>Now, there NO way it would go KABOOM upon touching atmosphere: at these

   No, he meant you'd have to search for rubble of the earth with a
   microscope. While this is something of an exaggeration, it is not much
   of one.

Well, A 500 Km rock is 250 Km radius, ans as 4/3 pi r^3 = V we have
V approx = 4.4 * ( 1/4 * 1000 km )^3 = 1.1/16 * 10e18 m^3 or approx 10e17
m^3, which means you have approx 10e20 to 10e21 kg of rocks in this small
moon. Now, 10e16 of rock is 1000 times smaller, so the moon "supposed to
destroy the earth to atoms" would be only 25 km radius? Come on, when a
small bird crashes in a 747 airplane, who gets killed? Who goes SPLAT?

I'll grant you even greater benefit: earth collides with earth... Hey, if a
measly small 25 km diam asteroid can completely destroy the earth, surely a
collision with another planet would insure complete and utterly
instantaneous disintegration up to the quantum level, no? NOT!

And you say that these two ball colliding at .9 c would be destroyed? SURE!
But VAPORIZED? NOOOOO! These are made of STONE, not PLUTONIUM. they don't
react in a mega-explosive way. Try throwing two blobs of mud at each other
in a gravity free environnemnt. Do they go fission style? No, they simply
do a plastic non-elastic collision. That would be the same thing with
planets.

Heck, NASA did even computer simulations
showing that two approximately identical STARS colliding each other at high
speed would sure lose some of their matter into deep space, but not go
nova! They may even kind of "pass through" each other (I did not say
"harmlessly, tough...").

You all seem to think that because it's bigger then it must be more
cataclysmic, more "reactive". NOT! Sure it's more cataclysmismic at *OUR*
level, but at the planet level it's only a small rock falling on it. It
doesn't go *KA-BOOOOM* like dynamite! It's ROCK, STONE, INERT SUBSTANCE
that when you heat it melts without exploding. Sheesh. How many times will
I have to repeat it?

I`ll say it again: Sure, it is perfectly possible that great cataclysm the
could even destroy civilisation on all of earth could result from a
collision with a 250 km diameter moon going at .9 c, but IN NO WAY will the
Earth go KAWOOOM like a nuclear reactor gone haywire...

   1E16 kg moving at even .5c contains 2.25E32 Joules of kinetic
   energy. To put some perspective on this, a 1 megaton nuclear bomb
   releases about 4E15 Joules. Let's see, the energy we're talking about
   is the rough equivilent of a 100,000 million million megaton bomb.

Sure. It's normal. the USED fissionnary part of the bomb is under ONE
kilogram. And at 1/2 c you have by Kinetic energy rule that says that
KE = m v^2 (and not KE = mv, sorry about the previous post) (maybe not
even that... but I manage)... Well you have at least 1/8 of the TOTAL
energy INSIDE the rock (using E = m c^2 = m v^2 and dividing by 2 just to
be safe in case its 1/2 m v^2. If it's directly "mv", then you have HALF
the energy in the rock. Pure nuclear energy enquivalent, but being put into
movement instead of fission reaction). Now, 100,000 millions millions is
10e17 times the bomb... but OH, look, the mass of the asteroid divided by
the mass of matter used in the nuclear reaction, is roughly 10e17, too!
Coincidence? NOT! matter = energy, simple as that.

   It has been conservatively estimated that a 100 gigaton (100,000
   megaton) bomb would have the effect of blowing out the atmosphere like
   a candle. The energy we're talking about it absofuckinglutely billions
   of times beyond anything that has ever happened to the earth or the
   moon to date.

The atmosphere isn't the earth. Like if you said that by passing a
marshmallow over oa fire you utterly and completely destroy it : proof of
it: it's "atmosphere" (it's surface) is burned! Look again: the center is
still white... Mass can pulverise more mass, but not "atomize" more. Only
in a matter-antimatter reaction would you achieve the fieat of utterly
destroying THE SAME amount, no more, of matter, than you used anti-matter.

Someone a previous poster talked about "such energy is 30 Joules for
everygram of the earth!" Well, let me tell you something:

1- You assume 100% efficiency in your "nuclear reaction", which it in fact
isn't: it's a NON-ELASTIC (or plastic) COLLISION.

2- You assume that the energy will be well spread like butter on a bread,
while actually most of it affect only the local area or go back into space
into another form.

3- You underestimate the capacity of energy absorption of mass just being
"shaped" into another form. Even following your "Nice well spread 30J per
gram, what does that means? Well, for water, it's ONLY an increase in
temperature of 4 degrees... (Okay, it fick an ecosystem, but the water wont
even evaporate!). Why do you assume that all that energy has got to STAY
under the form of Kinetic Energy??? It clearly does not!

