THE WASP-18b & Space Probes Mysteries

jsfisher said:
Then why did you state something to the contrary? If you K N O W the speeds must be different, why claim they are the same?.
Click to expand...

I DO NOT
I am ONLY asking about the FORCE involved...


It isn't so. It cannot be so. You already K N O W that, right?
Click to expand...
We are still not talking about what can be or not
ONLY about FORCE

No, it isn't. You have manufactured an impossible situation involving speed, distance, and F O R C E.
Click to expand...
I have NOT only asking a question about FORCE, - a poison one nobody likes to answer.
 
Reality Check said:
P.S. There is no linear relationship between the power consumption of the LHC and the speed of the particles. There is a little thing called special relativity that you have forgotten.
Also much of the power is used to create magnetic fields. The power consuption of the magnets is unlikely to be linearly related to the speed of the particles.
Click to expand...

You don't even need to include any of that stuff to see the problem. If you ignore all the incidental power consumption and assume all the power is due to keeping the particles in circulation, and if you ignore relativity entirely, it still isn't linear. The biggest cause of power loss from a beam of charged particles in a circulator accelerator is synchrotron radiation. The power loss scales with the 4th power of energy. If anyone expects linear relationships here, they're going to be sorely disappointed.

Bjarne said:
I K N O W that...

We are O N L Y - discussion the amount of FORCE involved if this was so.
Click to expand...

If you actually knew that then you would understand that your question is utterly nonsensical and stop asking it. The speeds are not, and can not, be the same. Therefore asking what would happen if they were the same makes as much sense as asking what would happen if the Moon were actually a pink rabbit called Fluffy.

Let say I am living at a moon in Andromeda and you on Mars, or what so ever.
Click to expand...

I suspect that that's a rather accurate description of where you are living, mentally at least.

This is a attempt to stepwise show that something is wrong
Click to expand...

There is no need to be stepwise about it. You have been shown to be wrong about every single thing you have said since you came here.

This is a pure F O R C E question…..
Click to expand...

Yet you still can't even understand what the units of force are. Perhaps you should consider learning that, and the rest of the remedial physics and maths courses you still need to take, before asking any more questions about it.
 
Last edited:
Bjarne said:
I DO NOT
Click to expand...

You do not what? Are you saying you do not know the speeds cannot possibly be the same, or are you denying you said they were?

If the former, well then now you know. If the latter, let's do a fact check, shall we?

Bjarne said:
Consider two objects orbiting the two orbits (inner and outer) shown above, both travelling with the same speed and mass.
Click to expand...


Why look, there it is right there, still bold, still underlined. So, yes you did state the speeds were the same, which is an impossible condition.

(I do notice you edited the weight thing -- well done!)


Why not now rephrase you question using a situation that would actually be possible under the laws of Physics. So far, you haven't.
 
Bjarne said:
You are NOT answering the question. Forget everything about orbit and only answer the hypothetical question below [/b]


Consider two objects orbiting the two orbits (inner and outer) shown above, both travelling with the same speed and mass.
Click to expand...

If they have the same speed, they can't both be in orbit.


I mean is it not logical that it requires 4 times larger force to keep the inner object in orbit compared to the outer, simply because the inner would accelerate inwards 4 times faster, when the force is 4 times larger..
If this is wrong, what is then the correct answer?
Try now to stick to that question.
Click to expand...

I HAVE answered it. You just don't seem to want to understand it.

Let's assume the outer object is in a stable orbit. Then the inner object, if it moves with the same speed, is NOT in a stable orbit. Since it is under 4 times the gravitational force, it will spiral inwards. Half of the lacking force is compensated by the increased centripetal force due to the reduced radius, but the rest has to be compensated by a higher speed.

YOUR problem is that you try to take only the speed into account.

(Answer #21) :rolleyes:

Hans
 
Bjarne said:
This is a pure F O R C E question…..
Click to expand...

That is silly, since several other factors are involved. It is like you were asking how the left front leg can hold a whole cow up, and we try to tell you about the other 3 legs, but you insist that you will only speak about the left front leg.

The F O R C E is produced by speed and angular displacement.

Hans
 
OK fellows.
I can see my question was so poison; nobody really wanted to touch it.
coworbit.jpg


Let's then ask the same question, - but this time let’s only replace the orbiting object with a orbiting bull.

Like an astronomic body, - also the bull only wants to move straight ahead.
But the Cowboy doesn’t allow the bull to do so. (See the image below.)

g1.jpg


How much stronge relative force would it take to keep the bull in an orbits / radius 4 times smaller.
The bull (mass) and the speed of the bull is the same.
 
Bjarne said:
OK fellows.
I can see my question was so poison; nobody really wanted to touch it.
Click to expand...

No, you question was not poisonous. It was an ignorant hypothetical of an impossible condition.

