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Why does rotation create gravity?


Kingpin1989

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the effects of acceleration and gravity are indistinguishable

 

well, expect that gravity acts in one direction and you could accelerate in any direction (incl. away from a source of gravity).

 

also gravity has (take into account ONE object e.g. a car) a fixed force, (as the mass of the car is the same) whereas a car can accelerate at different rates and forces (dependant on how hard the engine works) whereas gravity will always stay constant force.

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Rotation doesn't create gravity, BUT, if you build your spacecraft as a wheel, with crew quarters around the rim, and make the wheel/spacecraft rotate, centrifugal force will push everything to the outside edge and crew will think the outside edge is "down".

 

Scooter93

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A particle in non inertial rotating frame, will have the same equation of motion as if the particle were in in an inertial frame with 3 additional forces involving the angular velocity.

 

If in am intertial frame S, we have equation of motion

[math]m\ddot{\mathbf{r}}=\mathbf{F}[/math]

 

the equation oof motion is the rotating frame S'

 

is

 

[math]m\ddot{\mathbf{r}'}=-2m\pmb{\omega}\times\dot{\mathbf{r'}}[/math][math]-m\dot{\pmb{\omega}}\times\mathbf{r}[/math][math]-m\pmb{\omega}\times(\pmb{\omega}\times\pmb{r})[/math][math]+\mathbf{F}[/math]

 

 

These are the apparent forces on the particle due to the consequences of the non inertial rotating frame.

 

The first term on the right is called the "Coriolis Force". the best example is that it defines the way water goes down a sink, according to where you are on earth. Also a particle dropped from height H will not land on the point directly below from where you dropped it. unless you are in the specific lcation on earth. So the apparent gravity we face is the not the same thing as the real one. i.e it doesnt point straight down to the centre of the earth. it points somewhere below the centre.

 

the second term is 0 if angular velocity is constant

 

the third term is what we refer to as the Centrifugal Force. A combination of these apparent forces in a rotating frame will create apparent gravity.

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a particle dropped from height H will not land on the point directly below from where you dropped it. unless you are in the specific lcation on earth.

 

I'm pretty much an amateur but I find science fascinating. Would this location be the magnetic north and south poles? Thanks.

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i dont know, at least i dont think so.[edit]come to think of it, what does magnetism has to do with gravity, in the macroscopic world anyway[/edit]

explicit calculation gives

 

[math]\pmb{g'}=-(g-a\omega^2\cos^{2}\lambda)\mathbf{k}-a\omega^2\sin\lambda\cos\lambda\mathbf{j}[/math]

 

where i j k are right handed unit vectors fixed on the surface of the earth, radius a, such that j points to the north pole and k points straght up from the centre of earth. and lambda is the latitude. i.e angle from the equator. omega is angular velocity, whose direction is aligned with the north pole of the earth.

 

so as you can see, if u are at the poles then u exprerience true gravity. i.e [math]\mathbf{g'}=\mathbf{g}=-g\mathbf{k}[/math]

 

and at the equator u get [math]\mathbf{g'}=-(g-a\omega^2)\mathbf{k}[/math]

 

so at equator and poles , gravity pulls towards the centre of earth, but with less magnitude at the equator.

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In Einstein's general relativity,

the effects of acceleration and the gravity is the same.

Imagine that you are in a spaceship in space very far away from any star or planets moving with constant acceleration in a straight line equal to 9.81 m/s^2 = g, you will feel yourself being pulled to the floor of the ship, as if you can feel your weight similar to what you experience back on earth.

 

From the understanding of Newtonian mechanics, you know that acceleration can also be due to a change of direction, and you will experience a centripetal acceleration towards the centre of the circular path as you move in a circle. Hence rotation of an object can effectively produce the same effect of gravity due to the Einstein's equivalence of gravity and acceleration. In other words, the effects of accelerated frame and gravitational force due to presence of mass are indistinguishable, both will deform the space-time fabric in the vicinity.