   Assuming the same density you used to calculate the mass of the 500km
   rock, the same energy would be released by throwing the 13000km *earth*
   at (one hell of) a brick wall at roughly 1200 kilometers per *second*.

Duh? r Earth = 26     times r small moon.
thus v earth = 26^3   times v small moon approx = to 17500 times more.
thus m earth = 17500 m small moon.
If E = follow m v^2 then 17500 vEarth^2 = vAsteroid^2,
or v earth = v asteroid / 132.
Suppose 0.5 c for the v asteroid, then you effectively get a movemont of
1200 km/s for the earth (I like to know where the numbers come from).

But where does the "brick wall" comes from? What's it's size and mass?
Infinite? There's no such things as infinites. Not when you start out with
a finite number. And most of all, in space things always move relatively to
each other. One could as well say that it's the earth moving at half c
toward an imoobile asteroid than the reverse.`

The 1200 km/s you got as a result simply is the SPEED at wich the earth
would be kicked out of its orbit in a perfect ELASTIC COLLISION, the kind
of collision where all the KE berfore the collision stays KE after. In such
a scenario, the asteroid would perfectly "rebounds" perfectly elascticly on
the earth, and his speed would be far greater than 1/2 c. But in REAL LIFE,
you DON'T have that kind of collision. We have PLASTIC collision. And a lot
of the energy (and I MEAN a lot) is spent just to change the shape of the
other object, to change it's temperature, etc... THAT is why there is a
crater after the event...

   It wouldn't be dust, but there wouldn't be much left. Of course, the
   asteroid would probably puncture and go through, so not all of this
   energy would actually get expended on our erstwhile planet, but still.

Puncture and go through? If it's big enough, yes. That's why two colliding
stars (computer simulation) appear like two blobs that temporarily become
only one blob and then CONTINUE through the other (or "rebound" back, since
by that time their indivuduality has gone and they were two similar stars
anyway). It's kind of a weird effect, and I assure you that even though
stars are far more prone to nuclear reactions than planets ;-) that the two
stars DIDN'T go "KABLOOIIEE!" Much of the original mass stays after the
collision, even collision as drastic as these.

Come on: If even after a NUCLEAR reaction in a NUCLEAR bomb only *_less
than a few percent_* of the PLUTONIUM actually gets to fission, how do you
think ole rocky earth (or newbie suicidal asteroid) will lose of his mass?

   >But my final argument lays in space for all to see: Just look at the moon.
   >It has very big craters on it. Okay, the moons doesn't have life on it, but
   >it's not like it ever did. Most important is the craters that shows that
   >indeed the moon received MASSIVE "damage" in the past, from BIG craters.

   For one thing, those craters can not have been caused by anything
   moving at anywhere near relativistic speeds. The escape velocity of
   the *sun* is no more than a 100 km/sec or so, much less the earth.
   Energy goes up as velocity squared. (BTW, you forgot to square your
   "scale" factor in your example to get the same relative energy).

Following E = m v^2:
Ah. but then sqrt(mass) can compensate for speed. An asteroid 10 times
bigger has 10e3 more mass and so it can go 31.6  times slower and still
carry the same momentum. Thus if all other things being equals you "change
scale" from a factor like my "giant's descriptions" of 10e7 to have a 1.3
meters wide planet earth and a 30 m/s light speed, then the ration of
"augmentation" in Kinetic energy for a ball of fixed size (say 1 meter)
going at a fixed spedd (say 1 m/s) is 31.6^7 times better, IF you consider
yourself acting over a change in scale of 1^7 and that in fact the ball
always stays 1 meter big and goes at 1 meter per second. That means that
a 1e7 meter size 1*7 meter speed gicantic ball has 31.6^7 times more
momentum energy in it than a one meter size 1m/s speed ball RELATIVELY to
the scale change.
That's quite a punch.
But DENSITY augments with r^3 !!! So your the DENSITY of matter RELATIVELY
to a 1^7 scale change of a 1e7 meter size 1*7 meter speed gicantic ball is
10^21 times more than a one meter size 1m/s speed ball, considering that
it's you who change size and the ball stays 1 meter wide and still moves at
1 meter per second.

Now, if after doing the scale change you see that your increase in momentum
relative the scale chage (I specify because someone could mistake that with
the REAL increase in momentum) doesn't change is the SAME way as your
increase in density, then that means that objects don't react exactly the
same way when you change how big they are. That means that (arbitraryly
stooppid example) if at the true scale a collision between two of these 1
meter wide balls make a sound that goes "splat! kabooom!" that at the other
scale level the (surely not the same balls if you put one balls ...

read more »

Let me first say that I made an egregious error in my previous posting
regarding the momentum transferred...see below.

