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Centrifugal forces ' appear ' to act opposite to gravity . How is this possible?


Mike Smith Cosmos

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SwansonT,

 

Definitionally, how close to the Earth's equator can you be an object in geosynchronis orbit, before you are an object standing on the rotating surface?

 

Regards, TAR

The two situations do not overlap, seeing as the speed for a geostationary orbit is very different from the rotation speed of the surface of the earth. As Mordred's link shows, geosynch orbits are at an altitude of about 36k km.

 

IOW, the earth's radius would have to be that much bigger, with no change in mass, for that to happen. But that raises issues of planetary integrity

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Mordred,

 

OK, the two situations do not overlap...except...if you were to build a 35 thousand km high tower and release something from its highest point, would that something be in geosynchronous orbit?

 

Regards, TAR

 

Does an object at the top of number one World Trade have a different velocity than an object on the first floor? A different speed? A different acceleration?

If we built a replica of the tower on the equator, and another on the North pole and wheighed a standard mass on the first floor and the top floor, would there be a difference in the whieghts taken in the six locations?

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35,000 km is not quite high enough, but if we use 36,000 km, we see:

 

v = Circumference / 1 Day = 331280.image2.png

 

2*Pi*(36,000km + 6,378km)/24hr * 1hr/3,600s = ~3.08km/s

 

Orbital velocity of a geosynchronous satellite is only 3.07 km/s.

 

Note: We are assuming a perfectly rigid tower. Another way to look at it is as though a cable were attached between the satellite and the Earth's surface.

Edited by Endy0816
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OK, the two situations do not overlap...except...if you were to build a 35 thousand km high tower and release something from its highest point, would that something be in geosynchronous orbit?

Pretty much.

 

Does an object at the top of number one World Trade have a different velocity than an object on the first floor? A different speed? A different acceleration?

Yes.

 

If we built a replica of the tower on the equator, and another on the North pole and wheighed a standard mass on the first floor and the top floor, would there be a difference in the whieghts taken in the six locations?

That's four, but yes, they would be different, if measured with sufficient accuracy and if by weighing you mean measuring the normal force.

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SwansonT,

 

I was counting the the real number 1 World Trade as an example at around 45 degrees, the one on the equator as an example at 90 degrees and the one on the North Pole as an example of a tower on the axis of rotation, that should not have much of an orbital velocity at all...unless the change in orientation counts as an acceleration. That is 6 different places to consider what forces are on the wheighed mass.

 

The forces normal to the center of the Earth in each case, if different, are caused by centrifugal forces generated by the inertial momentum of the object due to angular velocity which is large at the equator, medium in NY and minimal at the North pole. Or so my speculation would go.

 

Regards, TAR


So, with our 36,000km tower...what if we built one of those on the equator, one in NY and one on the North Pole. What would happen to an object released from the highest point, in each case?

Edited by tar
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Have you any idea how that difference in weight could be measured?

For the equator vs pole, a good scale would suffice.

 

The forces normal to the center of the Earth in each case, if different, are caused by centrifugal forces generated by the inertial momentum of the object due to angular velocity which is large at the equator, medium in NY and minimal at the North pole. Or so my speculation would go.

 

But this idea doesn't work. The force pushing out is smaller for the rotating case and largest at the north pole.

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SwansonT,

 

Oh, I thought we were saying earlier that the mass would wheigh less at the equator, more at the axis. I remember some link or search I read wrong, then. I thought the ton of stuff wheighed at the equator would wheigh 2002lbs. at the North Pole, without the tangentially pointed inertia of the angular velocity caused by moving along with the surface of the Earth, "lifting" it away from the surface.

 

Regards, TAR


Thread,

 

So if a ton of stuff wheighs 1998 pounds on the equator and 2000 pounds on the North Pole, is it not a real force lifting the ton off the scale a little, on the equator, due to the spin of the Earthly carosel?

 

Regards, TAR

There was a brief exchange of thoughts as to "why" a thing wheighed less on the equator than on the pole, but it was never suggested that a mass does not wheigh one part in a thousand less on the equator, than on the pole.


SwansonT,

 

By my rough interpolation of what I thought were the facts, a ton wheighed in NY would wheigh 2001lbs. on the North Pole and 1999lbs. on the equator.

 

Is this not true?

 

Regards, TAR

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SwansonT,

 

Oh, I thought we were saying earlier that the mass would wheigh less at the equator, more at the axis. I remember some link or search I read wrong, then. I thought the ton of stuff wheighed at the equator would wheigh 2002lbs. at the North Pole, without the tangentially pointed inertia of the angular velocity caused by moving along with the surface of the Earth, "lifting" it away from the surface.

 

Regards, TAR

There was a brief exchange of thoughts as to "why" a thing wheighed less on the equator than on the pole, but it was never suggested that a mass does not wheigh one part in a thousand less on the equator, than on the pole.

SwansonT,

 

By my rough interpolation of what I thought were the facts, a ton wheighed in NY would wheigh 2001lbs. on the North Pole and 1999lbs. on the equator.

 

Is this not true?

 

Regards, TAR

 

weigh

 

An object at the equator weighs less, i.e. the normal force is smaller. The normal force pushes out. In the thread where we were discussing this it was more than suggested

http://www.scienceforums.net/topic/88621-nonspherical-earth-split-from-centrifugal-forces/?p=864186

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weigh

 

An object at the equator weighs less, i.e. the normal force is smaller. The normal force pushes out. In the thread where we were discussing this it was more than suggested

http://www.scienceforums.net/topic/88621-nonspherical-earth-split-from-centrifugal-forces/?p=864186

Google Normal Force

 

In mechanics, the normal force is the component, perpendicular to the surface (surface being a plane) of contact, of the contact force exerted on an object by, for example, the surface of a floor or wall, preventing the object to fall.

