force) in the rope to keep it going in a circle, because it's going twice as fast? No! The faster one needs 4 times as much tension in the rope to keep it going in a circle.

The right way to explain this is with the mathematics, as has already been explained. But if you're looking for a more "intuitive" explanation of why the faster bull needs four times the force instead of twice the force, consider this. The one that's going twice as fast has twice the momenutm; it's twice as hard to change its path (much the same as if it had twice the mass). By "twice as hard" I mean that in order to cause the same change in momentum (such as a certain change in direction) by applying a force to both bulls for the same amount of time, your cowboy would have to pull twice as hard on the faster bull.

In order to keep either bull going in a circle, the cowboy has to change its direction by 360 degrees for each circle the bull makes. It takes twice as much force to change the faster bull's course in the same amount of time. But the cowboy doesn't have the same amount of time for the faster bull. The faster one completes its circle in half the time. So the cowboy has to pull the faster one twice as hard again, or four times as hard altogether. One factor of 2 comes from having to apply twice the force to get the same effect in the same amount of time, the other factor of two comes from having to get the same effect (a 360 degree revolution around the center) in only half the time.

(Whether that helped or not, I don't know. For most people the math is clearer than such explanations, but sometimes a different way of explaining can be helpful.)

In any case, going back to the two planets with the inner having 1/4 the radius and 2x the speed, the inner one needs 4x as much force for the 4x smaller radius (as the example of the bulls at different radii at the same speeds shows) and 4x as much force for the 2x faster speed (as the examples of the bulls at the same radius at different speeds shows). 4 x 4 = 16, so all told the inner planet needs to experience 16x the force that the outer planet does, for both to stay in circular orbits. And 16x the force is exactly what gravity provides.

Respectfully,
Myriad

 

[Bjarne]

Right. At the same speed, with a radius 4x smaller, there would be 4x as much force (that is, tension in the rope).

Now consider two bulls that are circling with the SAME radius, but one of them is going at twice the SPEED of the other.

Does the faster one need twice as much tension (force) in the rope to keep it going in a circle, because it's going twice as fast? No! The faster one needs 4 times as much tension in the rope to keep it going in a circle.

The right way to explain this is with the mathematics, as has already been explained. But if you're looking for a more "intuitive" explanation of why the faster bull needs four times the force instead of twice the force, consider this. The one that's going twice as fast has twice the momenutm; it's twice as hard to change its path (much the same as if it had twice the mass). By "twice as hard" I mean that in order to cause the same change in momentum (such as a certain change in direction) by applying a force to both bulls for the same amount of time, your cowboy would have to pull twice as hard on the faster bull.

In order to keep either bull going in a circle, the cowboy has to change its direction by 360 degrees for each circle the bull makes. It takes twice as much force to change the faster bull's course in the same amount of time. But the cowboy doesn't have the same amount of time for the faster bull. The faster one completes its circle in half the time. So the cowboy has to pull the faster one twice as hard again, or four times as hard altogether. One factor of 2 comes from having to apply twice the force to get the same effect in the same amount of time, the other factor of two comes from having to get the same effect (a 360 degree revolution around the center) in only half the time.

(Whether that helped or not, I don't know. For most people the math is clearer than such explanations, but sometimes a different way of explaining can be helpful.)

In any case, going back to the two planets with the inner having 1/4 the radius and 2x the speed, the inner one needs 4x as much force for the 4x smaller radius (as the example of the bulls at different radii at the same speeds shows) and 4x as much force for the 2x faster speed (as the examples of the bulls at the same radius at different speeds shows). 4 x 4 = 16, so all told the inner planet needs to experience 16x the force that the outer planet does, for both to stay in circular orbits. And 16x the force is exactly what gravity provides.

Respectfully,
Myriad

That's right,
But still force is tranfered back on the cowboy, - so why not to the Earth's rotation ?

 

[Bjarne]

The answer is:
The approximated energy consumption of the LHC when the particles are moving with top speed is 120 MW relative to the 180 MW.

In post 517 you wrote:
Power consumption of the LHC: "between 220 and 300 megawatts of electricity".

So what do you mean?

How much energy (MW) is an accelerator using when keeping 600 trillion particles in orbits (by constant speed):

2.99999997 E8 m/s (or topspeed)
2.98000000 E8 m/s
2.97000000 E8 m/s
2.80000000 E8 m/s

Please mention the energy consumption in % relative to the top-speed (I believe it is - 2.99999997 m/s) or in MW relative to the numbers of particles

 

[Reality Check]

In post 517 you wrote:
Power consumption of the LHC: "between 220 and 300 megawatts of electricity".

So what do you mean?

I meant that the web page that I linked to stated that the power consumption of the LHC was between 220 and 300 megawatts of electricity.