In an article, Ran...@siegfried.vlsi.polymtl.ca (Patrick Rannou) writes:

Let's use something a little closer to analogous, try shooting a .223
caliber rifle bullet at a solid glass paperweight. Of course, this
isn't anywhere near the same relative energy as the 1e16 kg rock
moving at .5c in the example I was discussing, but it will have to do.
The paperweight doesn't "vaporize" exactly, which is exactly what I
said. It does, however, turn into a nicely scattered pile of rubble.

You're assertions seem to be based on some kind of human scale damage
combined with comparing damage done by meteor strikes. Maybe this will
be a vivid analogy (yes, I realize that there are some glaring
problems with this due to the change in energy and momentum scale):

Say it takes x amount of energy to rubblize and disperse v amount of
volume on the surface of the earth. There are meteor craters on earth
at least as large as 1km deep and 50km wide.  They were created by a
meteor moving at no more than 50km/sec (a very high guess, unless some
alien grabbed a rock and flung it at us at near c, that is :-).

If the *same sized* (much less a 500km) meteor hit the earth at .5c,
it would expend 27,000,000,000 times as much energy, and thus should
disperse something like 27,000,000,000 times as much volume as the
slow meteor.  The volume of the earth is something like
850,000,000,000 cubic km.  The volume of one of these craters is
something like 2,000 cubic km, but only half of that is probably
displacement.

Therefore, the same size meteor travelling at .5c, should do to the
mass of the earth about 200 times what that slow meteor did to the
matter in the crater it created. I.e. it should, *at least*, rubblize
and scatter much of the earth quite nicely.

Sort of like when you drop a water balloon into molten steel it heats
without exploding, right? (hint: wrong)

Of course it's not coincidence. In fact, if you look at the 2
equations E=mc^2 and KE = 1/2mv^2, you will see that a rock moving at
.5c contains 1/8 the kinetic energy that a solid 50/50
matter/antimatter rock of the same size contains in *mass* energy at
"rest". In other words, *if* all the energy of the 1e16 rock moving at
.5c is expended on the earth, it is the exact same amount of energy as
would be released if a 5e14 kg rock made of solid antimatter were to
combine with the earth. Note: this produces only hard gamma radiation,
which ultimately goes solely into *heat* of the surrounding matter. It
is the vaporization and expansion of *this* that causes an explosive
effect.

There are 3 choices for where the kinetic energy of the fast rock
could go:

1) Heat, which has exactly the same effect no matter how it is
produced.

2) Momentum. Consider that in a perfectly inelastic collision, the
earth will go from a relative standstill to a relative 8 km/sec (my
earlier assertion was wrong, see below) in a very short period of
time. I.e. if it takes 1 second, this is a 900 g acceleration, which
will do some interesting things to the planet. It can't be *much* more
than 1 second, because the rock would traverse the entire diameter of
the planet in about 1/10 of a second at .5c, which would imply an
acceleration of 9000 g.

3) The rock could simply rip right through and not expend all of its
energy on the earth. I'm not sure how likely this is. At first, I
would have guessed that it was very unlikely, because the rock is
likely to shatter into a billion pieces when it hits, but a shotgun
blast rips right through stuff quite nicely too, so my intiutive sense
is a bit confused.

It takes *much* less energy to reduce something to a cloud of rubble then
it does to completely annihilate it.

Again, if it is a perfectly inelastic collision, all of that energy is
transferred to the earth, mostly in the form of kinetic energy and
heat, probably...where else would it go? BTW: 30J/gram (whether that's
the right number or not) is enough kinetic energy to get that gram
moving at 8 km/sec, well on its way to escape velocity.

Of course not. But, if *all* of the momentum is simply stuck to the
earth, the resulting momentum of the combined system *has* to be the
same, otherwise you have violated CLM. Therefore, the earth (or parts
of it) *does* take off rapidly.

However, I miscalculated quite a bit because P=mv. Therefore, the
earth "only" takes off at 8 km/second. If it were a perfectly elastic
collision, it would take off at *twice* that speed. I'd say it's much
more likely that a huge fraction of the earth's mass (like, say, 3/4)
takes off at higher speed, and the rest coelesces back into a smaller
mass some time later. For a while, though, it's all likely to be
rubble.

It won't, of course. I mean, *some* of it will probably go into
fission of the heavier elements of the rock and the earth, but not
*that* much. Some of the hydrogen probably will fuse, but again, not
that much.

--
"When you're down, it's a long way up
 When you're up, it's a long way down
 It's all the same thing
 And it's no new tale to tell"                      ../ray\..

   Let's use something a little closer to analogous, try shooting a .223
   caliber rifle bullet at a solid glass paperweight. Of course, this
   isn't anywhere near the same relative energy as the 1e16 kg rock
   moving at .5c in the example I was discussing, but it will have to do.
   The paperweight doesn't "vaporize" exactly, which is exactly what I
   said. It does, however, turn into a nicely scattered pile of rubble.