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From what I've read, one of the possible application of this thread discussion is to create a third party planet like ring for people to live on, if technology allows.

 

If a large ring can be made to rotate at very high constant speed v in a circle, we know that an acceleration mv^2/r = mrw^2 is being directed towards the centre of the ring. Using the gravity-acceleration equivalence principle, this will produce a force in the other direction away from the centre of the circle. This may be confusing to some, so I use an analogy. If you are in a lift and the lift starts accelerating upwards from rest, you will feel heavier, just as if extra weight has been added and this extra weight is also directing downwards opposite to the direction of accleration of the lift. So simliarly, the force created by the accelerated frame will be opposite in direction to the acceleration.

 

If people are to live on this large ring, they will feel their own weight pulling them downwards away from the centre of the ring, and all will agree that their sky (upward) will be the centre of the large ring. They will not notice themselves rotating in a circle if the diameter of the ring is made large enough. What they will experience is like the same experience back on earth.

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If people are to live on this large ring, they will feel their own weight pulling them downwards away from the centre of the ring, and all will agree that their sky (upward) will be the centre of the large ring. They will not notice themselves rotating in a circle if the diameter of the ring is made large enough. What they will experience is like the same experience back on earth.

so for example the planet Halo (from the games halo and halo2) is a ring planet. people live on the inner side of the ring.

 

whilst this may seem impossible due to lack of gravity (and the fact such a formation is not natural) it is possible due to centrifugal there would appear to be gravity (although not actually).

 

obviously this object would need to be moving at high speeds to create enough centrifugal force to hold every-thing/one down.

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obviously this object would need to be moving at high speeds to create enough centrifugal force to hold every-thing/one down.

 

Well, if we want the acceleration to be one standard gravity (9.81ms-2), and have one revolution per day (just like the Earth has):

 

The acceleration is [math]a=\omega^2 r[/math] where [math] \omega = \frac{2 \pi}{60 \times 60 \times 24 s} = 0.727 \times 10^{4} s^{-1}[/math], then we need a radius of:

 

[math]r=\frac{a}{\omega^2}=\frac{9.81 \times 60^2 \times 60^2 \times 24^2}{4 \pi^2} m = 1.17 \times 10^{10} m[/math]

 

which for reference is about 1821 times the radius of the Earth....

 

 

If you want it to be the same radius as the Earth, [math]r=6.4 \times 10^6m[/math] then we need an angular velocity of:

 

[math]\omega = \sqrt{\frac{a}{r}} = \sqrt{\frac{9.81 ms^{-2}}{6.4 \times 10^6m}} = 0.001238 s^{-1}[/math] which corresponds to a speed on the surface of [math]v = \omega r = 7923.2 ms^{-1}[/math] which is roughly 24 times the speed of sound!

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I´d guess there are reasons for the numbers used in the Ringworld novels and I´d bet these numbers -obtainability of an artificial planet way bigger than earth left aside- work quite well. Can´t give them here as I don´t own the books and read them something like 10-15 years ago, though.

 

It´s a bit dissapointing that Kingpin didn´t bother explaining his/her question so far. My bet was that the question was about the different metrics for describing a rotational black hole (Kerr-metric) and a non-rotational one (Schwarzschild-metric). But as this is the Classical Mechanics section and not Relativity (hint@yourdadonapogostick), Gilded´s 1st post sums it up quite well (except that the distribution of the mass in space also plays a role).

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  • 3 weeks later...
Centrifugal force doesn't exist though right? Well it is not a force, we just call inertia of an object having centripedal acceleration...I think :confused:

 

Correct. The centrifugal force is a pseudoforce. It isn't real. It is e.g. the illusion that you are being thrown outward when you go around a circular path, when in fact you are really tending to go in a straight line. It's just that you are in a rotating reference frame and are still trying to process information as if you were in an inertial frame. Since you are trying to "subtract out" the centripetal force, you have to "add in" the centrifugal force for things to make sense.

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