on the "dropping rocks" weapons thread
we...@venus.ycc.yale.edu
large enough to cause significant damage masses about 1.0 x 106 metric tons.
Such a mass would be difficult to tow (even for the Enterprise-D), and the
limited velocities permitted by sheering forces on a tractor beam would would
severely reduce maximum velocity. With this in mind, it should be possible for
the "average" planet to destroy both ship and asteroid before "effective" range
has been achieved. Especially, since the asteroid doesn't need to be
destroyed, simply "redirected". Since a 50 megaton nuclear device will destroy
approximately 5.0 x 107 of matter (if detonated inside the mass), an external
detonation should be more than sufficient to both break the tractor, and change
the course of the asteroid. Therefore, strike weapons such as phasers and
(particularly)photon torpedoes, will be far more effective weapons. Assuming,
of course, that the "target" planet has technology even close to that of the
TOS enterprise, not to mention Enterprise-D.
Matt
Drau
First, you only have get the asteroid moving. Once it's accelerated to a fast
enough clip to reach the target within a few hours/days/whatever, you can leave
the rest of the work to gravity. Second, you can always get Scotty/Geordi/
Weasel to rig a shield generator for the asteroid, if you're really concerned
about not having to do things over. With your ship trailing the asteroid, using
it as a shield against fire, you can maintain tractor or immediately reengage it
if it's broken and keep the asteroid on target. Again, if you're really
concerned, you might even put a computer and control thrusters on it. Now
we're getting expensive, but it would have more bang for the buck than a photon
torpedo against a planet.
- Drau
All opinions expressed are facts. Your opininons are irrelevant. You will be
assimilated with the Borg. The Borg will be trashed by A-ko, but this is also
irrelevant.
buck...@woods.ulowell.edu
A 50 Megaton nuclear blast will not 'destroy' 50 million ( I
assume you meant to say 'tons' ) tons of matter, but merely give you
50 million tons of matter in smaller pieces.
The only problem you have with this is the idea of effective
planetary-based weapons. I thought the idea of shielding the rock
was a cute idea.
Without a planetary deflector system ( ala 'Paradise Syndrome' ),
not too much should be able to stop it. ( Short of a similalarly
teched-out space craft, that is. )
Just placing it on a collision course should be sufficient, since
gravity will take over. Moving ANYTHING interplanetary distances over a
reasonably short time will give you more than enough delta-v to give
you a quite respectable impact explosion.
( And yes, when I said 'just drop rocks', I thought it was fairly
obvious that I was reffering to masses in the 10^8 tons+ range. )
Jason Kim
>
>
> A 50 Megaton nuclear blast will not 'destroy' 50 million ( I
>assume you meant to say 'tons' ) tons of matter, but merely give you
>50 million tons of matter in smaller pieces.
I'm assuming you meant to add a smiley to that one.
>
> The only problem you have with this is the idea of effective
>planetary-based weapons. I thought the idea of shielding the rock
>was a cute idea.
> Moving ANYTHING interplanetary distances over a
>reasonably short time will give you more than enough delta-v to give
>you a quite respectable impact explosion.
>
I missed the beginning of this thread, but from what I've read of it,
the idea of dropping rocks on enemies (with all the shields and other
trappings attached) is not just cute, but totally ingenious. I think the
idea could be worked into an entire novel.
And yes, to address what somebody else said about this, an asteroid *is*
too much for the Enterprise to handle. In "Q2," an asteroid threatened
to mash a planet or something and they had a hell of a time dealing with it.
Q suggested that they simply "change the gravitational constant of the
universe."
Wait, was that episode what prompted this thread? If so, I'm sorry for
wasting bandwidth!
Jason Y. Kim
Peter Flugstad
>
>I missed the beginning of this thread, but from what I've read of it,
>the idea of dropping rocks on enemies (with all the shields and other
>trappings attached) is not just cute, but totally ingenious. I think the
>idea could be worked into an entire novel.
>
The idea has been used at least once that I know of, and very effectively.
Check out "The Moon is a Harsh Mistress" by Robert Heinlen (sp?). He uses
rocks with controlling rockets on them... to bomb Earth from the Moon.
I don't remember the size of the rocks, but the idea was to use the
gravitational well of the earth to accelerate the rocks, using the
rockets and the launch time (they used a catapult to launch them from
the moon) to control where they hit... the results (assumming Heinlen
did his research right, which I'll bet he did) were on the order of megatons
of TNT... easily the same as a H-bomb, with basically no radiation...
The book is very good and I highly recommend it...
Pete
sorry, no .sig
Boggles the MIND
I'm confused, maybe I'm missing something, BUT.....
