I understand how the ozone layer would be quickly depleted and the UV radiation from the sun would reach the ground...etc I understand all this, but what confuses me is where all this huge amount of energy would go if our atmosphere stopped it ? As far as I know, gamma rays will be stopped by the earth's atmosphere and the rays will not touch the ground, also, a typical GRB has about 10^43 to 10^45 joules of energy for the gamma photons, so where would this huge amount of energy go if our atmosphere absorbed it other than ozone depletion ? Would it heat the atmosphere to thousands of degrees ? Or am I just wrong about something here ?
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asked Dec 29, 2013 at 15:53
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3$\begingroup$ $10^{43}\,\text{J}$ of energy is enough to convert all of Earth into ultrarelativistic plasma. When talking about ozone layer depletion from GRB one usually means that the event happened within the stellar neighborhood, that is within tens to hundreds parsecs. $\endgroup$– user23660Dec 29, 2013 at 16:29
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$\begingroup$ But gamma rays don't lose energy while traveling, and correct me if I am wrong, GRB is so compact and the rays should not be scattered much so its energy should be lower by not too much when it hits the earth. $\endgroup$– Abanob EbrahimDec 29, 2013 at 16:49
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2$\begingroup$ The energy of GRB is emitted in a cone of 2 to 20 degrees. This means the energy per area decreases according to usual $1/r^2$ law, and over an interstellar distance this means much smaller energies are absorbed by the planet. $\endgroup$– user23660Dec 29, 2013 at 17:07
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$\begingroup$ I was mistaken then. Thank you for clarifying this. $\endgroup$– Abanob EbrahimDec 29, 2013 at 17:44
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1 Answer
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Using the values from the Wikipedia article, let's say we have two $10^{44}\ \mathrm{J}$ GRBs, each beaming their energy uniformly in two beams with half angles of $1^\circ$.
If the energy were not beamed, it would be distributed evenly over the area of the sphere centered on the GRB with radius equal to the GRB-Earth separation $d$. However the actual area $A$ of this sphere that is illuminated is instead $$ A = 4\pi d^2 \frac{2\cdot\pi (1^\circ)^2}{4\pi (180^\circ/\pi)^2}, $$ where the numerator is the angular area of the beams and the denominator gives the number of square degrees in the whole sphere.
Whatever the formula for $A$, the fraction of the burst energy received by Earth is simply $$ f = \frac{\pi R_\oplus^2}{A}, $$ where $R_\oplus = 6{,}371\ \mathrm{km}$ is the radius of Earth.
Suppose one GRB is rather close in our galaxy, at a distance of $10{,}000$ light years, $d_1 = 9.5\times10^{19}\ \mathrm{m}$, and another is at $7.5$ billion light years, $d_2 = 7.1\times10^{25}\ \mathrm{m}$. Then the fractions of the total energy received by the whole planet are $f_1 = 7.4\times10^{-24}$ and $f_2 = 1.3\times10^{-35}$ -- very small portions indeed.
The total energies received are $E_1 = 7.4\times10^{20}\ \mathrm{J}$ and $E_2 = 1.3\times10^9\ \mathrm{J}$, which sounds like a lot especially in the first case. However, the mass of the atmosphere is $5\times10^{21}\ \mathrm{g}$, and air has a specific heat capacity of something like $1\ \mathrm{J/(g\cdot K)}$. If all the energy is dissipated in the atmosphere, the nearby GRB will raise the temperature of the atmosphere by about $0.1^\circ\mathrm{C}$, while the distant one will raise it by about $(2\times10^{-13})^\circ\mathrm{C}$.
That takes care of temperature. Direct biological effects of radiation are much trickier, so I'll only briefly touch on them. In another answer of mine I go into detail about calculating how much radiation passes through material. The important points are that the atmosphere is equivalent to $1\ \mathrm{m}$ of lead, which has an attenuation coefficient of about $0.2\ \mathrm{cm^{-1}}$ for gamma rays. Thus the atmosphere blocks all but about $10^{-10}$ of the incoming gamma radiation.
Suppose the nearer GRB went off. The surface of Earth would receive $8\times10^{-4}\ \mathrm{J/m^2}$ in gamma rays. Say a person lying down is $2\ \mathrm{m}$ tall, $40\ \mathrm{cm}$ wide, and $75\ \mathrm{kg}$ in mass. If he (unrealistically) absorbs all gamma rays passing through his body, he will receive $9\times10^{-6}\ \mathrm{J/kg}$. This is about $1\ \mathrm{mrem}$, which is the typical daily background exposure.