Comparison NOT valid: characteristics of bullet very different from
characteristics of glass. Asteroid and Earth have same characteristics,
apart size and relative speed. Don't mix oranges with apples. If I said
"you shoot a glass marble at a steel paperweigth, then the paperweigth is
OK and the marble is destroyed, that would be doing the same thing...
reversed. You must have objects of the same density (approx).

   Therefore, the same size meteor travelling at .5c, should do to the
   mass of the earth about 200 times what that slow meteor did to the
   matter in the crater it created. I.e. it should, *at least*, rubblize
   and scatter much of the earth quite nicely.

Maybe, but I don't think so. I'll have to check on a physics teacher.

   >that when you heat it melts without exploding. Sheesh. How many times will
   >I have to repeat it?

   Sort of like when you drop a water balloon into molten steel it heats
   without exploding, right? (hint: wrong)

Comparison NOT valid: characteristics of ballon very different from
characteristics of steel. Asteroid and Earth have same characteristics,
apart size and relative speed. Don't mix oranges with apples. If I said
"you shoot a molten steel ballon into water, then the steel is
OK and the water is destroyed, that would be doing the same thing...
reversed. You must have objects of the same density (approx).

   2) Momentum. Consider that in a perfectly inelastic collision, the
   earth will go from a relative standstill to a relative 8 km/sec (my
   earlier assertion was wrong, see below) in a very short period of
   time. I.e. if it takes 1 second, this is a 900 g acceleration, which
   will do some interesting things to the planet. It can't be *much* more
   than 1 second, because the rock would traverse the entire diameter of
   the planet in about 1/10 of a second at .5c, which would imply an
   acceleration of 9000 g.

Since the collision is inelastic, a good part of the energy would go into
deformations. You acknoledge that. But then you say that the Earth would
get all this energy and start to move, like in a perfectly elastic
collision. Are you trying to put here more energy than there was at the
start? Some E will be dissipated as heat, some as deformation, melting,
vaporization, and nuclear reaction with the stone. *NO* "matter-antimatter"
reaction takes place. Other E would stay as momentum.

This is becoming far more complex than one might think, heh?

   3) The rock could simply rip right through and not expend all of its
   energy on the earth. I'm not sure how likely this is. At first, I
   would have guessed that it was very unlikely, because the rock is
   likely to shatter into a billion pieces when it hits, but a shotgun
   blast rips right through stuff quite nicely too, so my intiutive sense
   is a bit confused.

That may or may not be possible.

   >>still white... Mass can pulverise more mass, but not "atomize" more. Only
   >in a matter-antimatter reaction would you achieve the fieat of utterly
   >destroying THE SAME amount, no more, of matter, than you used
   >anti-matter.

   It takes *much* less energy to reduce something to a cloud of rubble then
   it does to completely annihilate it.

Granted. But try to calculate the energy contained within the Earth Gravity
well. I'm sure you'll see that a mere 30kJoule per kilogram of Earth's mass
isn't enough to blow all rubble into space. I say there Earth, while still
maybe damnaged at our level, would still look like a nice spjerical ball
even after the collision.

   Again, if it is a perfectly inelastic collision, all of that energy is
   transferred to the earth, mostly in the form of kinetic energy and
   heat, probably...where else would it go? BTW: 30J/gram (whether that's
   the right number or not) is enough kinetic energy to get that gram
   moving at 8 km/sec, well on its way to escape velocity.

BZZT! Wrong answer! Inelastic collision = Good chunk of energy used in the
deformations.

   >the earth, and his speed would be far greater than 1/2 c. But in REAL LIFE,
   >you DON'T have that kind of collision. We have PLASTIC collision. And a lot

   Of course not. But, if *all* of the momentum is simply stuck to the
   earth, the resulting momentum of the combined system *has* to be the
   same, otherwise you have violated CLM. Therefore, the earth (or parts
   of it) *does* take off rapidly.

SUPER BZZZT! PLASTIC = inelastic collision = momentum DOESN'T stay the
same. Your turn to go back to a physics book. Take to silly putty ball
moving toward each other at 1 meter per second. they each have momentum
energy, right? they scollide and merge into one bigger silly pputty ball.
movement: 0, so momentum 0. Were did all the energy go? into the
"fusing/merging" of the silly putty. It takes energy to change an objets's
shape, you know...

   However, I miscalculated quite a bit because P=mv. Therefore, the
   earth "only" takes off at 8 km/second. If it were a perfectly elastic
   collision, it would take off at *twice* that speed. I'd say it's much
   more likely that a huge fraction of the earth's mass (like, say, 3/4)
   takes off at higher speed, and the rest coelesces back into a smaller
   mass some time later. For a while, though, it's all likely to be
   rubble.

