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Question 1074917: During World War 2, the Manhattan Project developed the first nuclear weapon and detonated it in 1945 at the Trinity test. The explosive yield of the bomb was classified by the U.S Army; however, in 1950 British physicist Sir Geoffrey Taylor made a good estimate of the yield based on publicly available photographs and data. He developed an equation that linked several key aspects of the test and used it to calculate the yield:
R = S(Et^2/p)^1/5
R is the blast radius in meters; t is the time in seconds after detonation, p (the greek letter of rho) is the air density at the blast site; S is a constant describing the heat capacity of the air; and E is the energy released by the bomb in joules. ( 1 joule (j) = 1 kg x m^2/s^2)
From available atmospheric data Taylor concluded that S = 1.041 and p = 1.000 kg/m^3
By studding a picture of the blast taken at t = 0.025 he estimated that the blast radius was 140 meters. Explosive yield is often described by how much TNT would be required to produce a similar explosion. The explosive force of 1 kiloton (kt) of TNT is approximately 4.184 x 10^12 joules
1. Rearrange Taylors equation to give you a formula for the yield of the bomb in kilotons. Hint: remember that in the original equation E is in joules
2. Calculate the yield of the Trinity bomb to the nearest kiloton.
3. Taylors estimate was lower than the actual yield of 20.08 kilotons. Which quantities might he have underestimated when using them in his equations? Which might he have overestimated?
( the 1 is next to the kg x m^2/s^2 also the x means multiply just so you don't get confused and the 1000 is also next to kg/m^3)
( thank you for whoever answers this sorry its so long)
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! R = S(Et^2/p)^1/5
Solve for E
R/S=(Et^2)^(1/5)/p^(1/5)
(R/S)^5=Et^2/p
Et^2=p(R/S)^5
E=p(R/S)^5/t^2
----------------------------
E=(140/1.041)^5/0.025^2
=7.039 x 10^ 13
Divide by 4.184 x 10^12.
E=16.82 kilotons.
The heat capacity of the atmosphere might have been different.
The density of the air in New Mexico is likely less than 1, since the state is significantly higher than sea level. The heat capacity is closer to 1.006, which would give a value of almost exactly 20 kilotons. He could have made a mistake in estimating the blast radius, and a small error in time would appear to be easy to make at the level of a thousandth of a second.
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