180
5 Miscellaneous Calculations
would have appeared brighter than m = +6 for observers residing on all three of
these planets, although only Venus and Mars were above the horizon at the time.
Could astronomers have detected the light of Trinity as reflected from the Moon?
The Moon is seen by reflected sunlight, so the issue is how the flux of Trinity light
at the moon would have compared with that from the Sun. In his LA-6300 report,
Bainbridge remarks that the total radiant energy density received at 10,000 yards was
12,000 J m
–2 . If we presume that all of this light was emitted over one microsecond,
such an energy density corresponds to a flux of 6.8 W m
–2 at the distance of the moon.
The solar flux at the moon will be essentially the same as that at the Earth, about
1400 W m
–2 , some 200 times greater. A change of one part in 200 corresponds to
~0.005 magnitudes, which would have been difficult to detect with 1945-era observatory technology. The idea of reflected Trinity light being visible from Earth is literary
license.
Finally, we can estimate what fraction of Trinity’s yield was in the form of visible
light. Various estimates of the yield can be found in the literature; I use 20 kt TNT
equivalent. Explosion of one ton of TNT liberates 4.2 × 10
9 J of energy; 20 kt would
be equivalent to 8.4 × 10
13 J. If the energy of the explosion radiated uniformly in
all directions, an energy density of 12,000 J m
–2 at 10,000 yards corresponds to a
total energy of 1.26 × 10
13 J, or approximately 15% of the 20-kt total. The fraction
of a bomb’s energy emitted as thermal radiation was cited as ~35% in Sect. 2.8.3;
“thermal” would include photons over all wavelengths, so we should expect the
visible fraction to be smaller than this, as has been estimated here. Well over half of
Trinity’s energy release was in forms invisible to the human eye.
These brightness calculations have an interesting historical connection. In the
published version of the Los Alamos Primer, Robert Serber admits that he overlooked
the brilliance of the fireball as a potential source of damage (Serber 1992). Anybody
who has stood outside under the noonday Sun on a hot summer day knows how little
exposure is required to get a serious sunburn. A nuclear weapon briefly acts like a
small Sun, leading to what are known as “flash burns.”
Serber offered some figures of his own for estimating brightness. He adopted the
radius of the fireball as being about 425 feet (130 m) at about three-tenths of a second
after the explosion, at which time he estimated that it would have a temperature of
about 7,000 C (~7,270 K). [At t = 0.3, Eq. (2.130), drawn from Geoffrey Taylor’s
analysis of the Trinity explosion indicates a radius of ~360 m, but this will not
affect the calculations which follow, as the specific time is not involved.] Serber then
estimates that for an observer one mile away (1600 m), the fireball would cover an
area of the sky about 350 times as large as the Sun, and be about 3.5 times as bright.
As argued in what follows, his size estimate is accurate, but his brightness estimate
seems much too low.
The apparent size of the fireball can be estimated with trigonometry. For an
observer 1600 m away, the angular diameter of the fireball would be about (260/1600)
= 0.16 radians, or about 9.3°. The angular diameter of the Sun is about 0.5°, so the
fireball would appear about 18.6 times wider and thus cover an area 18.6
2 ~ 345
times that of the Sun, as Serber claimed.
5 Miscellaneous Calculations
would have appeared brighter than m = +6 for observers residing on all three of
these planets, although only Venus and Mars were above the horizon at the time.
Could astronomers have detected the light of Trinity as reflected from the Moon?
The Moon is seen by reflected sunlight, so the issue is how the flux of Trinity light
at the moon would have compared with that from the Sun. In his LA-6300 report,
Bainbridge remarks that the total radiant energy density received at 10,000 yards was
12,000 J m
–2 . If we presume that all of this light was emitted over one microsecond,
such an energy density corresponds to a flux of 6.8 W m
–2 at the distance of the moon.
The solar flux at the moon will be essentially the same as that at the Earth, about
1400 W m
–2 , some 200 times greater. A change of one part in 200 corresponds to
~0.005 magnitudes, which would have been difficult to detect with 1945-era observatory technology. The idea of reflected Trinity light being visible from Earth is literary
license.
Finally, we can estimate what fraction of Trinity’s yield was in the form of visible
light. Various estimates of the yield can be found in the literature; I use 20 kt TNT
equivalent. Explosion of one ton of TNT liberates 4.2 × 10
9 J of energy; 20 kt would
be equivalent to 8.4 × 10
13 J. If the energy of the explosion radiated uniformly in
all directions, an energy density of 12,000 J m
–2 at 10,000 yards corresponds to a
total energy of 1.26 × 10
13 J, or approximately 15% of the 20-kt total. The fraction
of a bomb’s energy emitted as thermal radiation was cited as ~35% in Sect. 2.8.3;
“thermal” would include photons over all wavelengths, so we should expect the
visible fraction to be smaller than this, as has been estimated here. Well over half of
Trinity’s energy release was in forms invisible to the human eye.
These brightness calculations have an interesting historical connection. In the
published version of the Los Alamos Primer, Robert Serber admits that he overlooked
the brilliance of the fireball as a potential source of damage (Serber 1992). Anybody
who has stood outside under the noonday Sun on a hot summer day knows how little
exposure is required to get a serious sunburn. A nuclear weapon briefly acts like a
small Sun, leading to what are known as “flash burns.”
Serber offered some figures of his own for estimating brightness. He adopted the
radius of the fireball as being about 425 feet (130 m) at about three-tenths of a second
after the explosion, at which time he estimated that it would have a temperature of
about 7,000 C (~7,270 K). [At t = 0.3, Eq. (2.130), drawn from Geoffrey Taylor’s
analysis of the Trinity explosion indicates a radius of ~360 m, but this will not
affect the calculations which follow, as the specific time is not involved.] Serber then
estimates that for an observer one mile away (1600 m), the fireball would cover an
area of the sky about 350 times as large as the Sun, and be about 3.5 times as bright.
As argued in what follows, his size estimate is accurate, but his brightness estimate
seems much too low.
The apparent size of the fireball can be estimated with trigonometry. For an
observer 1600 m away, the angular diameter of the fireball would be about (260/1600)
= 0.16 radians, or about 9.3°. The angular diameter of the Sun is about 0.5°, so the
fireball would appear about 18.6 times wider and thus cover an area 18.6
2 ~ 345
times that of the Sun, as Serber claimed.