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No More Leap Second?
WerewulfX writes "CNN reports: "In a phenomenon that has scientists puzzled, the Earth is right on schedule for a fifth straight year." Update yeah, this is a repost. Whatever- it's a holiday. Nothing else to post :)
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MORONS!
This is a dupe. Not only that, I emailed about this being a dupe, 10 minutes before it went live.
TEN MINUTES!
Not only does Slashdot not even bother to correctly check for dupes, they don't even read their email that says "YOUR UPCOMING STORY IS A DUPE!".
Just what is the "email the editor if you see a problem with this story" link for, then, anyhow?
Just for the hell of it?
BAH!
Scientists have known about this leap second since the early 70s. So there has been atomic clocks made that take leap seconds into account. If my calculations are correct, doesn't this mean that there may be atomic clocks around that contain 5 extra seconds?
Exactly how do they figure out where the Earth is supposed to be down to a given second?
I understand Atomic clocks and how they work - but I don't understand how scientists can deduce where the earth should be to the exact second and correct it as such.
Looking for hardware (Currently need: Large Etch-a-Sketch) Have one? See my journal!
How about instead of replacing slashdot, Taco just gives a hundred or so random subscribers the ability to mod a story as a dupe before it is officially posted for everyone else to read? It wouldn't be much harder to implement then mod points or metamoderation. If it wasn't a dupe then noone would have to do anything, but if it was a dupe then those "lucky" few could get someone's attention by clicking a "This is a dupe" link. Regards, Steve
I don't think this should be that suprising. It is widely known that the moon's orbit is increasing. The effects of this is slowly causing the tides to be less severe. It is also slowing the earth's rotation. Eventually the moon will be in orbit around the equator, and there will be no more tide. I think there might be a good chance that the loss of the leap second could be related to this.
Don't waste time... procrastinate now!
If the Earth is assumed to be a homogeneous sphere and the rotational axis is assumed to be the straight line passing through the north and south geographic poles, the moment of inertia of the Earth is I = MR^2 where M is the total mass of the Earth and R is its radius. The kinetic energy of a rotating Earth is given by K = 1/2 I w^2, where w is the angular velocity.
The energy associated with an angular velocity which is increased by 1 second over a year is equivalent to an extra 1.6e22 Joules of energy or 40 times the annual energy consumption of mankind (DoE 1999). A detailed analysis and matlab script are available here
Yeah, this is a rereply. Whatever. It's a holiday. Nothing else to reply to.
Michael.
Linux : Mac
But haven't you seen their new advertising campaign? They've reformed! Everyone likes AOL now, even Snoop Dogg!
I personally expected them to give up on their CDs a while ago, but there's no accounting for marketing...
True story.
Not going to suggest I'm an expert, I am a 4th year physics student however.
;)
Just using basic physics(I'm not going to try to solve the n body problem in GR) we have Newton's Law of gravity:
F = G m1Xm2/r^2, the famous inverse square law.
Now we know that the sun has been burning for about 4 billon years, and probably has about that left, but lets say that the sun dies tommorrow, so the sun lasts about 4 billion years.
Lets make a further assumption that all of the mass of the sun is consumed in that time, either by radiation, or by mass blasted of the surface, but at a constant rate(should be a fair assumption in this case as no radical sun behaviour seems to have happened in the last 1 billion years or so, sense were here to talk about it).
Okay, so in a year we have about 31.5M sec. So the one second correction is 1/31.5M of a year. The mass change(and the force is linearly dependant on the mass remember) is 1/4B ie:
3.17e-8 vs 2.5e-10.
But hold on, a force is an acceleration right. So its cumulative over time. Lets make a further assumption that the earth is in a circular orbit. The reduction of the gravitational force due to the loss in mass is going to make the earths orbit larger. The velocity of the earth around its tranjectory will remain constant(the gravity force is tangential to the direction of motion). Lets calculate the change in the orbit radius:
for circular motion:
F = Gm1m2/r^2 = m2v^2/r -- centripetal force equation, force from gravity
r = m2v^2/F
plugging in F = 1 and F = 0.99999999975 gives
delta r ~= 0.000000001
time in year is portional to perimeter
2pi(r) verses 2pi(1.000000001r) ie correction on the order of a hundredth of a second a year. And keep in mind that the sun doesn't lose all of its mass in 4 billion years so this is a great exageration.
So, nope mass loss of the sun isn't our solution.
There may be reasons to do with general relativistic corrections to the problem that I don't understand, but I don't see how the earth's core can play a factor either.
The earth and moon are 'almost' (except for there gravitational effects on nearby guys like the sun) a closed system. The center of mass of the system has to stay the same. So you can replace the whole thing with a point with the mass of the earth and moon combined at the center of mass and nothing should be changed. Say the whole core shrunk to the size of a pea and glued itself to the underside of China. Well some force would have to have moved it from on average the center of the earth to the surface. That force would have to have come from either the earth or the moon. But 'for every action there's an equal and opposite reaction' right? The forcing object would get pushed back and our center of mass point would stay the same.
We've taken a radical case with the sun radiating away its mass and got a negligible effect, I really don't see how a tidal like situation with the sun and moon/earth combination could make that much of a difference. After all, similar research has shown that the earth's tides are getting less severe because the moon apparently is getting farther away from the earth. So tidal effects have to fall of with distance pretty quickly. They'd have to fall off at least as fast as the basic gravitational force creating them ie 1/r^2.
But as I said, I'm an undergrad in physics, not a professor in planetary dynamics. Hope the excercise in physics was as much fun for you guys as it was for me. It's been at least a year since I last used Newton's law of gravitation and I'm in physics
> It takes half a second to scan the headline and realize that its a repeat
Did you notice this is the same kind of logic some spammers use when they try to convince others that spamming is harmless ? Something along the line of "It just takes a second to read, and if you don't like, just ignore it.".
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Roses are #FF0000, Violets are #0000FF, find / -name '*base*' |xargs chown -R us && mv zig greatjustice
When (greater) Sweden was going to change from Julian to Gregorian calendar they stuffed it up, got one day out of sync with all Julian countries, so to get back in sync they added an extra leap day, creating the only 30th Feb in history.
They eventually made the change from Julian in 1753 by having (gregorian) 1st Mar 1753 after (julian) 17 Feb 1753 removing ten days.
Makes a leap second seem a bit insignificant....
If you have no idea about Julian (as in Ceasar) and Gregorian (as in pope) calendars, have a look here