Wouldn't the rocks in question reach terminal velocity- like a skydiver.
When the force of the air friction is equal to the accelleration due to
gravity, the object falling will eventually reach a certain velocity, and
stay there. (Unless you change the gravitational constant of the universe 8^).)
^^^^^ (I mean stay at that velocity)
So I can't picture how a rock, unless it is VERY large, like miles,
could impact the earth with a huge force.
Were the rocks HUGE or am I missing something???
Jim
Drau
The rocks are in space most of the time, outside earth's atmosphere. By
the time they reach the atmosphere, they're at a high enough velocity that they
zip through before friction slows them much.
- Drau
James Gordon Currie
From what I have read, people are *not* taking into account the delta-
V of the "rock" If you take a rock of about 2 or 3 hundred kilos, accellerate
it up to about .9999 C in the dirction of a planet, then leave it alone to
follow it's merry way, it will do a *great* deal of damage to said planet.
Why? for the simple reason that it will have already impacted with the crust
(and likely set-up a shock-wave that will play havock with the planet's
innards) before the atmosphere can affect the thing. Likely the planet's core
and magnetic field (at the least) will be directly affected. At worst, said
planet would break in half... (lots of SF stories deal with precicely that)
Ciao
J G Currie
C
C
C
C
B
Peter Flugstad
>Wouldn't the rocks in question reach terminal velocity- like a skydiver.
>When the force of the air friction is equal to the accelleration due to
>gravity, the object falling will eventually reach a certain velocity, and
>stay there. (Unless you change the gravitational constant of the universe 8^).)
>^^^^^ (I mean stay at that velocity)
> So I can't picture how a rock, unless it is VERY large, like miles,
>could impact the earth with a huge force.
>
> Were the rocks HUGE or am I missing something???
>
> Jim
Not quite... the rock, either big small or whatever will surely reach
terminal velocity, and it will start to burn up, hence the falling stars
at night. But most falling stars that we see at night are very small, on
the order of centimeters, they never get to the ground. If the rock
in question was several hundreds or thousands of tons, there is no way
that it would burn up before hitting the earth. And by the time it hit
the earth, it would be going very fast, and probably still have lots of
momentum, and make a very big explosion...
Also, with a little bit of control, you could arrange for the rock to fall
at an angle to the atmosphere, like the shuttle does when it lands... then
even though you'd still lose a lot of mass to burning off, it'd still make
a big hit.
In thinking about this some more, I believe that Heinlen wrapped his rocks
with some metal to shield it and make it last longer on reentry. I'll have
to go look for sure..
Pete
Max Pandaemonium
> Wouldn't the rocks in question reach terminal velocity- like a skydiver.
> When the force of the air friction is equal to the accelleration due to
> gravity, the object falling will eventually reach a certain velocity, and
> stay there. (Unless you change the gravitational constant of the universe 8^
Not necessarily. Terminal velocity is caused by the friction of the
atmosphere rubbing up against the intruder. This heats the rock and
tends to slow it down.
If you drop an object from a great height (within the atmosphere), it
will reach terminal velocity and go no faster. If you throw an object down
from a great height, it will slow to terminal velocity and go no slower.
If the body is sufficiently large and dense enough, the effect of the
atmosphere will have little effect. Assume that the object is falling at
a speed of 50 km/s. The bulk of the atmosphere lies well below 100 km.
For the rock to fall from 100 km to 0 takes only 2 s. That's not much
time for the viscous force of the atmosphere to slow the body down much.
What you have to worry about here is not the atmosphere slowing the rock
down, but the atmosphere tearing the rock to pieces.
----------
Max Pandaemonium Omnia quia sunt, lumina sunt. Coming soon: UNIVERSE _ | _
USmail: 1070 Oakmont Dr. #1 San Jose CA 95117 ICBM: 37 20 N 121 53 W _>|<_
UUCP: ..!apple!uuwest!max Usenet: m...@west.darkside.com 464E4F5244 |
buck...@woods.ulowell.edu
Once you reach a mass of a few hundred tons ( I believe ), it
really doesn't matter too much about the trip through the atmoshere.
Use a figure of about 100 kilometers for the depth of the
atmoshere. Be generous and say your rock is moving about 10kps.
That gives you about 10 seconds to travel that distance. With a
large enough mass, you pretty much splash into the air.
Sure, you heat the surrounding air to plasma and have some minor
scorching of the rock, but probably very little of the velocity has
been bled off.
Have a bigger or faster rock, then it means less that there is
a trip through air.
I think Lunar escape velocity is just under 3kps and terrestrial
is, what? 20kps?
