How far from ground zero does a Nuclear Explosion effect?

Question

When a nuke explodes it releases extremely large amounts of energy, how far does this energy (how far is the destruction) expand past ground zero, and how far does the radiation carry from ground zero as well. I know the whole world feels this energy in some small way, but how far does the destruction carry.

Answer 1

This is not impossible to answer, but the answer has to be general since your question was general.

Lethality (L) effects vary as the 1/3 power of the yield Y. Thus, for example, suppose a Y = 100 KT weapon has a 50% lethality range of l = 1 mile. What would the 50% lethality range (L) of a 1000 KT (1 MT) weapon be?

Set up a ratio [Y(1,000)/Y(100)]^1/3 = (L/l) = 10^1/3; so that L = l (2.15). Thus, by increasing the yield by a factor of ten, we increase the 50% lethality range just a bit over two times. Since we assumed l = 1 mi, then L = 2.15 miles radius around ground zero. For the most part, we are talking about lethality from blast here.

Radiation is a horse of a different color. There are two kinds: direct or radiant radiation consisting of high energy neutrons and gamma rays for the most part; and indirect or induced consisting of irradiated particles coming from whatever surface the bomb exploded over or on and pulverized.

The direct radiation also includes all that heat and visible light that's also created by the blast. Gamma rays, travel at light speed (they are a part of the light spectrum in fact); so they reach the victim the same time he or she sees the blast. The high energy neutrons are fast, but slower than light speed; so they'll be the second thing to reach the vic. [Blast, which travels at the speed of sound, comes third.]

The good news, if there is any here, it that the direct radiation energies diminish by the inverse of the square of the distance from ground zero (i.e., ~ 1/R^2). For example, if E(1 mi) is the energy a victim endures at one mile from ground zero then E(2 mi) = E(1 mi)/2^2 = E(1)/4 is the energy the victims at two miles will be exposed to.

The gamma rays tend to disappear almost immediately; at the speed of light so to speak. But the neutrons, being mass and traveling much slower, can stay around for hours, depending on the yield. As they float around, they lose their unnaturally high energies gained from the atomic explosion; and they become ordinary free neutrons that might be captured by atomic nucelii.

The irradiate particles can circle the Earth if they are small and light enough. In fact, they have been known to reach the stratosphere and circumnavigate the Earth several times. As they pick up dust, water, and whatever in their travels around the globe, they eventually get heavier enough to fall to the Earth. And that's when "the whole world feels this energy in some small way."

The heavier particles begin to drop to Earth almost immediately; so they end up close in to ground zero. And everything in between weight and size spreads out, generally in the direction of the prevailing winds. And then they fall, sometimes rained out...and that's the fallout you've probably heard about.

Depending on the chemical nature of the particles, the induced radiation can last a matter of minutes all the way up to decades. Some of the ground around our bomb test sites is still hot even though the last bomb tested was back in the 1960s or 70s. [You can look this up on the web.]

Answer 2

1. The human body would be vaporized. 2. People at or near ground zero would not feel anything. 3. That is why there were shadows burned into the ground. There would not be anything left. 4. If the metal tomb was not lead-lined, you would not survive because of gamma rays and radiation.

Answer 3

That is dependant on yield of the weapon, type of burst, (Air, surface, or sub-surface) Altitude if an air-burst, type of target-structures affected, etc. You gave NO data to figure effects with. Your question cannot be answered as-asked. The link wil help you determine effects when you specify a yield.

Answer 4

as stated by another user it is dependent of the size of the device, another aspect to consider is the height at which the detonation occurs. Detonations that are at ground level do not result as well as detonations that are above ground level.

Answer 5

It is totally directly dependant upon the size of the explosion. Since you have neglected to include this bit of information in your question, it is impossible to answer.

Answer 6

(Rad level at distance1) (Distance1)^2

=

(Rad level at distance2) (Distance2)^2




Example:

Rad level at 1 mile = 1,000,000 R/hr

What is it at 10 miles?

10^6R/hr (1mile)^2 / (10miles)^2

10^6R/hr (1m^2) / (100m^2)

10^4R/hr

10,000 R/hr


What about 50 miles?

10^6R/hr (1mile)^2 / (50miles)^2

10^6R/hr (1m^2) / (2,500m^2)

400 R/hr


And 100 miles?

10^6R/hr (1mile)^2 / (100miles)^2

10^6R/hr (1m^2) / (10,000m^2)

100 R/hr