9 September 2008—The Large Hadron Collider (LHC) at the European Organization for Nuclear Research (CERN), set to start up tomorrow, is the largest physics experiment in history, and it’s probably the most power hungry. Spanning the border between Switzerland and France, the 27-kilometer accelerator ring with its accompaniment of radiation-hardened integrated circuits, feeder accelerators, computers, and supercooled superconducting magnets will, according to varying estimates, draw between 220 and 300 megawatts of electricity—enough to power the city of Geneva twice over. Keeping the power flowing reliably takes a good bit of ingenuity, as a sudden loss of power could mean serious damage to the machine and months of lost work.
Once all of these accelerators are fully operational in 2009, CERN’s estimated annual electricity consumption could approach 1000 gigawatt-hours, IEEE Spectrum learned on a visit to the lab in July. The massive LHC will account for about 60 percent; less than 15 percent of the total will go to mundane functions like keeping the lights on; and the other accelerators in the complex will account for the rest. A big part of the consumption is the hundreds of enormous superconducting magnets, though they draw much less power than equivalent conventional magnets would. The superconductors must be cryogenically cooled to temperatures between 1.8 and 4.5 kelvins (colder than outer space). If the temperature creeps even a fraction of a kelvin above that, the magnets stop working and lose control of the beam. An uncontrolled beam can melt 500 kilograms of copper in an instant, causing serious damage and halting the experiment for months. So it is crucial to keep power flowing into CERN at all times.

(emphasis added)

How much energy (MW) is an accelerator using when keeping 600 trillion particles in orbits (by constant speed):

2.99999997 E8 m/s (or topspeed)
2.98000000 E8 m/s
2.97000000 E8 m/s
2.80000000 E8 m/s

The LHC has never been run at "top speed" (7 TeV per beam).
It has been run at a total of 1.18 TeV for the 2 beams lately.
The first circulations (20 Nov) and collisions (23 Nov)were at 450 GeV per beam.

If you want to know the power consumption for those runs then contact CERN.

Then all you have to do is learn

• The basic mathematics that you hvae not displayed yet.
• The basic physics that you hvae not displayed yet.
• Particle accelerator design and engineering.
• The specific engineering used at the LHC, e.g. the power consumption of each type of magnet used to create the magnetic fields used to keep the protons in orbit.
• etc.

 

[Reality Check]

That's right,
But still force is tranfered back on the cowboy, - so why not to the Earth's rotation ?

Strictly force is never "tranfered".

The bull and the cowboy experience equal and opposite forces. Otherwise the bull will not be at a constant distance from the cowboy.

No force is transferred to the Earth's rotation by the orbit of the Moon.
Tidal forces however have changed the rotation of the Moon so that it is now tidal locked. Tidal forces have also (and are still) changing the Earth's rotation.

 

[Bjarne]

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No force is transferred to the Earth's rotation by the orbit of the Moon.

“Right” it doesn’t “seem” to happen.

But we have (as already mention in this thread) many reasons to believe that this could be the case.

It’s not only a nut’s idea, - if this was the case we also should believe that employers by NASA etc… are nuts too > here

NASA (and everyone else) also knows it is a BIG problem that it doesn’t "seem" that the planets are affected by this or the opposite accelerating force...

Matters requirement of space connect the Moon and the Earth.
Because of between these object space is lacking / stretching/ (bends).

Much point to the assumption that the “mass-attraction connection” also makes it possible to transfer rotation energy, (centrifugal force) to the Moon.
If so the same “connection” must transfer energy back to the Earths rotation, because of the moons orbit, (as already illustrated earlier in this thread here

So far I have demonstrated that it is possible that the very small Moon: Probes, (due to the fast and close orbit), can transfer enough energy to overcome Mars resistance against rotation.

Today I found the energy source that can explain how this is possible.

Last week we where discussing: how it was possible for the force of centripetal and mass attraction (at a certain point) automatically to equalize each other. The two strange forces are able to do nearly exacts the same.

The decelerating force, that affects the Moon because of :

1. Resistance against orbit motion
2. That the Moon pulls the resistance against the Earth rotation

- must off course be affected by an opposite force.

The opposite force is the (centrifugal force) powered from the Earth’s rotation.
These two forces automatically equalize each other, as discussed previously.

The energy source to overcome the total “unified resistance” of both the Earth rotation and the Moons orbit, - comes from ADG…

Notice that there is (off course) no deceleration of the Moon or the Earth, simply because these two forces equalize each other. All what is required for the Earth / Moon system to overcome the lost of energy is a tine little extra amount of gravity (ADG).

The total resistance require (about) 4E-11 m/s stronger ADG (67.5 millionth parts more gravity). This mean that the Moon just need to be 3 meter closer to the Earth,(as we believe it is), - and the energy supply would be there..

The resistance require to keep the Earth in orbit is about 5E-9 m/s.
This means that the Earth must be about 60 kilometres closer to the Sun as we believe it is.

I think you already have guessed that this also mean that the speed of light is wrong.
-------

 

[The Man]

-------


“Right” it doesn’t “seem” to happen.

