Anti gravity room on earth?

[xxdutchmasterxx]

So a object gains mass when in motion, and the gravitational pull of a object is based on its mass.....so I am curious as to how fast must we spin a solid metal disk (say it weight is 500 pounds)  before its mass increases enough to a point in which its gravitational pull matches the earths pull of gravity perfectly? I am imagining this disk being stationed above a room, in theory, if we can spin it fast enough, then it should be possible for its gravity to cancel out earths gravity, thus, we would just float around in the space between the earth and this disk. so how fast does a 500 pound disk needs to spin to do this?....I am pretty sure its impossible to spin a disk fast enough to match the earths gravity because the disk would probably break before we get any where near the needed speed, how ever, I dont care if its possible or not, I just want to know how fast it needs to spin for it to work.

 

not sure if this helps but I think you can solve this by finding out how much of a gravitational pull a object has in respect to its velocity, once you know how much mass a object gains when in motion then you can calculate how much of a gravitational pull that object has while moving at any given speed....maybe, I wouldnt know, im horrible at math.

Edited July 25, 2018 by xxdutchmasterxx

[Bender]

Let's assume the disk has a diameter of about 1 m and nearly all its mass is located at the outer rim.

The outer rim would have to move slightly below light speed.

So about 300 000 000 m/s divided by a circumference of about 3, gives a rotational speed of 100 000 000 /s or 6 billion rpm

And yes, it will break long before that point. It will also require many orders of magnitude more energy than humanity ever produced.

Note that its "relativistic mass" could be much smaller than the earth's, because you would be much closer to its centre. The exact mass is not relevant, since it has absolutely no influence on my estimate above.

Edited July 25, 2018 by Bender

[xxdutchmasterxx]

interesting, thanks. 6 billion rpm is insane, guess we will never have anti gravity rooms using a spinning disk

[Janus]

A couple of things:

If you place your spinning disk just above the room, then you could only exactly cancel out the Earth's gravity at one exact point in the room.  For example, at a spot on a direct line between the disk and center of the Earth and halfway between floor and ceiling.   Gravity falls off by the square of the distance.  Thus if your room was  3 meters high, you could arrange it that at 1.5 meters above the floor, you would cancel out Earth's gravity.  But at 1 meter from the floor, you have increased the distance from the disk by a factor of 50% and the effect of the disk's gravity to 44%.  The Earth gravity will be stronger and anything placed there would fall to the floor.  At a distance of 2 meters above the floor, you have decreased the distance to the disk, so that its gravity effect will increase by a factor of 2.25and an object placed there would "fall" towards the ceiling.

In addition, to produce a 1g effective gravity at a distance of 1.5 meters would require an effective mass of 3.3e11 kg ( about the equivalent of a small asteroid), which would in turn be attracted to the Earth and have to be supported in some way.

In order to get a fairly uniform ( but still not perfect) zero gravity effect over the whole room, you would have to remove the disk to some great height above the room.  You'll get the closest to uniformity at all points in the room if you put your disk 1 Earth radius above the floor of your room.  But this also means that you will have to increase the effective mass of your disk up to the same as the Earth's.   To do this by pumping energy into the disk, would require an amount of energy equal to the entire output of the Sun over 44 million years.

Other problems would include:

  The effect would not be limited to just the room but would be felt to various degrees over the entire surface of the Earth.

An Earth equivalent mass, located just 1 Earth radius above the Earth would produce enough tidal force to rip the Earth apart.

Even ignoring the tidal force problem, the only way to maintain that object at that distance would be to put it into orbit  (actually the Earth and object would orbit a common barycenter.) Such an orbit would have a different period than the rotation of the Earth, and thus your "sweet spot" for  the location of the room would move across the surface of the Earth

[John Cuthber]

1 hour ago, Janus said:

Such an orbit would have a different period than the rotation of the Earth, and thus your "sweet spot" for  the location of the room would move across the surface of the Earth

That's easy to deal with. Put the "second Earth" above the North or South pole.

Obviously, that's going to screw up the Earth's orbit and so on.

But, since this thread is in cloud cuckoo land already, that hardly matters.

This is a rather odd example of a general problem

https://en.wikipedia.org/wiki/Lagrangian_point

 

[MigL]

Not quite a room...
This is currently done by NASA using an empty passenger plane to approximate weightlessness for training purposes
The effect won't last as long, but buying and modifying a plane for such a purpose is immensely cheaper than your scheme..

[J.C.MacSwell]

1 hour ago, MigL said:

Not quite a room...
This is currently done by NASA using an empty passenger plane to approximate weightlessness for training purposes
The effect won't last as long, but buying and modifying a plane for such a purpose is immensely cheaper than your scheme..

As would renting the International Space Station for a "relatively" small fee...

[Janus]

8 hours ago, John Cuthber said:

That's easy to deal with. Put the "second Earth" above the North or South pole.

Obviously, that's going to screw up the Earth's orbit and so on.

But, since this thread is in cloud cuckoo land already, that hardly matters.

This is a rather odd example of a general problem

https://en.wikipedia.org/wiki/Lagrangian_point

 

How are you going to keep it over the pole? it still has to be in orbit.  You can have polar orbits that pass over the poles, but they make a "orange slice" pattern over the Surface of the Earth. At least an equatorial orbit would keep the sweet spot following the equator.

The Lagrange points are too far away for this particular problem.

[Markus Hanke]

12 hours ago, xxdutchmasterxx said:

So a object gains mass when in motion, and the gravitational pull of a object is based on its mass

Firstly, what objects in relative motion gain is relativistic mass, which is a very different concept from rest mass. Secondly, relativistic mass is not a source of gravity - otherwise things could be either normal objects or black holes at the same time, depending on who observes them, which is an obvious contradiction.

This being said, placing a massive disc spinning at relativistic velocity into a pre-existing gravitational field will of course have some effect - but it is not going to be anything like a neat “counterforce” that balances out earth’s gravity. In fact this is a very challenging problem, mathematically speaking. Solutions to Einstein’s equations for spinning discs do exist, but they generally assume the absence of background curvatures. Here are some examples, to give you an idea how complicated this gets.

It is very difficult to predict what happens when you add background curvature, but I’d say you’d get a whirlpool-like frame dragging effect, not unlike the Kerr metric itself. In fact, if the background curvature is small enough, the Kerr disc would be a good approximation.

 

[DrP]

9 hours ago, MigL said:

This is currently done by NASA using an empty passenger plane to approximate weightlessness for training purposes
The effect won't last as long, but buying and modifying a plane for such a purpose is immensely cheaper than your scheme..

 They call it 'The Vomit Comet'!  iirc. lol.

[t686]

On 7/25/2018 at 4:11 PM, John Cuthber said:

https://en.wikipedia.org/wiki/Lagrangian_point

 

It's interesting that the angle in the L1, L4, and L5 lagrange points is the same or very similar to the angle in the Collatz Fractal