To begin with I go with David, you are talking the altitude of geostationary satellites. However, the actual answer is not so simple. It may be true in the central belt of the planet where most geostationary satellites are positioned but a tower at either North or South axial pole (not the magnetic poles but the true axis of rotation) would not experience a centrifugal effect as it would be spinning around its own axis and therefore and gravity would always win. You could only theoretically get balanced forces over the equator in any case.
2007-09-04 02:54:35
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answer #1
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answered by oldhombre 6
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Anywhere. I am currently on the third floor of a building, and there are two forces acting on me. First: gravity, acting as a centripetal force, pulling me inwards towards the Earth. Second: a normal force from my chair acting as a centrifugal force, pushing me outwards. Since I am not accelerating, the two are equal. The same argument applies to any floor of a tower.
2007-09-04 02:33:01
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answer #2
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answered by ZikZak 6
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No. Centrifugal rigidity and gravity are thoroughly diverse. Centrifugal rigidity is a rigidity created by using flow. This rigidity only happens while something is spinning. If the spinning stopped, the rigidity could end besides. Gravity is a rigidity simply by mass. whether the Earth stopped spinning, we would nonetheless sense the comparable gravity because of the fact the mass of the Earth continues to be the comparable.
2016-12-12 17:44:23
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answer #3
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answered by boven 4
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At 22,240 miles up, the altitude of geostationary satellites. For a tower at the equator that is. A tower anywhere else would feel side forces, because centrifugal force is perpendicular to the axis of rotation but gravity is perpendicular to the surface of the Earth and the only place on Earth where those two are parallel is at the equator.
2007-09-04 02:21:02
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answer #4
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answered by campbelp2002 7
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the siphon - which uses the centrifugal force generated in a high rigid tower to raise a 250,000 km train of tethered vehicles. As vehicles are released from the top of this tower, new vehicles are fed in at the base, so that the siphon can pump vehicles continuously into space. Apart from the energy required to build the tower and to prime it with a train of connected bodies, the energy source for the siphon is the rotational energy of the earth: no other power source is required. In this tower theory, the answer is: Nowhere , close at the top. Most of these answers are "inside the box".
2007-09-04 02:07:18
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answer #5
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answered by Anonymous
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Doctor John:
centrifugal force:
–noun; an outward force on a body rotating about an axis, assumed equal and opposite to the centripetal force and postulated to account for the phenomena seen by an observer in the rotating body.
http://en.wikipedia.org/wiki/Centrifugal_force
Whereas the centripetal force is seen as a force which must be applied by an external agent to force an object to move in a curved path, the centrifugal and coriolis forces are "effective forces" which are invoked to explain the behavior of objects from a frame of reference which is rotating.
http://hyperphysics.phy-astr.gsu.edu/hbase/corf.html
2007-09-04 02:07:42
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answer #6
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answered by Anonymous
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This point is called geostationary orbit, and it's usually high in orbit around the earth. You'd have to build a tower tall enough to poke out of the atmosphere.
Of course, if you did build such a tower, it would provide a cheap way to get into space. But you'd need materials so strong that such material has not yet been invented.
2007-09-04 02:03:02
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answer #7
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answered by David S 3
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Hello. In a boat it is called M don't know how or why, Some-thing about capsizing.(toppling over) ps in a boat it has something to do with the keel maybe it's in the foundations. only guessing. Star question .
2007-09-07 21:30:23
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answer #8
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answered by Anonymous
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F(cent) = m*omega^2*r
where m is the mass of the object
omega is the radial velocity
and r is the distance of the object from the center of rotation
F(grav) = m*g
where m was previously defined
and g is the acceleration (relatively downward) due to gravity
F(cent) = F(grav)
m*omega^2*r = m*g
r = g/omega^2
2007-09-04 02:06:18
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answer #9
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answered by civil_av8r 7
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I would expect it to be about 130,000 feet as this is the upper stratosphere or the 'edge of space'. As its the gravitational pull that keeps our atmosphere from spinning off the planet then your hypothetical tower would need to be roughly that high for weghtlessness to become apparent. Just an educated guess, though.
2007-09-04 01:59:46
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answer #10
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answered by Anonymous
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