Say your rock starts at 3kps and travels a straight path of about
390000km. That gives you a travel time of about 37 hours. Now factor
in acceleration. ( I'll have to play with this to figure it out, no
tme now. )
Fun, huh?
And that's a short-range weapon.
M.D. O'Leary
Joshua Susser
>
> In article <1992May10.2...@acsu.buffalo.edu> v128...@ubvmsd.cc.buffalo.edu (Boggles the MIND) writes:
> > I'm confused, maybe I'm missing something, BUT.....
> >Wouldn't the rocks in question reach terminal velocity- like a skydiver.
> > So I can't picture how a rock, unless it is VERY large, like miles,
> >could impact the earth with a huge force.
> >
> > Were the rocks HUGE or am I missing something???
>
> terminal velocity, and it will start to burn up, hence the falling stars
> at night. But most falling stars that we see at night are very small, on
> the order of centimeters, they never get to the ground. If the rock
> in question was several hundreds or thousands of tons, there is no way
> that it would burn up before hitting the earth. And by the time it hit
> the earth, it would be going very fast, and probably still have lots of
> momentum, and make a very big explosion...
>
> Also, with a little bit of control, you could arrange for the rock to fall
> at an angle to the atmosphere, like the shuttle does when it lands... then
> even though you'd still lose a lot of mass to burning off, it'd still make
> a big hit.
>
> In thinking about this some more, I believe that Heinlen wrapped his rocks
> with some metal to shield it and make it last longer on reentry. I'll have
> to go look for sure..
Nope, In "The Moon is a Harsh Mistress," the rocks were cased in metal so
the linear accelerator would have something to grab magnetically.
A rock that was even a few dozen feet across would probably make it to the
surface just fine, and it would hit going at least a few miles a second.
Just remember that escape velocity is about seven miles a second, so
anything falling from beyond orbit would be coming in about that fast -
probably not much faster or it would just miss the Earth entirely.
Atmospheric friction might slow it down a bit, but just think: travelling at
5 miles a second through about 100 miles of noticable atmosphere, there's
only about 20 seconds for the atmosphere to brake the meteor. And remember,
the air is pretty thin way up there. I'm not an aeronautical engineer or
anything, but that doesn't seem like much time to slow something that
massive down that much.
Joshua Susser, Object Percussionist
Apple Computer, Advanced Technology Group
inet: sus...@apple.com | link: susser.j | phone: 408/974-6997
Peter Flugstad
>In article <1992May11....@wuecl.wustl.edu>, pe...@wucs1.wustl.edu (Peter Flugstad) writes:
>>
>> In thinking about this some more, I believe that Heinlen wrapped his rocks
>> with some metal to shield it and make it last longer on reentry. I'll have
>> to go look for sure..
>Nope, In "The Moon is a Harsh Mistress," the rocks were cased in metal so
>the linear accelerator would have something to grab magnetically.
>
Thanks, I couldn't remember the reason for the metal casing...
>Joshua Susser, Object Percussionist
>Apple Computer, Advanced Technology Group
>inet: sus...@apple.com | link: susser.j | phone: 408/974-6997
Pete
sorry, no .sig
LUCIFER
have any of you heard of the siberia explosion circa 1908?
a largish asteroid was falling to eath, slowly burning up, when enough
holes were burnt through it for it to combust all at once (sort of like
gunpowder). It exploded in the air, leveling a 10 mile radius, and even
derailing a train 400 miles away.
Max Pandaemonium
> The rocks are in space most of the time, outside earth's atmosphere. By
> the time they reach the atmosphere, they're at a high enough velocity that th
> zip through before friction slows them much.
A quick calculation, using an equation I derived [rubknucklesonshirt],
shows that a body falling from Moon orbit to the Earth's atmosphere would
achieve a velocity of around 11 km/s.
Hmm. Not bad.
Mike Sellers
> >
> >
> > A 50 Megaton nuclear blast will not 'destroy' 50 million ( I
> >assume you meant to say 'tons' ) tons of matter, but merely give you
> >50 million tons of matter in smaller pieces.
>
> I'm assuming you meant to add a smiley to that one.
I'm assuming you both know that the "megatons" refers to the TNT
equivalent, not the matter destroyed.
> I missed the beginning of this thread, but from what I've read of it,
> the idea of dropping rocks on enemies (with all the shields and other
> trappings attached) is not just cute, but totally ingenious. I think the
> idea could be worked into an entire novel.
See Niven & Pournelle's (I think it was them) _The Mote in God's Eye_.
This is idea is used, though not in the present tense. Heck, then read
Niven's _Lucifer's Hammer_ for what _one_ good size comet could do...
> Jason Y. Kim
Don't read this.