Okay. I got you HERE. Say 1/10 or the earth's mass reaches JUST under
escape velocity (i.e. not enough for this mass to go into space: it will
fall down again). Calculate the momentum energy contained into THAT. I'm
pretty sure it's THOUSANDS of times bigger than the total momentum energy
that was initially into the asteroids, thus PROVING without a doubt that
the asteroid CANNOT send even a small part of earth flying into deep space,
since it doesn't have enough energy to make that feat. Like Beverly would
ssay: HA, got you there.

Try to get out of THAT one.

Paradak.

In an article, Ran...@siegfried.vlsi.polymtl.ca (Patrick Rannou) writes:

The momentum of the entire system *has* to be conserved. This is a
fundamental law. Remember, however, that momentum is a vector.

The momentum of that entire system is exactly 0 before the collision.
It is exactly 0 after the collision. The momentum (P=mv->) of the
first ball exactly cancels (in a vector addition sense) the momentum
of the second ball (P=mv<-) In the case of the earth and the asteroid,
I am trying not to make any assumptions about which direction the
asteroid hits from (not that it makes *that* much of a difference in
overall effect). Therefore, I'm simply looking at the relative
accelerations involved.

The momentum of a 500 km diameter asteroid at .5c (which is what was
discussed in the original post, though that poster got the mass, and
thus the momentum wrong by 4 orders of magnitude...it's about 1e20 kg,
compared to the earth's 1e24 kg or so) is the same as that of the
earth moving at about 8km/sec, assuming similar density.

Mind you, that's one fucking huge asteroid. However, for smaller
rocks, see below.

Actually, it isn't, in fact the momentum is considerably less (about a
factor of 16 or so). Proof:

   m(earth) = 1e24 kg or so.
   m(500 km asteroid) = 1e20 kg or so (assuming similar density).
   Momentum (p=mv) of asteroid at .5c = 1e20 kg * .5c = 1.5e28 kg m/s
   Momentum of 1/10th m(earth) at 11 km/sec = 1e23 kg * .00003 c = 9e26 kg m/s
   KE(asteroid) = 1/2 * 1e20 kg * .5c ^ 2 = 1e36 Joules
   KE(1/10th earth) = 1/2 * 1e23 kg * .00003c ^ 2 = 4e30 Joules

Actually, it doesn't. If you take the entire momentum of the asteroid
and transfer it to the earth, *every single bit* of the earth's mass
will contain enough momentum (different from kinetic energy) to move
at 8 km/sec (slightly lower than earth's escape velocity, somewhat
higher than the escape velocity of the sun from 1AU).

If you transfer the momentum into 1/10 of the earth's mass, that mass
will be moving 10 times as fast, because p=mv is *conserved*. In fact,
if you apply the momentum to 8/11ths of the earth's mass, every bit
will be moving at 11km/sec, which is earth's escape velocity (of
course, the escape velocity is a funny concept when 8/11th of the
earth's mass is no longer in the general vicinity).

The earth's orbital velocity is about 3km/sec, just to give you a
comparison (1AU * pi * 2 / 3e8 seconds, since you like to know where
the numbers come from).

Even if we use the mass (~1e16 kg) erroneously calculated by the original
poster, I'm curious, what do you think would happen if a 1e15 kg chunk
of pure anti-matter hit the earth? I'll give you a hint.
Matter/antimatter annihilations are 100% gamma rays that cause
explosions *solely* by heating up the surrounding material. In fact,
by and large, the same is true of nuclear bombs...they don't "explode"
in the way that a stick of dynamite does...they just heat the
surroundings via gammas and neutrons. Exactly the same amount of
energy would be expended on the earth (via heating and momentum
transfer) by a 8e15 kg rock hitting the earth at .5c.  Proof:

   Energy expended by 1e15 kg antimatter blob:
      E = mc^2
      m = 2e15 (1/2 matter and 1/2 antimatter)
      E = 1.8e32 Joules (mostly in heat)

   Energy expended by 8e15 kg rock at .5 c:
      KE = 1/2 mv^2
      m = 8e15
      v = 1/2 c
      KE = 1.8e32 Joules (in heat and other kinetic energy).

Whether this goes into heat or into kinetic energy is somewhat
irrelevant. It's still about 1e17 megatons-tnt of energy. It's got to
go somewhere. Again, that's 100,000 million million 1-megaton-bombs
worth of energy. And it has an attitude.

--
"When you're down, it's a long way up
 When you're up, it's a long way down
 It's all the same thing
 And it's no new tale to tell"                      ../ray\..

   In an article, Ran...@siegfried.vlsi.polymtl.ca (Patrick Rannou) writes:
   >SUPER BZZZT! PLASTIC = inelastic collision = momentum DOESN'T stay the
   >same. Your turn to go back to a physics book. Take to silly putty ball

   The momentum of the entire system *has* to be conserved. This is a
   fundamental law. Remember, however, that momentum is a vector.