Let's then ask the same question, - but this time let’s only replace the orbiting object with a orbiting bull.
Click to expand...

First off, you do realize that a bull is not a cow, right? I don't want you shifting analogies improperly.

That aside, no, it is not the same question. In the orbiting moon case, the force, mass, and distance relationships are well-established by the laws of gravity. For your lassoed bull and cowboy situation, there is no telling just how much the bull will dig in (or not) to resist the pull imposed by the rope.

The two questions are completely different, and you have not provided sufficient information or constraints for this new question to be in any way meaningful.
 
Bjarne said:
I still do not see how you know or have calculated this.
The link also not explain how you have come that result.

Anyway lets start the attack.
If you compared to the to speed (2.99999997 m/s) and the energy consumption (220-300 MW) – it shows that if you decreases the speed about 1 % the energy consumption decreased about 50 % . – How do you explain that?




And that’s in fact a problem..


Off course, - but the problem is that the force used, is two times that force it seems to be necessary! - WHY?
It’s not enough to know how to calculate the force required, - the point is what happens?
Why is the speed only the haft of that it should have been?


[qimg]http://www.science27.com/forum/g2.jpg[/qimg]​

Consider two objects orbiting the two orbits (inner and outer) shown above, both travelling with the same speed (and weight).

Radius factor is 1 for the inner and factor 4 for the outer object.
How big force (factor) would be necessary to keep these two objects in these orbit.

I mean is it not logical that it requires 4 times larger force to keep the inner object in orbit compared to the outer, simply because the inner would accelerate inwards 4 times faster, when the force is 4 times larger..
If this is wrong, what is then the correct answer?
Try now to stick to that question.

That’s a poison question right?

Don’t deflect the discussion with personal attack and equations that in fact only confirms that when you compare these to the point above, you’ll easy see something is wrong…
Now just stick to the simple question above. This is what this is about ….

How much force does it require to keep the inner object in that orbit, compared to the force it requires to kept the outer object in orbit..? – And please explain why you conclusion is like it is.
Click to expand...

The calculated force "required" is the force “necessary” for rotational motion. It is all right there in that F = MV2/R relationship. That your guesses at what “it should have been” are simply wrong do not change the force that is required and necessary for rotational motion.
 
Bjarne said:
How much stronge relative force would it take to keep the bull in an orbits / radius 4 times smaller.
The bull (mass) and the speed of the bull is the same.
Click to expand...
If the bull/cowboy follows the same laws as the moon/planet system, you are still asking for an impossible situation: the bull would not be in an orbit, if the speed is the same.
 
Bjarne said:
You are NOT answering the question. Forget everything about orbit and only answer the hypothetical question below [/B]

[qimg]http://www.science27.com/forum/g2.jpg[/qimg]​

Consider two objects orbiting the two orbits (inner and outer) shown above, both travelling with the same speed and mass.

Radius factor is:

  • Factor 1 for the inner
  • Factor 4 for the outer object.

I mean is it not logical that it requires 4 times larger force to keep the inner object in orbit compared to the outer, simply because the inner would accelerate inwards 4 times faster, when the force is 4 times larger..
If this is wrong, what is then the correct answer?
Try now to stick to that question.

Don’t deflect the discussion with personal attack and equations that in fact only confirms that when you compare these to the point above, you’ll easy see something is wrong…
Now just stick to the simple question below. This is what this is about….

How much force does it require to keep the inner object in that orbit, compared to the force it requires to kept the outer object in orbit..? – And please explain why you conclusion is like it is.

That’s a poison question right?
Click to expand...


I specifically answered your question before, using the values you provided in your diagram, as others have.


http://www.internationalskeptics.com/forums/showthread.php?postid=5416769#post5416769


Again the reason “you conclusion is like it is” is simply circular motion.


Obviously that point is not getting through to you yet.
 
For entertainment value only

LAMPOON said:
I K N O W that...

We are O N L Y - discussion the amount of MONKEYS involved if this was so.

Let say I am living at a moon in Andromeda and you on Mars, or what so ever.

This is a attempt to stepwise show that something is funny

This is a pure MONKEY question…..
Click to expand...
 
Bjarne said:
OK fellows.
I can see my question was so poison; nobody really wanted to touch it.
[qimg]http://www.science27.com/forum/coworbit.jpg[/qimg]

Let's then ask the same question, - but this time let’s only replace the orbiting object with a orbiting bull.

Like an astronomic body, - also the bull only wants to move straight ahead.
But the Cowboy doesn’t allow the bull to do so. (See the image below.)

[qimg]http://www.science27.com/forum/g1.jpg[/qimg]

How much stronge relative force would it take to keep the bull in an orbits / radius 4 times smaller.
The bull (mass) and the speed of the bull is the same.
Click to expand...

