| MadSci Network: Astronomy |
All highly-energetic energy blasts (like asteroid impacts) tend to behave the same way. In particular, the effects of a surface nuclear weapons test are very similar to those of an asteroid impact. Much of the following information comes from measurements of nuclear explosions.
The most important factor in studying the effects of an asteroid impact is the kinetic energy of the asteroid. This is given by the formula
E = 1/2 M V2where V is the velocity of the asteroid (typically 20-30 km/sec). To compare this with the strength of nuclear blasts, note that an explostion of 1 kiloton of TNT corresponds to 4.185*1012 joules. So an object moving at 3 km/sec has a kinetic energy equal to its weight in TNT; an object moving at 30 km/sec has an energy equal to 100 times its weight in TNT. The following is copied from the sci.space Frequently Asked Questions page (the calculations section)
COMPUTING CRATER DIAMETERS FROM EARTH-IMPACTING ASTEROIDS
Astrogeologist Gene Shoemaker proposes the following formula, based on
studies of cratering caused by nuclear tests. Units are MKS unless
otherwise noted; impact energy is sometimes expressed in nuclear bomb
terms (kilotons TNT equivalent) due to the origin of the model.
D = Sg Sp Kn W^(1/3.4)
Crater diameter, meters. On Earth, if D > 3 km, the crater is
assumed to collapse by a factor of 1.3 due to gravity.
Sg = (ge/gt)^(1/6)
Gravity correction factor cited for craters on the Moon. May hold
true for other bodies. ge = 9.8 m/s^2 is Earth gravity, gt is
gravity of the target body.
Sp = (pa/pt)^(1/3.4)
Density correction factor for target material relative to the
Jangle
U nuclear crater site. pa = 1.8e3 kg/m^3 (1.8 gm/cm^3) for
alluvium,
pt = density at the impact site. For reference, average rock on the
continental shields has a density of 2.6e3 kg/m^3 (2.6 gm/cm^3).
Kn = 74 m / (kiloton TNT equivalent)^(1/3.4)
Empirically determined scaling factor from bomb yield to crater
diameter at Jangle U.
W = Ke / (4.185e12 joules/KT)
Kinetic energy of asteroid, kilotons TNT equivalent.
Ke = 1/2 m v^2
Kinetic energy of asteroid, joules.
v = impact velocity of asteroid, m/s.
2e4 m/s (20 km/s) is common for an asteroid in an Earth-crossing
orbit.
m = 4/3 pi r^3 rho
Mass of asteroid, kg.
r = radius of asteroid, m
rho = density of asteroid, kg/m^3
3.3e3 kg/m^3 (3 gm/cm^3) is reasonable for a common S-type
asteroid.
For an example, let's work the body which created the 1.1 km diameter
Barringer Meteor Crater in Arizona (in reality the model was run
backwards from the known crater size to estimate the meteor size, but
this is just to show how the math works):
r = 40 m Meteor radius
rho = 7.8e3 kg/m^3 Density of nickel-iron meteor
v = 2e4 m/s Impact velocity characteristic of asteroids
in Earth-crossing orbits
pt = 2.3e3 kg/m^3 Density of Arizona at impact site
Sg = 1 No correction for impact on Earth
Sp = (1.8/2.3)^(1/3.4) = .93
m = 4/3 pi 40^3 7.8e3 = 2.61e8 kg
Ke = 1/2 * 2.61e8 kg * (2e4 m/s)^2
= 5.22e16 joules
W = 5.22e16 / 4.185e12 = 12,470 KT
D = 1 * .93 * 74 * 12470^(1/3.4) = 1100 meters
More generally, one can use (after Gehrels, 1985):
Asteroid Number of Impact probability Impact energy as multiple
diameter (km) Objects (impacts/year) of Hiroshima bomb
------------- --------- ------------------ -------------------------
10 10 10e-8 1e9 (1 billion)
1 1e3 10e-6 1e6 (1 million)
0.1 1e5 10e-4 1e3 (1 thousand)
The Hiroshima explosion is assumed to be 13 kilotons.
What about the other quantities you've mentioned? We can get a rough idea
of the size of an earthquake generated by an asteroid strike by comparing
with seismic waves generated by nuclear blasts. The equation for the
magnitude of an earthquake is
M = 0.67 log10 E - 2.9where E is the energy in joules. This is only the energy that ends up as seismic waves: for earthquakes, I think this accounts for most of the energy, but for nuclear blasts and asteroid impacts, lots of energy will be released as heat. I have had no luck figuring out or looking up the fraction of impact energy that ends up as heat, seismic waves, water waves, sound, and flying ejecta. I suspect almost all the energy goes into heat and seismic waves. I also suspect (but can't prove) that all large explosions have a similar fraction of their energy going into the various forms, so you could look up these data for common large explosions (such as surface nuclear blasts and volcanic eruptions) and scale them up proportional to the energy of the asteroid strike. An underground nuclear test is probably not a good model: since it's buried, much more of the blast's energy goes into seismic waves than for a surface explosion.
As for the height of tsunamis, I answered a question about that last year. There's also an article of mine on Usenet that might be helpful.
As for the ejecta, you can safely assume that the volume of rock blasted away from the impact equals the volume of the crater left over afterward. Now, suppose you know the mass M of the ejecta and the energy released into the ejecta, Ee Then using the kinetic energy equation above, the average velocity of the ejecta is roughly
V = sqrt(2 Ee/M)Using simple ballistics, the maximum range of an object launched at velocity V is
X = V2/g = 2 Ee g/M.Now, if you know the fraction fe of the total energy E which ends up in the ejecta, you can easily calculate Ee = fe E and use that to calculate the range of the ejecta. (Keep in mind this neglects atmospheric drag, which is a serious issue for small impacts). Even better, you could look at craters on the moon which have obvious ejecta patterns (Tycho is a good one), compute the ejecta mass by using the size of the crater, and find the ejecta energy Em. Then by running the crater-size calculation (given above) backwards, you can find the total energy E, and then compute fe. Then by assuming fe is constant for all impacts, you can find ejecta distances for other impacts on other planets.
A similar technique with other data sources might let you compute the energy fractions for sound, earthquake energy, etc. The sound wave from the Krakatoa eruption was measured 'round the world: you should be able to compute sound energy fraction from that data.
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