When a strong explosion takes place, a shock wave forms that propagates in a spherical manner away from the source of the explosion. The shock front separates the air mass that is heated and compressed due to the explosion from the undisturbed air. In the picture below you can see the shock sphere that resulted from the explosion of Trinity, the first atomic bomb ever detonated.
Using the concept of similarity solutions, the physicists Taylor and Sedov derived a simple formula that describes how the radius r (in m) of such a shock sphere grows with time t (in s). To apply it, we need to know two additional quantities: the energy of the explosion E (in J) and the density of the surrounding air D (in kg/m^{3}). Here’s the formula:
r = 0.93 · (E / D)^{0.2} · t^{0.4}
Let’s apply this formula for the Trinity blast.
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In the explosion of the Trinity the amount of energy that was released was about 20 kilotons of TNT or:
E = 84 TJ = 84,000,000,000,000 J
Just to put that into perspective: in 2007 all of the households in Canada combined used about 1.4 TJ in energy. If you were able to convert the energy released in the Trinity explosion onetoone into useable energy, you could power Canada for 60 years.
But back to the formula. The density of air at sealevel and lower heights is about D = 1.25 kg/m^{3}. So the radius of the sphere approximately followed this law:
r = 542 · t^{0.4}
After one second (t = 1), the shock front traveled 542 m. So the initial velocity was 542 m/s ≈ 1950 km/h ≈ 1210 mph. After ten seconds (t = 10), the shock front already covered a distance of about 1360 m ≈ 0.85 miles.
How long did it take the shock front to reach people two miles from the detonation? Two miles are approximately 3200 m. So we can set up this equation:
3200 = 542 · t^{0.4}
We divide by 542:
5.90 ≈ t^{0.4}
Then take both sides to the power of 2.5:
t ≈ 85 s ≈ 1 and 1/2 minutes
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Let’s look at how the different parameters in the formula impact the radius of the shock sphere:

If you increase the time sixfold, the radius of the sphere doubles. So if it reached 0.85 miles after ten seconds, it will have reached 1.7 miles after 60 seconds. Note that this means that the speed of the shock front continuously decreases.
For the other two parameters, it will be more informative to look at the initial speed v (in m/s) rather the radius of the sphere at a certain time. As you noticed in the example, we get the initial speed by setting t = 1, leading to this formula:
v = 0.93 · (E / D)^{0.2}

If you increase the energy of the detonation 35fold, the initial speed of the shock front doubles. So for an atomic blast of 20 kt · 35 = 700 kt, the initial speed would be approximately 542 m /s · 2 = 1084 m/s.