   >moving toward each other at 1 meter per second. they each have momentum
   >energy, right? they scollide and merge into one bigger silly pputty ball.
   >movement: 0, so momentum 0. Were did all the energy go? into the

   The momentum of that entire system is exactly 0 before the collision.
   It is exactly 0 after the collision. The momentum (P=mv->) of the
   first ball exactly cancels (in a vector addition sense) the momentum
   of the second ball (P=mv<-) In the case of the earth and the asteroid,
   I am trying not to make any assumptions about which direction the
   asteroid hits from (not that it makes *that* much of a difference in
   overall effect). Therefore, I'm simply looking at the relative
   accelerations involved.

   The momentum of a 500 km diameter asteroid at .5c (which is what was
   discussed in the original post, though that poster got the mass, and
   thus the momentum wrong by 4 orders of magnitude...it's about 1e20 kg,
   compared to the earth's 1e24 kg or so) is the same as that of the
   earth moving at about 8km/sec, assuming similar density.

OK.

   Mind you, that's one fucking huge asteroid. However, for smaller
   rocks, see below.

Maybe we should more properly call it a small moon, or simply a moon
(considering that many moons in the solar system are smaller than that).

   >   mass some time later. For a while, though, it's all likely to be
   >   rubble.
   >
   >Okay. I got you HERE. Say 1/10 or the earth's mass reaches JUST under
   >escape velocity (i.e. not enough for this mass to go into space: it will
   >fall down again). Calculate the momentum energy contained into THAT. I'm
   >pretty sure it's THOUSANDS of times bigger than the total momentum energy

   Actually, it isn't, in fact the momentum is considerably less (about a
   factor of 16 or so). Proof:

      m(earth) = 1e24 kg or so.
      m(500 km asteroid) = 1e20 kg or so (assuming similar density).
      Momentum (p=mv) of asteroid at .5c = 1e20 kg * .5c = 1.5e28 kg m/s
      Momentum of 1/10th m(earth) at 11 km/sec = 1e23 kg * .00003 c = 9e26 kg m/s
      KE(asteroid) = 1/2 * 1e20 kg * .5c ^ 2 = 1e36 Joules
      KE(1/10th earth) = 1/2 * 1e23 kg * .00003c ^ 2 = 4e30 Joules

   >that was initially into the asteroids, thus PROVING without a doubt that
   >the asteroid CANNOT send even a small part of earth flying into deep space,

   Actually, it doesn't. If you take the entire momentum of the asteroid
   and transfer it to the earth, *every single bit* of the earth's mass
   will contain enough momentum (different from kinetic energy) to move
   at 8 km/sec (slightly lower than earth's escape velocity, somewhat
   higher than the escape velocity of the sun from 1AU).

OK. OK....

   If you transfer the momentum into 1/10 of the earth's mass, that mass
   will be moving 10 times as fast, because p=mv is *conserved*. In fact,
   if you apply the momentum to 8/11ths of the earth's mass, every bit
   will be moving at 11km/sec, which is earth's escape velocity (of
   course, the escape velocity is a funny concept when 8/11th of the
   earth's mass is no longer in the general vicinity).

   The earth's orbital velocity is about 3km/sec, just to give you a
   comparison (1AU * pi * 2 / 3e8 seconds, since you like to know where
   the numbers come from).

   Even if we use the mass (~1e16 kg) erroneously calculated by the original
   poster, I'm curious, what do you think would happen if a 1e15 kg chunk
   of pure anti-matter hit the earth? I'll give you a hint.
   Matter/antimatter annihilations are 100% gamma rays that cause
   explosions *solely* by heating up the surrounding material. In fact,
   by and large, the same is true of nuclear bombs...they don't "explode"
   in the way that a stick of dynamite does...they just heat the
   surroundings via gammas and neutrons. Exactly the same amount of
   energy would be expended on the earth (via heating and momentum
   transfer) by a 8e15 kg rock hitting the earth at .5c.  Proof:

      Energy expended by 1e15 kg antimatter blob:
         E = mc^2
         m = 2e15 (1/2 matter and 1/2 antimatter)
         E = 1.8e32 Joules (mostly in heat)

      Energy expended by 8e15 kg rock at .5 c:
         KE = 1/2 mv^2
         m = 8e15
         v = 1/2 c
         KE = 1.8e32 Joules (in heat and other kinetic energy).

   Whether this goes into heat or into kinetic energy is somewhat
   irrelevant. It's still about 1e17 megatons-tnt of energy. It's got to
   go somewhere. Again, that's 100,000 million million 1-megaton-bombs
   worth of energy. And it has an attitude.

OK, OK... sheesh. I was wrong, OK?

