How come the Tunguska fireball reached the ground before the shockwave?

In this post on his Bad Astronomy blog, Phil Plait describes the Tunguska event as having had a fireball which was followed by a shock wave:

A chunk of rock (or possibly ice) about 30 meters across—the size of a house—barreled in at a speed probably 50 times that of a rifle bullet. Ramming through the Earth's atmosphere, incredible forces compressed it, crumbled it, and when it reached a height of just a few kilometers above the ground, those forces won. In a matter of just a few seconds the energy of its immense speed was converted into heat, and it exploded.

[...] The fireball created a huge forest fire over hundreds of square kilometers of the Podkamennaya Tunguska River region of the Siberian forest ... but then the immense shock wave from the blast touched down. It blew the fire out and swept down those trees like a rolling pin, knocking down untold millions of them.

I guess the forest-fires-then-shock-wave makes sense as an explanation for the large number of scorched and partially scorched trees found at the site. However, I'm confused by the explanation - wouldn't the shockwave be the first thing to hit the ground? Can anyone in the know comment on the anatomy of this explosion?

• In analogy of nuclear explosions I would venture to guess that what is meant here is that the thermal radiation from the fireball reached the ground, first. Since this happens almost at the speed of light, it will definitely outrun the shockwave. Given that the fireball was supposed to be just a few km above ground, the delay between the source and the absorbing ground was only on the order of $10\mu s$. Sep 23 '15 at 11:53
• @CuriousOne It could be (just a guess) that the nuclear explosion analogy is being wrongly applied here: my understanding is that the fireball in a nuclear blast forms pretty much instantaneously (i.e. with only $c$ delay) as X rays are absorbed by the atmosphere, which almost simultaneously becomes more opaque to them owing to chemical changes. So you have a fireball area heated up in the order of $10\mu s$ and thus the fireball would touch the ground first (nuclear blasts of this size give fireballs big enough to reach through the estimated altitude of the Tunguska bolide) ...... Sep 23 '15 at 12:16
• ..... But I'm not sure what form of EM radiation is given off by a disintegrating bollide: maybe it wouldn't be as short wavelength and as strongly absorbed by the atmosphere as nuclear blast X rays: it seems to me that knowledge about the bolide's demise would determine the answer to Emilio's question. Sep 23 '15 at 12:18
• @WetSavannaAnimalakaRodVance: The fireball in a nuclear explosion develops in a matter of microseconds (but it still takes some time after that to peak in total radiated power, I believe), so it would be much faster than the meteor fireball which should develop on the order of $r/v \approx 100m/10km/s = 10ms$. In that sense the delay of the radiation between the meteor and the ground probably won't matter, it's basically instantaneous. One can probably estimate the peak temperature with a black body model. I doubt that it's much more than a few 10,000K and it would cool very fast. Sep 23 '15 at 12:21
• Apologies if you have already seen this, but if you look at this video, (most of it is drama,) but scroll to 40 mins on and there is a simulation of the event that might be related slightly to your question youtube.com/watch?v=HXfvhJoNi90
– user81619
Sep 23 '15 at 13:50

As illustrated HERE on Astronomy SE, since the velocity of the meteor or comet nucleus is going to be about 40 km/s (essentially the escape velocity from the sun at at 1 AU) and the velocity of Earth is about 30 km/s, the relative velocity of the two can be anywhere from 10 km/s to 70 km/s (varying by a factor of 7) depending on the direction the object is coming from.

The kinetic energy of the object is proportional to the square of the velocity, so it varies by a factor of 50! All of the kinetic energy is converted, in milliseconds, to heat in the air (which quickly become visible and UV light) - or to forming a crater if it reaches the ground. Examining the damage, something of an estimate of the energy can be made, but this could be a smaller object coming in relatively fast, or a larger object coming in relatively slower. This means the mass of the object is very poorly constrained.

The Tunguska aerial explosion is thought to be from an object coming in at a shallow angle. This matters for two reasons: objects coming in at such an angle is more likely to explode before impact, and when they do, they leave a butterfly shaped pattern of damage under them owing to the explosion happening along a path segment rather than at a point.

Light from the explosion, not the fireball itself, reaches the ground essentially instantaneously. The explosion itself takes milliseconds. The light from the plasma created by the explosion fades quickly. Most of the light happens within a small fraction of a second. This light pyrolyzes, rather than burns, the vegetation, carbonizing some of it. Only in the case of a fireball barely above the surface would thr fireball itself interact with the surface.

The shock wave, which knocks over the trees and cools the trees to the point they will not burn, reaches the ground directly under the blast travelling at the speed of sound, in 1 to 3 seconds, depending on height. Ground some distance away from the center is hit a little later, also depending on distance. Trees directly below lose branches but stay standing, similar to Genbaku Dome under the atomic bomb at Hiroshima.