Just considering the aspects of circular motion (ignoring gravity) we can calculate the force on the rope when the bull is at a different radius with the same tangential speed.



w represents angular speed in radian per second

V1 = 1 Meter / Second
R1 = 4 Meter
M1=1 kg
Thus
F1 = V21M1/R1= 0.25 Newton
w1 = V1/R1 = 0.25 Radian / Second

V2 = 1 Meter / Second
R2 = 1 Meter
M2 = 1 kg

Thus
F2 = V22M2/R2= 1 Newton
w2 = V2/R2 = 1 Radian / Second


So

V2 / V1 = 1/1 = 1
w2 / w1 = 1/0.25 = 4
F2 / F1 = 1/0.25 = 4

Since mass is the same in both cases and F = MV2/R as well as w = V/R we can rewrite the F2 / F1 proportion as…

F2 / F1 = V2w2 / V1w1 = (1 * 1) / (1 *0.25) = 1/0.25 = 4

Or F2 / F1 = (V2/V1) * (w2/w1) = 1 * 4 = 4




You will note that although the forces are different the tangential speed remains that same (as you requested), however the angular speeds are still different due to the difference in the forces.

You keep ignoring angular speed, which is a fundamental aspect of rotational motion.

Is the point starting to get through to you yet?


ETA:

Pleas note also, just as before, the ratio of the forces equals the ratio of the tangential speeds times the ratio of the angular speeds.


Also ETA:

Sorry I misspoke myself before, the angular speeds are different due to the tangential speeds being the same and as a result the forces are different.


Yet another ETA:

It might help Bjarne if I put the angular speeds in terms of rotations per second (RPS). In those units w1 = 1.57 RPS and w2 = 6.28 RPS.

Ok time to eat lunch now so hopefully no more ETAs
 
Last edited:
That aside, no, it is not the same question. In the orbiting moon case, the force, mass, and distance relationships are well-established by the laws of gravity. For your lassoed bull and cowboy situation, there is no telling just how much the bull will dig in (or not) to resist the pull imposed by the rope.
Click to expand...

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

It might help Bjarne if I put the angular speeds in terms of rotations per second (RPS). In those units 1 = 1.57 and 2 = 6.28 RPS.
Click to expand...

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.

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


( What is ”radian” ? – 1 orbit)..
 
Bjarne said:
Yes it is – speed is the same factorm - like´you want...
Click to expand...

First off, same speed was your factorm [sic], not mine.

Second off, you still haven't provided enough information and/or constraints for the question to be meaningful. In one "bull orbit" the bovine might yield to the gentle pull of the weight of the rope alone, diverting it from its preferred path. In the other "bull orbit", it might actively oppose the rope with all its might.



( What is ”radian” ? – 1 orbit)..
Click to expand...

Oh, dear!!
 
Bjarne - F must be proportional to : Circle radius * speed

Bjarne said:
Only words
F must be proportional to : Circle radius * speed - it's so simple math....
Click to expand...

First asked 17 december 2009
Bjarne.
Please show the simple math that ends up with that for a body in circular motion:
F = k*r*s
where
F = magnitude of the force on an object in a circular motion.
k is the proportionality constant between Circle radius * speed.
r = the radius of the circle.
s = the orbital speed (the magnitude of the velocity vector v).
​
So far all we have seen is pretty pictures where you have arbitrary assigned values of orbital speed to various orbits according to some idea of yours.
 
jsfisher said:
First off, same speed was your factor [sic], not mine.

Second off, you still haven't provided enough information and/or constraints for the question to be meaningful. In one "bull orbit" the bovine might yield to the gentle pull of the weight of the rope alone, diverting it from its preferred path. In the other "bull orbit", it might actively oppose the rope with all its might.
Click to expand...

As I thought
Still a poison question ?
Come on man this is bull, cow and cowboy logic, isn’t it ?


Oh, dear!!
Click to expand...
I do not undertand that "radian" english word...
 
Bjarne asks "( What is ”radian” ? – 1 orbit)"

Bjarne said:
( What is ”radian” ? – 1 orbit)..
Click to expand...

Bjarne,
Why you do not know about radians?
The radian is a unit of plane angle, equal to 180/π (or 360/(2π)) degrees, or about 57.2958 degrees, or about 57°17′45″. It is the standard unit of angular measurement in all areas of mathematics beyond the elementary level.
Click to expand...

ETA:
But I see that it is a question of your bad English.
 
Last edited:
So far all we have seen is pretty pictures where you have arbitrary assigned values of orbital speed to various orbits according to some idea of yours.
Click to expand...

Just ask the question...
One touch at a number key, that's all you need to do...
 
You must log in or register to reply here.

Share this page

ISF - Join now!

Every member here is approved by hand. No bots, no spam, just people who care about evidence and honest debate.

Membership is free!

Create your free account

Back
Top Bottom