This is line fodder
for our draconian
P
N
E
W
S
program
G
R
R
R
!
--
Mike Sellers mi...@isgtec.com
User Interface Team Leader
ISG Technologies, Toronto, Canada
"Actum ne agas" -- Do not do what has been done.
buck...@woods.ulowell.edu
> In article <1992May9.0...@husc3.harvard.edu> ki...@husc10.harvard.edu (Jason Kim) writes:
>> In article <1992May8...@woods.ulowell.edu> buck...@woods.ulowell.edu writes:
>> >
>> >
>> > A 50 Megaton nuclear blast will not 'destroy' 50 million ( I
>> >assume you meant to say 'tons' ) tons of matter, but merely give you
>> >50 million tons of matter in smaller pieces.
>>
>> I'm assuming you meant to add a smiley to that one.
>
> I'm assuming you both know that the "megatons" refers to the TNT
> equivalent, not the matter destroyed.
>
HA! Thanks for pointing that out!
When I replied to the original message I only wrote it as a 50 Mt
warhead v. a 50 Mt mass. You are quite correct that the mass of the pieces
that remain is the same as that of the original target, just in a different
form.
Mechanical Eng consultant
>In article <1992May9.0...@husc3.harvard.edu> ki...@husc10.harvard.edu (Jason Kim) writes:
[megaton misinformation removed]
>> I missed the beginning of this thread, but from what I've read of it,
>> the idea of dropping rocks on enemies (with all the shields and other
>> trappings attached) is not just cute, but totally ingenious. I think the
>> idea could be worked into an entire novel.
>
>See Niven & Pournelle's (I think it was them) _The Mote in God's Eye_.
>This is idea is used, though not in the present tense. Heck, then read
>Niven's _Lucifer's Hammer_ for what _one_ good size comet could do...
>
>> Jason Y. Kim
>
> Mike Sellers mi...@isgtec.com
For an even better use of "mass x velocity= lethal killing force" I
would refer to Robert Heinlein's _The_Moon_is_a_Harsh_Mistress_:
[THERE BE HEINLEIN SPOILERS BELOW]
Mycroft did a wonderful job bashing the stuffings out of Earth targets
with small (~1 km x 1 km x 1 km) rock projectiles encased in a
ferrous jacket and fired from the lunar surface using a magnetic
accelerator. I believe the author maintained that each of these
rocks equalled the explosive force a large nuke, sans radiation.
The beauty of the situation was that it only took a small amount
of energy to reach lunar escape velocity, and from there it was
all downhill into the Earth's gravitational well; conservation of
energy said that the rock would just trade its potential energy
(inherent from sitting at the top of a gravitational well) for
kinetic energy in the form of velocity.
Your day has never really been shot until you've had millions of tons
of superheated gaseous and liquid rock raining down on your head.
--
Tola H. Marts "The same sun that melts the wax will harden clay;
Mech E. at Minnesota and the same rain that drowns the rat will grow the
email flames, etc. to hay; and the mighty wind that knocks us down, if we
TMA...@VX.ACS.UMN.EDU lean into it, will drive our fears away" -Amy Grant
Craig D. Berry
me30...@ux.acs.umn.edu (Mechanical Eng consultant) writes:
>In article <25...@isgtec.isgtec.com> mi...@isgtec.com (Mike Sellers) writes:
>For an even better use of "mass x velocity= lethal killing force" I
>would refer to Robert Heinlein's _The_Moon_is_a_Harsh_Mistress_:
> [THERE BE HEINLEIN SPOILERS BELOW]
>
>Mycroft did a wonderful job bashing the stuffings out of Earth targets
>with small (~1 km x 1 km x 1 km) rock projectiles encased in a
>ferrous jacket and fired from the lunar surface using a magnetic
>accelerator. I believe the author maintained that each of these
>rocks equalled the explosive force a large nuke, sans radiation.
I think you've massively overestimated the size of Mike's rice buckets
(the accelerator ordinarily delivered grain products to Earth). I don't
have the novel handy, but I have the feeling that the buckets were
cylinders around 20m long by 10m wide. This gives a volume of around
1600 cubic meters. Assume you load them with rock and soil with a
density sufficient for a net density of the whole pod of three times
that of water (this is conservative). The mass of each bucket is thus
4.8 million kg, or rougly 5,000 metric tons. Move this up to escape
velocity by dropping it a long way through earth's gravity field, and
you get a kinetic energy at impact of 290 trillion Joules (!!!), which,
for you Star Trek fans, is equivalent to banging together 1.6 grams
each of matter and antimatter (!!!!!). This, for me at least, is a
very satisfactory *ka-boom*.