So on an inclined slope the normal force isn't equal and opposite the gravitational force. So if the normal force is different at the two places where the weight is placed on a horizontal surface, what is going on?

Edited by Robittybob1
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Google Normal Force

So on an inclined slope the normal force isn't equal and opposite the gravitational force.

 

Yes, and this is completely irrelevant to the discussion.

 

So if the normal force is different at the two places where the weight is placed on a horizontal surface, what is going on?

In the case under discussion, I explained it in the other thread, to which I linked.

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Yes, and this is completely irrelevant to the discussion.

 

 

In the case under discussion, I explained it in the other thread, to which I linked.

I was just defining what the normal force is. I thought it might be equal and opposite to the G force but I was wrong. It had something to do with the slope as well. It is a force that I don't think about much.

So if on a horizontal surface the same mass weighs differently the force of gravity must have changed or on a rotating system the centripetal force alters.

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So if on a horizontal surface the same mass weighs differently the force of gravity must have changed or on a rotating system the centripetal force alters.

The presence of a centripetal force is all that's required. As I have explained.

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But if the only centripetal force is caused by gravity, the gravity will not change with speed of rotation but the normal force will.

 

Yes, that's right. What happens to the normal force as the speed increases, and why?

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That reduction in the normal force is called the centrifugal force.

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This is what I have been 'bleating on about ' for the last two years , but been told ' Centrifugal Force , it does not exist ' , or at least that is what I thought people were saying to me. Maybe they were not , and I just did not understand what people were saying to me!

 

Mike

Edited by Mike Smith Cosmos
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.

.

This is what I have been 'bleating on about ' for the last two years , but been told ' Centrifugal Force , it does not exist ' , or at least that is what I thought people were saying to me. Maybe they were not , and I just did not understand what people were saying to me!

 

Mike

It is something called the centrifugal force but this is still a fictitious force. http://en.wikipedia.org/wiki/Fictitious_force

 

 

A fictitious force, also called a pseudo force,[1] d'Alembert force[2][3] or inertial force,[4][5] is an apparent force that acts on all masses whose motion is described using a non-inertial frame of reference, such as a rotating reference frame.

The force F does not arise from any physical interaction between two objects, but rather from the acceleration a of the non-inertial reference frame itself.

Edited by Robittybob1
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Your misunderstanding the meaning of force, yes there is 4 basic forces. Gravity, weak, electromagnetic, strong. There is also mechanical force, force due pressure etc.

 

Look at how force is defined. In this case it follows under moments of inertia. Newtons three laws still hold

 

"force definition. In physics, something that causes a change in the motion of an object. The modern definition of force (an object's mass multiplied by its acceleration) was given by Isaac Newton in Newton's laws of motion."

 

Regardless of source anything that causes a change in motion is a force.

 

Much of this thread is been about how to calculate the net sum of force, whether via gravity or moments of inertia. Change in movement requires a change in the net sum of forces with a direction ( vector)

Quite basically high school physics.

 

They covered mechanical and inertial force

Edited by Mordred
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O.k.

Although , somebody has defined it as a 'Fictitious Force', there is still, as far as I can make out a force (centrifugal force ) wheather it exists as an 'accelerating inertial frame ' , or whatever . It is ' there 'you can feel it , and it does some useful stuff in keeping satellites in orbit , exciting skaters on the end of a skating chain, and I do think it could be used more , if we could get rid of the ' fictitious ' bit.

 

After all surely the reasoning by Einstein in one of his major ideas was that by the law of equivalence " linear (gravity and accelerating in a lift )and curved (centrifugal acceleration ) " , all , feel the same so by the law of equivalence are the same !

 

Or have I got hold of the ' wrong end of the stick ' again ? That is " my accelerating inertial frame , upside - down bucket of water " stick ?

 

Mike

Edited by Mike Smith Cosmos
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In particular moments of inertia which applies to centrifugal acceleration.

Look at Newtons three laws of inertia in detail. Key note change in momentum.

If an interaction regardless of classification causes a change in momentum or direction . It is a force

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In particular moments of inertia which applies to centrifugal acceleration.Look at Newtons three laws of inertia in detail. Key note change in momentum.

.

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I appreciate the ' Inertia ' bit as Inertia is the key to it all , but is this not the point that when we try to tamper with the inertia , in the straight linear direction , taking it into a curve , we take the inertia with us so to speak , into a rotational or angular momentum , so that everything is conserved. But we get this reaction appearing as ' Centrifugal ..acceleration ..force ' Is this not so ?

 

 

Mike

Edited by Mike Smith Cosmos
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Think change in direction requires a change force.

 

"Newtons three laws of inertia.

First law: When viewed in an inertial reference frame, an object either remains at rest or continues to move at a constant velocity, unless acted upon by an external force.[2][3]

Second law: The vector sum of the external forces F on an object is equal to the mass m of that object multiplied by the acceleration vector a of the object: F = ma.

Third law: When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body."

 

Then look at the definition of velocity. "Velocity is the rate of change of the displacement, the difference between the final and initial position of an object. Velocity is equivalent to a specification of its speed and direction of motion, e.g. 60 km/h to the north. ( Key note direction)

Now look at moments of inertia and it's applications to rotation.

 

http://en.m.wikipedia.org/wiki/List_of_moments_of_inertia

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