But my original intention was that I don't think the surrounding earth
material, being ROCK, which doen't "burn" or "fuse" or "react" or whatever
easily, would not go "KABOOOOM!" when the asteroid hit the heart. Okay, I
agree that a 500km asteroid at 0.5 c is pretty much like throwing a
baseball at a good speed into a meter-wide globe of water floating in free
space... we would all most likely die. But the earth wouldn't simply go
nova!!! It's not a chain reaction!!! I'd say that even if the wave effects
are enough to crush/melt the eaerth's crust, the overall effect would be
more like  "SPALOOOSH" than "KABOOOM". Why? Because I say that the BIGGER
part of the energy would go into the shape altering, the destruction of
bonding between each molecule of cristalline rock in both the asteroid and
the earth.

And at these scales, you DON'T have a clear "KABOOM" from the asteroid the
moment it touches the ground. The time it takes for the pressure wavefront
to propagate in it, it already has reached -inside- the planet. It may even
simply "pass through" the earth, doing (relatively) minor damage (we would
still die, but the earth would live...).

Paradak.

That's your definition.  I personally consider "relativistic" to mean
"where relativity becomes significant."  In other words, when it starts
changing your numbers.  Assuming that you are using three significant
figures (1.02 x 10^2), I consider relativistic to mean that it will
_change_ this value.  In other words, when the gamma factor is 1.01.  
(The gamma factor is that portion of the relativity equations that
determines how different things act.)

At 0.8c (that's a whopping 80% the speed of light), gamma = 1.67, which
is getting a little past "significant."  A quick calculation shows that
for gamma = 1.01, your speed is v = 0.14c, or 14% lightspeed.  I consider
that quite significant.  For precise measurements, 0.01c would still be
considered relativistic.

[entire rest of quoted but unused post deleted]

----------
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In article <ga5iLB5w1...@west.darkside.com>, m...@west.darkside.com (Erik Max Francis) writes...

     This is kind of off of the subject, but while everyone is talking about
collisions.  On r.a.s.c the topic has been about the fact that people don't
explode when they get blown out into a vaccuum.  (EEWWWWW!!)  Anyway, asssuming
its pretty chilly in space, what kind of damage would a frozen Romulan to
if it impacted a space ship that was travelling and full impulse.  I know
that the navy did some research and that a frozen chicken can take out the
best engines and windscreens at about 400 miles an hour.  And that a plastic
pellet can leave a hole in steel that is six inches across if it impacts at
12,000 mph or more.  
     I know that the shuttle lost tiles when it hit paint chips from previous
missions.  Does the Enterprise run with some kind of shields up when its doing
full impulse?  I realize in warp, it is protected by a warp bubble, but what
kind of protection does it have when doing impulse?  

   Just curious...

Sarah J. Simons
Arizona State University CS
"generic .sig here"    
[B

collo...@envmsa.eas.asu.edu (Upsilon Pi Epsilon) writes:

I don't think you'd see exploding corpses a la "Total Recall."
Decompression of a humanoid body would result in an untidy appearance,
but I don't think you'd have an explosive decompression.

Actually bodily fluids boil away in space rather than freeze.  The
resulting solids and the liquids do freeze, but exposure to vacuum is
a bit more complex than a cloud of expanding entrails.

I've seen these films.  Gross.

I don't think the warp bubble protects the ship in a shield-like
fashion.  But to address the question, starships have a twofold
navigational deflector system.  One phase is a field that deflects
small particles, and the other is a directed beam used to clear
comparatively large particles (e.g., asteroids, debris, frozen
chickens) from its path.
--------------------------------+---------------------------------------
Jay Windley                     | jwind...@asylum.utah.edu
Department of Computer Science  +---------------------------------------
Univ. of Utah, Salt Lake City   | Propter causam veritatis non tacebo.

In the referenced article, m...@west.darkside.com (Erik Max Francis) writes:

For back of the envelope calculations, most physicists I know only use
1 significant digit (if that, actually, most of them seem to use only
the order of magnitude). If your talking about general behaviors, a
factor of 2 is pretty insignificant, and you don't start getting that
until somewhere greater than .8c.

But, I mean, really, this is pretty silly. I know plenty of people
that use 4 or 5 significant digits for many calculations, does that
make .01c relativistic? Well, ok, maybe it does. I know people that
like to use single precision floats in calculations. Does that make
.000001c relativistic? Well, ok, maybe it does.

On the other hand, I think we will have no trouble agreeing that .9c
is much more relativistic than .8c, which is much more relativistic
than .1c (and .99c is much more relativistic than .9c).

While we can bicker back and forth for a long time, I'd argue that
relativity becomes interesting for casual conversations only at that
point where it starts to affect the order of magnitude of the
resulting number. Ok, maybe a factor of 2 is enough.
--
"When you're down, it's a long way up
 When you're up, it's a long way down
 It's all the same thing
 And it's no new tale to tell"                      ../ray\..

-> ... Anyway, asssuming its pretty chilly in space,

Arrrghh. Space is *not* cold. It's a *vacuum* - (ignoring the odd atom
here and there).

