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7 answers

♠ lets calculate! The mass of relativistic coin would be
m= m0/(1-(v/c)^2) =m0/√(1-0.999^2) =m0*22.4, where m0 is mass of a non-relativistic coin (a coin in your pocket); that is surplus to Earth’s mass is negligible!
coin’s kinetic energy = 0.5m*v^2 = m0*10^18 joules, this is equivalent to a mass M=m0*10^15, moving at 100 miles per hour; the result is impressive!
So here I might agree with other answerers and put a full stop.
♣ yet let’s analyze! The size of relativistic coin will not exceed that of a non-relativistic one; thus the area of contact with atmosphere and ground is rather small; the coin will pass through atmosphere in less that 0.002s, then plunge through Earth’s crust into mantle, then stuck most likely!
♥ The phenomenon may look like Tungus meteorite, 1908 – not much a disaster; the whole pattern may be compared with that of a red-hot bead plunging into a big piece of butter, and not like you’d sit down on the butter!

2007-06-11 01:41:17 · answer #1 · answered by Anonymous · 2 0

Well, the kinetic energy of the quarter would be comparable to a very large hydrogen bomb. The atoms within the coin would start scattering off atoms in the atmosphere, causing the number of high energy particles to multiply by scattering and particle-antiparticle pair production. As the number of involved particles multiply, their average energy would decrease proportionately, and they would start to spread out until all the particles in the effected volume, and perhaps some of the ground too, were involved. Imagine a 50 mile high Christmas tree made of hot gas. There would be a shock wave and lots of burning things for miles around. Sort of like a nuclear blast.

2007-06-10 23:08:26 · answer #2 · answered by Dr. R 7 · 1 0

Gobs. Because as moving objects approach the speed of light, due to relativity they also become much more massive. So it wouldn't just be a quarter moving with a huge amount of kinematic kinetic energy, it would be a very heavy quarter moving at near light speed.
In addition to leaving, I suppose, a rather large and probably unusually shaped crater, at that speed there would also be a lot of atoms smashed, so I would suppose a fair amount of radiation would also be released.

2007-06-10 21:52:23 · answer #3 · answered by Anonymous · 1 0

First off, at that speed, the coin would occupy most of the volume of the universe. Second, it would almost never get there. Third, what do you mean by damage?

2007-06-10 21:52:22 · answer #4 · answered by bullwinkle 5 · 0 2

If you were under it, doing an EVA above the atmosphere, the most ultimate amount of death possible. Your subatomic particles would be seperated from each other.

If you were thousands of miles away, probably nothing.

2007-06-10 21:53:12 · answer #5 · answered by anonymous 4 · 0 0

First thing that comes to mind would be the incredible momentum it would have. But funny things happen as objects get closer to the speed of light.

2007-06-10 21:52:45 · answer #6 · answered by darkhydra21 3 · 0 1

It might cause a shock wave as it entered the atmosphere and the vaporize.

2007-06-10 22:01:31 · answer #7 · answered by James 4 · 0 0

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