--
- Craig Berry
"This sentence is true" : Epiminedes' Paradox -- NOT!
".surivorter erutangis a ma I"
Patrick Rannou
[lines deleted on how to make a bomb of a dropped rock]
4.8 million kg, or rougly 5,000 metric tons. Move this up to escape
velocity by dropping it a long way through earth's gravity field, and
you get a kinetic energy at impact of 290 trillion Joules (!!!), which,
for you Star Trek fans, is equivalent to banging together 1.6 grams
each of matter and antimatter (!!!!!). This, for me at least, is a
very satisfactory *ka-boom*.
--
I think that the "KA-BOOM" would NOT be on impact with the earth, but with
earth's atmosphere. Remember that the faster you go, the more resistance
the air provides. So at some big enough speed, entering atmosphere should
be like slamming hard into a wall...
And a big KABOOM into the high atmospheric isn't my idea of the
"devastating" effects of a nuclear explosion.
Also, the "energy" you calculated would be spread over some falling
distance: how high is the atmosphere? 10km, 60? I don't know.
But I think that BEFORE talking about "mega-bomb falling
rocks", one should do A LITTLE research and get some info on how big a
meteorite caused that or such crater. In know there are some good craters
in the USA.
Personnally, I don't think a thousand ton meteorite, accelerated, would do
much damage.
Suppose the surface of your swimming pool is the surface of
the earth. No wind troubles the pool. The water in the pool is the lava in
the earth. Now take some fine blue, green and brown powder and "draw"
continents and oceans on your pool. This fine powder must be able to float
on the waters surface, but if you drop a little water on it, it should
sink. So it floats, but barely. The water is perfectly still...
Nowm take a brick and with all your strength, from a 10 story high
building, let a friend drop it... while with a high-speed camera, you watch
the effects of your "mega asteroid bomb".
Evaluation of the damage:
All the places where the powder has sunk are now exposed lava. All the
places where the pattern is modified are major fault lines, and exposed
lava too. All the places where the pattern is intact but not in the same
position anymore, or where the pattern keeps moving high and down on
small "waves", are areas where MAJOR earthquake happens. All the other
places are unaffected. Notice that the brick will not pierce the bottom of
the pool, and stop without damaging the "other side" of the earth.
This would give you a good idea of the damage of giant 500 km asteroid,
accelerated to near lightspeed, would to to the earth. On the worst case
scenario: nearly half the earth is now in lava turmoil. Many storms and
eartquakes rage on the earth, and everybody dies from the noxious gases
from the open wound in the earth. Least worst case scenario: an area
1000x1000 kilometers is now lava (approximately double the asteroids radius
= the craters radius. Half the asteroids radius = the craters deepness,
which is more than erath's crust so lava would flow right up).
Since this is VERY approximately 1/6*13*13 = 1/954 = 0.1% of earth's
surface, then that means that the noxious gases from the open wound would
create very somber days for earth's inhabitants, but they may survive...
--------- Any rock big enough can kick another one ---------
Paradak.
buck...@woods.ulowell.edu
[ Stuff deleted ]
> This would give you a good idea of the damage of giant 500 km asteroid,
> accelerated to near lightspeed, would to to the earth. On the worst case
> scenario: nearly half the earth is now in lava turmoil.
I beg to differ.
First, I don't think anyone has been referring to masses on quite
that large a scale.
Second, few of the 'rock droppers' have mentioned any relative
velocities greater than a few to a few hundred kilometers/second.
Third, your scenario wildly underestimates the damage caused in
such a cataclysm.
Slam a 500km MARSHMALLOW, let alone asteroid, doing relativistic
speeds into any planetary mass and you'd probably have to search for the
rubble with a microscope.
Patrick Rannou
I beg to differ.
I beg to differ. What do you define by 'relativistic' speeds? Even at .5 c,
time distortion is: to / sqrt( 1 - v^2/c^2 ) = to/sqrt(.75) = to/0.866 =
115% of normal time. That's a 15% distortion in time, so that's
relativistic.
Now suppose a 10 km cubic rock (or a spherical one of same mass), with 10
tons per cubic meter (or change the size a little to acount for real rock
density). That means 1*10E16 kg of rock.
Now, there NO way it would go KABOOM upon touching atmosphere: at these
size, solids don't react the same way. Try to imagine yourself as a BIG
giant floating in space. What is the earth next to you? A big 1.3 meter wide
floating ball (with a scale of your 1 meter being actually 10000 klicks).
That's a scale of 1:1*10E7
Now suppose you move you finger to poke the earth's surface? Your finger is
like a long asteroid 100 klicks from side to side, but maybe 600 klicks
long. So that's too big, you say. Okay, take a penball point from a pen. 1
millimeter to you, 10 kilometers to those on the surface.