• Nice answer, I only want to add that speed of Sound is 340 m/s and the speed of fireball is 10 000 -70 000 m/s Nov 22 '15 at 9:27

I know this has already been answered and accepted, and I agree with the answer by Eubie Drew. However, there are several things in the answer and comments that can be misinterpreted or are misleading. I just wanted to clarify a few things here.

The absolute speed of a shock wave is determined by the piston/driver causing it, not the speed of sound, $$C_{s}$$. The sonic boom heard on the ground does travel at the speed of sound.

So imagine this object is moving parallel to the surface of the Earth at a height $$h$$ and a speed $$V$$. The Mach cone that defines the leading edge of the sonic boom is defined by the angle $$\alpha$$, given by: $$\sin{\alpha} = \frac{ C_{s} }{ V } \tag{0}$$

If the object passes directly over your head, it would move a lateral distance $$\Delta x$$ before you heard the sonic boom. This distance is given by: \begin{align} \Delta x & = \frac{ h }{ \tan{\left( \sin^{-1}{\left( \frac{ C_{s} }{ V } \right)} \right)} } \tag{1a} \\ & = \frac{ h \ V \ \sqrt{ 1 - \left( \frac{ C_{s} }{ V } \right)^{2} } }{ C_{s} } \tag{1b} \end{align} where we have used a trig identity for the tangent of the arcsine of a real argument.

The amount of time it takes for this object to move $$\Delta x$$ is given by the following: \begin{align} \Delta t & = \frac{ \Delta x }{ V } \tag{2a} \\ & = \frac{ h \ \sqrt{ 1 - M^{-2} } }{ C_{s} } \tag{2b} \end{align} where $$M$$ is the Mach number.

So what's the point of this response?

There were several comments or insinuations that non-light-speed events reached something before a shock wave, which seemed to provoke confusion.

If the object had been normally incident on Earth's surface then nothing would reach the ground before the shock wave except light and potentially some ionized particles accelerated by the plasma shock.

If the object is on a shallow angle of approach, as suggested by Eubie Drew's answer, then the caveat here is that directly below the path of the object, yes, the shock wave would be slower than the heat it generated (well this is true everywhere for radiative heat transfer). If the object ignited and the shock generated enough heat to ionize the atmosphere, then again, directly below the path of the object it is possible that the ionization reached the ground before the shock wave (assuming some rather extreme heat generation).

A "fireball," generated by combustion not ionization, would not reach the ground first if the source was the comet, however. If the object were fast enough to generate enough heat to ignite the atmosphere, there would have been a lot more damage.

Now if we look at things in the object's path, again the first thing to reach the ground is electromagnetic radiation. If the object is exceedingly fast, then it's possible it's shock wave ionized some of the atmosphere. Those ionized particles can move upstream of the shock and possibly reach the ground first or recombine and generate electromagnetic radiation that could also reach the ground first. However, again if there were an actual fireball caused by combustion, that would not precede the shock wave.

Analogy to explosions

Similar to a chemical explosion, the fireball is behind the shock wave. It must be because the piston/driver of the shock wave is the rapid release of chemical energy resulting in combustion and projections ejected from the origin. This results in a special type of shock wave called a blast wave, which is usually a thin (roughly) spherical shell of highly compressed and heated gas. The "fireball" does not move ahead of the blast wave. There can be recoil shock waves associated with really strong explosions that give the appearance of a shock wave trailing the fireball or initial explosive elements, but this is a secondary effect.

A similar thing happens with nuclear explosions. You can find all sorts of example videos on YouTube if you so wish, but those detonated above ground/water typically show a common series of events. There is an initial blinding flash of light (i.e., electromagnetic radiation released from the nuclear reactions and the heat generated by this radiation interacting with the atmosphere) followed by an expanding, opaque cloud. If the bomb is detonated over or near the surface of the ocean, you can see a white ring expanding ahead of this opaque cloud. That outer edge of the white ring indicates the leading edge of the shock wave and the opaque cloud is condensation due to the rapid pressure changes. Eventually the cloud appears to dissipate/disappear and there remains a mushroom cloud rising into the sky, in the middle of which is a fireball of intense heat. If you watch closely, you can sometimes see the recoil shock wave that occurs after the atmosphere tries to recover from the first blast wave and vacuum-like conditions behind that led to the condensation cloud. But again, the shock wave leads here and is only preceded by electromagnetic and particle radiation, i.e., a fireball would trail not lead the shock wave.

Okay, what's the point?

The point being is that the only thing which beats the shock wave is electromagnetic and particle radiation. The shock wave is driven by a piston and the speed of the shock is determined by the piston, not the speed of sound. The sonic boom or sound wave generated by a small object moving faster than the speed of sound will travel at the speed of sound radially away from the source. The local Mach number on the boundary of the shock depends upon the angle of incident of the flow relative to the local shock normal. For instance, at the nose of a bow shock, the Mach number is highest, while on the flanks it is lower and decreases with increasing distance from the nose (i.e., bow shocks generate a shock surface that is roughly one side of a hyperboloid of two sheets).