It has no temperature but tends to insulate like a Thermos (tm) or
Dewar flask.

How does the Enterprise get rid of surplus heat?

--
 Regards,                         p...@x.co.uk           IXI Ltd.
        Paul Davey                p...@ixi.uucp          62-74 Burleigh St.
                                  ...!uunet!ixi!pd     Cambridge,  U.K.
 "These _are_ interesting times"  +44 223 462 131      CB1  1OJ

 The same way the sun  RADIATES its' heat to the earth.
--
*************************************************************************
*            Sig. Test                                                  *
*************************************************************************

   -> ... Anyway, asssuming its pretty chilly in space,

   Arrrghh. Space is *not* cold. It's a *vacuum* - (ignoring the odd atom
   here and there).

   It has no temperature but tends to insulate like a Thermos (tm) or
   Dewar flask.

   How does the Enterprise get rid of surplus heat?

VERY EASY. When you have perfected heat converter like they do, you have:

(I'm sure theyu do it in many less steps...)

1- convert heat to make the ship cooler, and the water in a machine VERY
hot.
2- Use the energy in the steam to power up an rotor. That makes the water
work, losing heat and thus you get the water back into the converter again.
3- Use the energy in the rotor to produce eletricity. That makes the rotor
go slower and you still can use the same rotor to receive energy form the
water.
4- use electricity to make an halogen lamp go on. OK, so a little heat is
back, but so what? You still have some light energy.
5- Now use a photon collector from ST:IV TVH
6- Pump the photons ino the dilithium cristals
7- The energy is used to propel the ship.

So, if you have enough people on board, you just created perpetual
movement! ;-) (OK, it's only a joke!)

Paradak.

   -> ... Anyway, asssuming its pretty chilly in space,

   Arrrghh. Space is *not* cold. It's a *vacuum* - (ignoring the odd atom
   here and there).

Space *IS* cold: about -270 'C, or 3 K. But you don't fel it cold...

   It has no temperature but tends to insulate like a Thermos (tm) or
   Dewar flask.

...Because rate of heat exchange is dependent not only on the difference in
temp (here instead of body 27 'C and air 22'C, whch is 5 'C difference, you
have "space" -270'C, which is 297 'C, or about 60 times more), but also on
the current number of collisions, which is directly proportinnal to
pressure.

And since pressure is THOUSANDS of times lower in space, heat exchange will
be at least 50 times slower than on earth, effectively making space a very
good "thermos"... And producing heat inside a thermos can have devastating
results...

Paradak.

--text follows this line--

   -> ... Anyway, asssuming its pretty chilly in space,

   Arrrghh. Space is *not* cold. It's a *vacuum* - (ignoring the odd atom
   here and there).

Space *IS* cold: about -270 'C, or 3 K. But you don't fel it cold...

   It has no temperature but tends to insulate like a Thermos (tm) or
   Dewar flask.

...Because rate of heat exchange is dependent not only on the difference in
temp (here instead of body 27 'C and air 22'C, whch is 5 'C difference, you
have "space" -270'C, which is 297 'C, or about 60 times more), but also on
the current number of collisions, which is directly proportinnal to
pressure.

And since pressure is THOUSANDS of times lower in space, heat exchange will
be at least 50 times slower than on earth, effectively making space a very
good "thermos"... And producing heat inside a thermos can have devastating
results...

Paradak.

collo...@envmsa.eas.asu.edu (Upsilon Pi Epsilon) writes:

Yeah, it's way off the subject.  :-)  No problem.

This is true.

The Air Force was having trouble with geese crashing through jet fighter
cockpits and seriously injuring pilots.

The Space Shuttle also had some problems with particles hitting the
"wind"shield.  They guessed that it was a fleck of paint from an Atlas
(? or other old rocket) rocket, and it did several thousand dollars in
damage.

It's not a matter of compactness or denseness; when you get up there,it's
just speeds.  To destroy a satellite, all you need to is throw a little
bit of gravel in a retrograde intersecting orbit.  The satellite will be
completely neutralized.

Might be the meaning of "deflectors" . . . who knows.

----------
Erik Max Francis   Omnia quia sunt, lumina sunt.  Coming soon:  UNIVERSE _ | _
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Well, it depends on the person and what calculation they're performing.  
I consider relativistic to be when it would change your otherwise
Newtonian calculation.  That, of course, depends on how many significant
digits you're using and how much you really care about the accuracy of
you're answer.  (Scientifically, if you're not being accurate, you should
use only 1 or 2 significant digit . . . but that's another matter.)

I understand your point, though.

----------
Erik Max Francis   Omnia quia sunt, lumina sunt.  Coming soon:  UNIVERSE _ | _
USmail:  1070 Oakmont Dr. #1 San Jose CA 95117  ICBM:  37 20 N  121 53 W _>|<_
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