The atmosphere at your size is less than a centimeter thick, but is
concentrated on the few first millimeters. The
earth's crust is likewise only millimeters thick.
Now, at these size, the whole thing is much more brittle, because the
elasticity coefficient is, for you, 1*10E7 times smaller. But at the same
time density also increase. Now, throw the penball point at a speed of 10
meters per second (1/3 of c, which IS relativistic). I don't see how this
meteorite can "destroys the earth". They're nothing compared to it.
Okay, there will be major local earthquakes, the earths crust will be
pierced, but the pen point ball wont go through to the other side. It has a
speed of 10 meters per second (for you) and it has 1 meter to stop, high
density material vs high density material.
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.
Now, it is only logical to suppose that the earth would have suffered the
same things. Was the earth destroyed in the past? I think not. The worst
that could happe if a big asteroid fell on us would be:
1- big eartquakes
2- big storms
3- earth crust cut right through in the area of the blast, causing
atmosphere poisoning
4- Major climatic changes
We would NOT, as a race, die, we would ADAPT.
David Paigen
>In article <1992May20...@woods.ulowell.edu> buck...@woods.ulowell.edu writes:
>> This would give you a good idea of the damage of giant 500 km asteroid,
>> accelerated to near lightspeed, would to to the earth. On the worst case
>> scenario: nearly half the earth is now in lava turmoil.
> I beg to differ.
> First, I don't think anyone has been referring to masses on quite
> that large a scale.
> Second, few of the 'rock droppers' have mentioned any relative
> velocities greater than a few to a few hundred kilometers/second.
> Third, your scenario wildly underestimates the damage caused in
> such a cataclysm.
> Slam a 500km MARSHMALLOW, let alone asteroid, doing relativistic
> speeds into any planetary mass and you'd probably have to search for the
> rubble with a microscope.
>I beg to differ. What do you define by 'relativistic' speeds? Even at .5 c,
>time distortion is: to / sqrt( 1 - v^2/c^2 ) = to/sqrt(.75) = to/0.866 =
>115% of normal time. That's a 15% distortion in time, so that's
>relativistic.
I call anything above 1% of C as relativistic.
>Now suppose a 10 km cubic rock (or a spherical one of same mass), with 10
>tons per cubic meter (or change the size a little to acount for real rock
>density). That means 1*10E16 kg of rock.
.
Patrick, read more carefully. We are NOT talking about 10km asteroids,
nor 500km asteroids. We are talking about rocks in the hundreds of
metric tonnes. That's 1e5 kg. NOT 1e16 kg.
But sure, let's take a look at this. At 1% of C, 1e16 kg has kinetic energy
equal to 4.5e28 Joules, or 1.8e29 calories. This equates to over
30 calories for every gram of matter in the earth.
>Now, throw the penball point at a speed of 10
>meters per second (1/3 of c, which IS relativistic). I don't see how this
>meteorite can "destroys the earth". They're nothing compared to it.
Patrick, read more carefully. We are NOT talking about relaticistic
speeds. The object is not to vaporize the planet, it is to subjecate (sp?)
the population. :-) Throwing rocks at relativistic speeds is wasteful
of energy, you can get adaquate results with just a few klicks per second.
No reason to go above 3,000 klicks/sec (1% C).
But sure, let's take a look at this. At 33% of C, 1e16 kg has kinetic energy
equal to 5e31 Joules, or 2e32 calories. This equates to about
35,000 calories for every gram of matter in the earth. I think that
will probably be enough to turn the planet into its constituent subatomic
particles.
>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.
None of these craters were caused by objects moving at relaticistic speeds.
Objects inside our solar system move at relative velocities on the order
of 50 klicks/sec (high). This is no where NEAR the 3000 klicks/sec of 1% C.
However, I suspect that if you through a rock (like 1e6 kg, Patrick) at
the Earth at 33% of C, the rock would pass through the Earth making a
tunnel about 10-20 m wide. I would not expect planet-wide upheaval.
This whole thread is starting to get silly. Patrick, run, not walk, to
your nearest bookstore and buy a copy of "The Moon is a Harsh Mistress"
by Robert Heinlein. While you are at it, pick up a beginning physics book.
====|====|====|====|====|====|====|====|====|====|====|====|====
David Paigen TRW Financial Systems pai...@tfs.com
"Lay on, McDuff, and damn'd be he who first cries, 'hold, enough!'"
Ray Trent
> Slam a 500km MARSHMALLOW, let alone asteroid, doing relativistic
> speeds into any planetary mass and you'd probably have to search for the
> rubble with a microscope.