But we have (as already mention in this thread) many reasons to believe that this could be the case.

It’s not only a nut’s idea, - if this was the case we also should believe that employers by NASA etc… are nuts too > here

NASA (and everyone else) also knows it is a BIG problem that it doesn’t "seem" that the planets are affected by this or the opposite accelerating force...

[qimg]http://www.science27.com/forum/MoonEarthss.jpg[/qimg]
Matters requirement of space connect the Moon and the Earth.
Because of between these object space is lacking / stretching/ (bends).

Much point to the assumption that the “mass-attraction connection” also makes it possible to transfer rotation energy, (centrifugal force) to the Moon.
If so the same “connection” must transfer energy back to the Earths rotation, because of the moons orbit, (as already illustrated earlier in this thread here

So far I have demonstrated that it is possible that the very small Moon: Probes, (due to the fast and close orbit), can transfer enough energy to overcome Mars resistance against rotation.

Today I found the energy source that can explain how this is possible.

Last week we where discussing: how it was possible for the force of centripetal and mass attraction (at a certain point) automatically to equalize each other. The two strange forces are able to do nearly exacts the same.

The decelerating force, that affects the Moon because of :

1. Resistance against orbit motion
2. That the Moon pulls the resistance against the Earth rotation

- must off course be affected by an opposite force.

The opposite force is the (centrifugal force) powered from the Earth’s rotation.
These two forces automatically equalize each other, as discussed previously.

The energy source to overcome the total “unified resistance” of both the Earth rotation and the Moons orbit, - comes from ADG…

Notice that there is (off course) no deceleration of the Moon or the Earth, simply because these two forces equalize each other. All what is required for the Earth / Moon system to overcome the lost of energy is a tine little extra amount of gravity (ADG).

The total resistance require (about) 4E-11 m/s stronger ADG (67.5 millionth parts more gravity). This mean that the Moon just need to be 3 meter closer to the Earth,(as we believe it is), - and the energy supply would be there..

The resistance require to keep the Earth in orbit is about 5E-9 m/s.
This means that the Earth must be about 60 kilometres closer to the Sun as we believe it is.

I think you already have guessed that this also mean that the speed of light is wrong.
-------

Nope, just that your guesses continue to be wrong. Again stop guessing, learn some math and physics, use them.

 

[steenkh]

The opposite force is the (centrifugal force) powered from the Earth’s rotation.

Bjarne, the centrifugal force is not dependent on the rotation of the central body, and it is a fictitious force ... This has been known for the past 200 years, but apparently it was not something that you learned about in school.

 

[Bjarne]

Bjarne, the centrifugal force is not dependent on the rotation of the central body, and it is a fictitious force ... This has been known for the past 200 years, but apparently it was not something that you learned about in school.

I am talking about this one > here

 

[steenkh]

I am talking about this one > here

The centrifugal force is not mentioned in that article.

So you just added your own bovine logic to the article, right?

 

[Reality Check]

I am talking about this one > here

There is no mention of centrifugal force this article.

There is the fact that the flyby anomalies seem to be related to the angle with the axis of the Earth and perhaps rotation:

That is because the angles in the formula are defined with respect to Earth's equator, which is perpendicular to the planet's spin axis. "That suggests to me that the effect is related to the Earth's spin direction, or perhaps to the rotation of the Earth," Anderson told New Scientist.

This has nothing to do with your delusion.

 

[Bjarne]

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The centrifugal force is not mentioned in that article.
So you just added your own bovine logic to the article, right?

That’s right
What Anderson (etc..) have discovered is that there is a mathematical connection to the rotation of the Earth.
And as you wrote, it must be pretty bovine logic that the connection link; “the mass attraction-connection” (M/r^2) + of course the rotation is involved. = (not a centripetal force), but rather something that only can be expressed as ”centrifugal force” .

This is the same as what i meant by ”space wind” – BUT centrifugal-force is a better expression.

Anderson
That is because the angles in the formula are defined with respect to Earth's equator, which is perpendicular to the planet's spin axis. "That suggests to me that the effect is related to the Earth's spin direction, or perhaps to the rotation of the Earth," Anderson told New Scientist.

This has nothing to do with your delusion..

It has a lot to do with what I claim.
But it is only reveals a small fragment of a much larger truth, - that guaranteed also within few years will with bring new light to the holly omniscient fanatic intolerant science, - down under.

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[steenkh]

Bjarne, you do not know what the centrifugal force is, do you?

 

[Dancing David]

I am not sure Bjrane knows what is is.

 

[Bjarne]

Bjarne, you do not know what the centrifugal force is, do you?

This force continues from the planet out in space. Or to avoid confusion call it: space wind..

 

[The Man]

Bjarne, you do not know what the centrifugal force is, do you?

This force continues from the planet out in space. Or to avoid confusion call it: space wind..

How’s that for a confirmation steenkh, and you even provided a link to the wiki article?

Bjarne, I would be more than happy to see you “avoid confusion” for once.