January 4, 2014 Planetary Alignment Decreases Gravity

www.newshound.net ^ | Dec 16, 2013 | news hound

[Wingy]

The theory part is that no rotational kinetic energy is lost in the ‘everybody jumps up at the same time’ scenario.

Not true! I have it on Congressional integrity (whatever that is..) that Guam will flip over and sink.

[Rembrandt]

And on that day, many light in the loafers guys, some on Fox News in the evening, will never touch ground.

[21twelve]

“I think even a tenth of a second is total BS. More like .00000000001 seconds.”

It really doesn’t matter - isn’t the world supposed to end when all of these planets align? (At least that is what I recall the one time it happened and I didn’t do my 4th grade homework!)

[ReaganÜberAlles]

The two planets are too miniscule, in relation to the Sun, to have any effect whatsoever.

[Billthedrill]

Hooboy. Uh, Pluto is a very, very, very, very long way away. And tiny.

Nevertheless, it's all coming clear now. This is why scientists took Pluto's planetary status away! It was to prevent this from happening. So if you jump in the air at that particular moment and you don't hover for three seconds, you'll have proof that the scientists are all in on a gigantic conspiracy. I blame Carl Sagan.

[DoughtyOne]

I’ve had some late nights where I swear it too me at least three seconds to hit the bed. And then I fell off... LOL

[dr_lew]

There are no tidal forces involved in the phenomenon described.

Au contraire, it's the only thing that COULD be involved, because of the free fall principle. The earth and everything on it, could free fall at any rate with out our noticing it. It is only the differential rates of fall, or attraction, that could cause objects on the earth to separate from each other.

This phenomenon is discussed in detail in treatments of objects falling into black holes. Of course, the claim here is supremely ridiculous, because the magnitude of the effect claimed would imply very strong tidal forces.

The -2 in the -2dr/r is not a tensor.

I said, "The "-2" in -2dr/r is the -2 in the tidal tensor ..."

... IN the tidal tensor, you see. That tensor is represented by the matrix:

1 0 0
0 1 0
0 0 -2

Recall the tensor maps a vector to a vector, and this tensor maps a displacement vector ( from earth center ) to a force. The -2 maps a z displacement to a force in the z direction, and derives from the differential that you wrote down.

I know this stuff! Believe me!

[DoughtyOne]

This just in: Pluto has pulled out of the deal because it didn’t get the billing rights it was seeking. Downgraded from planetary status, it will not be participating in this event.

[headstamp 2]

I’ll be able to drive my car on two wheels.

[FredZarguna]

I taught this stuff for years. There are no tidal forces involved. It is strictly classical gravitation that's being described and nothing more.

[Kickass Conservative]

Sounds like a good time to go for that Pole Vaulting World Record.

[FredZarguna]

Das is nicht richtig nur falsch. Das ist nicht einmal falsch.

[ViLaLuz]

I thought Pluto was demoted.

[SamAdams76]

If you think you will be able to float around your house then you will be mistaken. BUT if you jump in the air at 9:47AM local time on January 4th 2014, it should take you about 3 seconds to land back on your feet instead of the usual 0.2 seconds.

I'm a white guy so most likely the best hang time I can hope for is 0.33 seconds (white men can't jump).

Also, note the difference between an astrologer and an astronomer. It's like the difference between a blonde bimbo and Margaret Thatcher.

[loungitude]

Conservation of rotational kinetic energy -- which I agree doesn't change -- requires the angular velocity of the Earth to decrease when the jumpers jump. The moment of inertia of the system composed of the Earth and jumping people will change. In order to conserve angular momentum -- which must be conserved because there is no external torque -- the angular velocity, ω, must decrease. [In your alternative formulation, there are no external sources of energy, therefore the angular velocity must decrease when I increases in order for the kinetic energy to remain constant: Iiωi2 = Ifωf2] The differential dI would (at most, assuming the people ball themselves up into spheres when they jump) would be 2m * r * dr, where r is the radius of the earth, dr is the number of meters they jump, and m is the mass of six billion people. The number is insignificant compared to the moment of inertia of the Earth, which is roughly 2/5 M r2, where M is the mass of the Earth. And it is clearly this which dominates. =================================================================================================== Fred this sounds good, however I struggle with the fact that the jumpers become detached from the earth part of the system, momentarily creating two systems, each possessing a fraction of the initial energy of the earth/people system. I suggest that the incremental change of MOI of the people part of the system would cause their omega to change. This would result in their landing in a different spot than they jumped up from. And the angular velocity of the earth stays the same.

[FredZarguna]

I was speaking perhaps too loosely. When I say angular "velocity of the Earth," in that post, I am always talking about the whole system of Earth + people. So even though the people "jump up" I am still drawing a box around the whole thing including the jumpers. If instead you draw two boxes, one around the Earth, and one around the people, it is clear that the rotational kinetic energy of the "Earth Box" never changes, as you say, and in that box (which conveniently excludes the people) the moment of Inertia of Earth alone doesn't change when they jump, so ω of Earth alone doesn't change.

Is is correct that the people do not quite return to where they were before. This does not change the fact that (absent air drag) both angular momentum and rotational kinetic energy -- for the whole system and for the individual components -- have to be conserved.

The way to reconcile the two pictures dynamically is that the center of mass, and therefore center of rotation in my picture -- I can have only one because I have only one system -- is actually moving. It moves slightly away from the center of the earth and rotates in a (very very small) ellipsoid of some kind until the jumpers return to Earth, at which point it returns to the center of the Earth again, so mine is more complicated.

In a rigid body (the people can only jump up and down in elevator shafts or tubes or pole-dancing or something) my picture is largely the same, but simpler because the center of mass doesn't move. Your's actually will become the more complicated one, because now there are external torques being exerted by the people in the tubes on the Earth, and the Earth on the people in the tubes, which keeps their ω the same -- and less -- until they return to earth.

[loungitude]

Yes, I think we agree. Thanks for the explanation.