>
>I beg to differ. What do you define by 'relativistic' speeds? Even at .5 c,
Normally, "relativistic" is informally considered to start somewhere
around .8c. Your analysis is correct, but so is one using v=100
km/sec, it just distorts time less.
>density). That means 1*10E16 kg of rock.
>
>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.
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.
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.
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*.
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.
>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).
Of course, as I've said before, this is all silly. If you have the
energy available to accelerate an asteroid to "relativistic" speeds,
you have the energy available to use it against the planet in the
first place. Comparatively minor effects like the 11 km/sec gravity
well of the earth are minor.
--
"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\..
Patrick Rannou
>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 from each
scale side to side: there IS, after, a change in scale of 1^7...
relativistic effects being ignored unless you really go fast enough to
create a sizeable time/space distortion, which isn't the case here, since
1^7 = 0.033 c which create a distortion like 0.5% of time and space). Then
having different energy level characteristics at each scale level, it
follows that at the (now 1^7 Giant scale level) two one meter (relative to
you, Giant Size, they appear one meter: that's the idea behing "changing
scales") balls each moving at one meter per second and colliding are, for
you, the same experiment as before: Relative to you, the balls of each
experiment appear the same size, to move at the same speed in the same
fashion. The *ONLY* thing that changed is the change of scales. Now, you
look at the collision, and instead of doing "splat! kabooom!" the balls
instead do "splooosh! sboing!". WHAT? You say, bewildered...
But the sol;ution is very near. Surely you don't expect me to believe that
the way two human size colliding balls react is exactly thew same as the
way giant 1^7 meters colliding balls react during their collision. You all
agree with that. If not, then you can ALWAYS simulate the way bigger or
smaller objetcs will react simply by doing tests on human size objects, and
we all know that this is not the way things work.
And the variation in energy levels relatively to a change in scale is just
that: the differences between the way objects will react at each given
scale!!! We now have our solution very near We will now know how differently mega-big
colliding objets reacts. Human size objects, in my previous arbitrary
example, did "splat! kabooom!" when colliding. Now, will Giant scale level
ball do "splooosh! sboing!" or will they do "WHAAAAAAM! SKRAKAKAKOOOW!" ?
We just have to COMPARE the two values we had: the available energy inside
the object, its MOMENTUM that will be released on impact, and how "tough",
resistant to damage, the object has become, its DENSITY.
Human Size Scale Level Giant Size Scale Level
Momentum 1 ==> 31.6^7
Density 1 ==> 1*10e21
And we can now compute a new value, the "Blast Effect", or "Deformation
Sensibility" if you like. Blast effect will augment with Momentum energy,
but will diminish with density since the object with higher density is more
resistant to deformations. So we have:
Blast Effect at this scale effect should measure up
to <value> "blast units" per scale unit:
Human Scale Level: 1
Giant Scale level: ( 31.6 / 1*10^3 )^7 = (3.16*10e-2)^7 = 3.16*10e-14
= NEAR ZERO.
So we can conclude with a BIG ASSURANCE (just compare the two numbers...)
THAT collisions on the Giant Scale level have an effect that is far far far
more like "splooosh! sboing!" than "WHAAAAAAM! SKRAKAKAKOOOW!". We could
even say that the effects of collisions are dtermined like:
Object is of this Scale Level Blast Effect is like
Giant Scale Level "splooosh! sboing!"
Human Scale Level "splat! kabooom!"
Miniature Scale Level "WHAAAAAAM! SKRAKAKAKOOOW!"
Examples:
Stars Collid "splooosh! sboing!"
Rock falls on brick "splat! kabooom!"
Combustion of Paper "WHAAAAAAM! SKRAKAKAKOOOW!"
It should be noted that the described effects are concerning the spectators
who happen to be at the same scale level.
A Giant scale human seing two planets colliding won't have much of a show,
but the human size human on the planet will have the (last) time of his
life.
A Human Scale human TOUCHING paper will hurt himself, but a Miniature Scale
human 1 meter OVER the sheet (not ON it) will still get the hell beaten out
of him just by exposure to the convection air currents to sheet creates.
So two planets who collide will act more like two floating blobs of clear
liquid, fusing themselves (their shape: not the nuclear reaction) together
then continuing their movement as if passing through each other. Sure,
after that you have no more crust on the planets, but you don't have a
vaporisation of the planets either.
The "vaporisation" comes about/can be seen not at the Giant Scale level but
at the Human Scale level. There Every linear effect the Humans observe of a
Giant scale effect is 1e7 times stronger, so Globally we note that:
We are at Human Scale. We have FOUR balls.