The density behaves in the exact opposite way. If you increase it 35fold, the initial speed halves. So if the test were conducted at an altitude of about 20 miles (where the density is only one thirtyfifth of its value on the ground), the shock wave would propagate at 1084 m/s
Another field in which the TaylorSedov formula is commonly applied is astrophysics, where it is used to model Supernova explosions. Since the energy released in such explosions dwarfs all atomic blasts and the surrounding density in space is very low, the initial expansion rate is extremely high.
This was an excerpt from the ebook “Great Formulas Explained – Physics, Mathematics, Economics”, released yesterday and available here: http://www.amazon.com/dp/B00G807Y00. You can take another quick look at the physics of shock waves here: Mach Cone.
Estimate of the yield of the Nagasaki bomb.
Oak Ridge National Laboratory ORNL data, critical mass versus natural uranium reflector thickness ( dr ) for 93% U 235 sphere.
The following is an empirical formula to a few % accuracy :
Rc(dr) = [ 0.92 +{ 0.48 / (1+ dr/4.25) } ] Rc(dr => 21.6 cm.)
Fat Boy ( Nagasaki )
Uranium tamper 2.5” thick, 6.35 cm, estimate 115 Kg
Aluminium Pusher 4.5”, 11.43 cm, “ 118 Kg
Plutonium 6.2 Kg
Mt approx.. 240 Kg
ORNL Plutonium 16.5 gm/cm3 critical bare mass 16.25 Kg, infinite reflector 5.8 Kg.
Rc(dr = 0) = 6.17 cm, Rc(dr = infin.) = 4.37 cm. R at 6.2 Kg = 4.47 cm.
Rc (dr = 6.35 cm) = 4.87 cm, Mc(dr) = 8.0 Kg.
ORNL compression factor Kc_ “twice ”and “2.5 possibly somewhat less”. The numerical mean is 2.25 and the phrase “possibly somewhat less” would put this figure in the region of 2.2 to 2.1.
Pu fission MFP = 14.2 cm giving t = 8.5 x 10^9 sec. Neutron yield n = 3.
E = Mt/2 { [R Kc^2/3 – Rc][1 ( Rc/(R Kcom^2/3))^2 ][n – 1 ]/4t }^2 ergs.**
Energy equivalent 4.2×10^16 ergs/ton TNT.
It is estimated that 10% of the energy released was due to the fission of the tamper.
Kc = 2.1 yield 18,584 x 10/9 = 20,644 tons TNT
2.15 21,398 23,766
2.2 24,405 27,116
Dr Penney observed the Nagasaki explosion from an escort plane and estimated from an on the ground survey a yield of 22+ 2 Kilo tons TNT.
** Based on a lecture given by Werner Heisenberg 1945.
Estimate of the yield of the Hiroshima bomb.
U 235 enrichment 80 %, q = 0.8 Mf = 64 Kg,
Tungsten carbide tamper Mr = 322 Kg. Mt = 372 Kg.
Over the range of q = 0.8 to 1, the critical radius Rc to a first approximation is given by :
Rc = ( pi Aw) / (2 p Av) { 1/[(n – 1) Gf* Gs* ]}^1/2.
Aw = atomic mass, Av = Avogadro constant, p = density, n = neutron yield, and Gf*, Gs* are the effective neutron crosssection for fission and scatter respectively.
Gf* = Gf235q + Gf238( 1– q ) and Gs* = Gs235q + Gs238 ( 1 –q ).
________ Gf_____ Gs.
U235____ 1.24____ 4.56 x10^24 cm2.
U235____ 0.3_____ 4.8.
Gf* = 1.052×10^24cm2, Gs* = 4.6 x 10^24cm2, n = 3, p =18.95 gm/cm3.
Rc =6.63cm , Mc = 23 Kg.
M =64 Kg, R = 9.3 cm, n = 2.64, t = 1 x108 sec.
Et = Mt/2 { ( R – Rc )( 1 – (Rc/R)^2)( n – 1)/4t }^2 **.
Et = 5.39 x10^20 ergs and at 4.2 x10^16 ergs/ton TNT : yield = 12.800 tons TNT.
Dr. Penney estimated from a ground survey a yield of 12+1 Kilo ton TNT.
** Based on a lecture given by Werner Heisenberg 1945.
Estimate of the yield of Ivy King
.
Little information is available on this device other than it consisted of 60 Kg HEU and a natural uranium tamper. The fission of the later was estimated to contribute from 10% to 20% of the final yield.
When compressed to the natural density of 18.95 gm/cm3, the radius would be 9.1 cm. Assuming a natural uranium tamper of 2.5 inches or 6.35 cm thickness, the estimated critical radius would be 6.22 cm.
Using a neutron yield ‘n’=2.64 and assuming a compression factor Kc = 2.2, Mt = 375 Kg, t = 1×10^8 sec.
E = Mt/2 { [R Kc^2/3 – Rc][1 ( Rc/(R Kc^2/3))^2 ][n – 1 ]/4t }^2 ergs
= 1.865×10^22 ergs or 444 K tons TNT
For a 10% contribution due to the fission of the U238 tamper gives 493 K tons TNT and a 20% contribution 533 K tons TNT. The official yield was put at about 500 K tons TNT
RUSSIA