The first two do one collision: one is 1 meters wide and immobile, the
other 1 cm and moving at 1 meter per second, and are colliding face to
face. The two other balls are an immobile planet 1e7 meters wide and an
asteroid 1e5 metrs wide, moving at 1e7 meters per second and
collidong face to face. Now we notice the damage caused on each ball by
their respectives collisions.
The Human Scale Blast Effect on the smaller balls is 1, because it is our
reference (see above)
The Human Scale Blast Effect of the bigger balls is 1e7 times the Giant
Scale Blast Effect of the same. That is, if a Giant Scale human see a 1
millimeter dent into to (to him 1 meters wide) balls, to us at human scale
this damage caused by the collision will appear 1e7 times bigger, or 10 km.
That's not a dent, but an incredible chasm. So the Human Scale Blast Effect
of the Giant Balls is: 3.16*10e-14 * 1*10e7 = 3.16*10e-7 (If we consider
that MOMENTUM and DENSITY were the only two elements to be considered.
What do that means? That means that the Noticeable Blast Effect Caused by
the Giant Scale Collision is 3.16*10e-7 less than on the one on the Human
Scale Collision. In other words, if on our 1 meter rock ball we notice a
"crater" (or a hit-spot)
31.6 millimeters diam, then *PROPORTIONATELY*, the same "crater" (or
equivalent damage) will be smaller on the planets, because momentum
increases slower than density when you change scales.
But then again, Momentum and density aren't the only things that counts...
stability of propagation of vibrations, heat-transmission, elasticity, all
of these factors change with the scale, and many may affect the "Blast
Effect" in a way as to bring back the damage done in the cataclysmic
"WHAAAAAAM! SKRAKAKAKOOOW!" range. I don't know. The precedint was just a
theory, and I clearly see that the obtained value of 3.16*10e-14 is
completely wrong. One can't address so complex an issue as this with just
two variables. These don't even account for the fact that the "same" 1
meter rock ball stay rocky at human, pressure being the same everywhere,
but melts it's center at planet scale...
For a final word, using "common sense", I would say that all other things
being equals, I feel the destruction would be more disastrous with bigger
asteroids... for US. But there no way earth would be "blown to
bits/instantly vaporized"
Of course, as I've said before, this is all silly. If you have the
energy available to accelerate an asteroid to "relativistic" speeds,
you have the energy available to use it against the planet in the
first place. Comparatively minor effects like the 11 km/sec gravity
well of the earth are minor.
Yes.
Ray Trent
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:
>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.
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.
>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)
>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.
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.
>>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.
>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.
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.
>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.
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.
>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?
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.
Patrick Rannou
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.
Ray Trent
>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.
Mind you, that's one fucking huge asteroid. However, for smaller
rocks, see below.
> 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).
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.
Patrick Rannou
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....
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.
Erik Max Francis
> Normally, "relativistic" is informally considered to start somewhere
> around .8c. Your analysis is correct, but so is one using v=100
> km/sec, it just distorts time less.
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]
----------
Erik Max Francis Omnia quia sunt, lumina sunt. Coming soon: UNIVERSE _ | _
Upsilon Pi Epsilon
>r...@erg.sri.com (Ray Trent) writes:
>
>> Normally, "relativistic" is informally considered to start somewhere
>> around .8c. Your analysis is correct, but so is one using v=100
>> km/sec, it just distorts time less.
>
>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.
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
Jay Windley
> 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.
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 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.
I've seen these films. Gross.
> 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?
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 | jwin...@asylum.utah.edu
Department of Computer Science +---------------------------------------
Univ. of Utah, Salt Lake City | Propter causam veritatis non tacebo.
Ray Trent
>> around .8c. Your analysis is correct, but so is one using v=100
>
>"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.
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.
Paul Davey
-> ... 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
Francis P. Cyran
>>>>>> On 29 May 92 23:27:00 GMT, coll...@envmsa.eas.asu.edu (Upsilon Pi Epsilon) said:
>-> ... 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?
>
--
*************************************************************************
* Sig. Test *
*************************************************************************
Patrick Rannou
-> ... 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.
Patrick Rannou
-> ... 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.
Patrick Rannou
In article <PD.92Ju...@herts.x.co.uk> p...@x.co.uk (Paul Davey) writes:
-> ... 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
Erik Max Francis
> This is kind of off of the subject, but while everyone is talking about
Yeah, it's way off the subject. :-) No problem.
> 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, asssumi
This is true.
> 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.
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.
> 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 doin
> full impulse? I realize in warp, it is protected by a warp bubble, but what
> kind of protection does it have when doing impulse?
Might be the meaning of "deflectors" . . . who knows.
Erik Max Francis
> 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.
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.