If the Earth all of sudden was instantaneously transmuted into pure light, what would happen?
Such explosion is of course practically impossible in reality, but from the point of view of modern physics this would not broke any absolute conservation laws:
Energy released would be $E = m_E c^2 = 5.97219 × 10^{24}\,\mathrm{kg} × (299792458 \,\mathrm{\frac{m}{s}})^2 = 5.36753 × 10^{41}\,\mathrm{J}$. Earth volume is $1.08321 × 10^{21}\,\mathrm{m^3}$, so energy density would be $U = 4.95520 × 10^{20}\,\mathrm{\frac{J}{m^3}}$. Assuming that the photons will form Planck radiation, from relation $U = \frac{\pi^2}{15} \frac{k_B^4}{\hbar^3 c^3} T^4$ we get $T = \sqrt[4]{\frac{15}{\pi^2} \frac{\hbar^3 c^3}{k_B^4} U} = 8.996 × 10^{8} \,\mathrm{K}$.
From Wien's displacement law, $\nu_\max = 5.879 × 10^{10} \,\mathrm{\frac{Hz}{K}} × T = 5.289 × 10^{19} \,\mathrm{Hz}$. So typical photon has energy of $E = h \nu_\max = 3.504 × 10^{-14} \,\mathrm{J} = 0.2187 \,\mathrm{M}e\mathrm{V}$. This is less than the rest energy of electron ($m_e c^2 = 0.510998910 \,\mathrm{M}e\mathrm{V}$), so pair creation is not important. (This comes from mean density, $5.5 \,\mathrm{\frac{g}{cm^3}}$. Actually we should take into account density of different parts. The density of the atmosphere is $0.0012 \,\mathrm{\frac{g}{cm^3}}$ or less, and of the inner core—up to $13\,\mathrm{\frac{g}{cm^3}}$. But this gives no qualitative difference—atmosphere still gives gamma rays or hard X-rays (and total energy from it is neglegible), and core has no pair creation.)
The flash would last $2 r_E/c$ or $0.04259 \,\mathrm{s}$ (the difference between time of arrival of the light from the nearest and furthest points of Earth).
Space-time will not be destructed as this is relatively low energy. Around Planck energy, about $10^{22}$ MeV, or even few orders of magnitude lower, strange things might happen, for example transition to other vacuum, maybe with other number of observable dimensions.
Gravitational binding energy is of the order of $E_g \sim \frac{G m^2}{r}$ (details depend on inner structure). For example for Earth $E_g \sim 3.728 × 10^{32}\,\mathrm{J}$. Earth would be totally destroyed by explosion of another Earth for $E_g \sim E s / S = \frac{E \pi r_E^2}{4 \pi R^2}$ ($s = \pi r_E^2$ — surface of Earth cross section, $S = 4 \pi R^2$ — surface of a sphere on which energy is dispersed for given distance) or for stream of the order of $I_1 = \frac{E_g}{\pi r_E^2} \sim 3 × 10^{18}\,\mathrm{\frac{J}{m^2}}$ or for distance not more then $R_1 \sim \sqrt{\frac{E}{4 \pi I_1}} \sim \frac{1}{2} \sqrt{E/E_g} r_E \sim 1.2 × 10^{11}\,\mathrm{m} \sim 400 \,c\cdot\mathrm{s}$. Moon and other inner planets have less binding energy. This means that Moon (distance $3.844 × 10^{8}\,\mathrm{m}$) would most definitely be destroyed and inner planets should partially survive (distance comparable to distance to the Sun, $1.496 × 10^{11}\,\mathrm{m}$, and probably some energy would be lost). Jupiter will survive. It is further and much bigger. Sun is also big enough to survive. Calculations for individual planets, moons, planetoids... would take too much time.
Planetary scale destruction is something of the order of $I_2 = \frac{1000\,\mathrm{\frac{kcal}{kg}} × 1000\,\mathrm{km}}{1000\,\mathrm{\frac{kg}{m^3}}} \sim 4 × 10^{9} \,\mathrm{\frac{J}{m^2}}$ (1000 km of water like-material heated by 1000 K or something equivalent). $R_2 \sim \sqrt{\frac{E}{4 \pi I_2}} \sim 7 × 10^{15}\,\mathrm{m} \sim 0.3 \,\mathrm{ly}$
Medium irradiation is something of the order of $I_3 = 1 \,\mathrm{\frac{J}{m^2}}$ or 0.1 Sv to the depth of 1 cm (Gamma ray absorption coefficient is of the order of 1/cm, so most energy will be absorbed by the surface 1 cm). $R_3 \sim \sqrt{\frac{E}{4 \pi I_3}} \sim 2 × 10^{20}\,\mathrm{m} \sim 2 × 10^{4} \,\mathrm{ly}$
Typical [real] gamma-ray bursts are observed to have a true energy release of about $10^{44}$ J, or about 1/2000 of a Solar mass energy equivalent—which is still many times the mass energy equivalent of the Earth (...) [and focus their] energy along a relatively narrow beam.
(From Wikipedia)
So they are a few orders of magnitude stronger. Probably they release substantial part of the rest energy of stars in something connected with black holes. You can read about gamma-ray burst to get idea of how it would work in the area, where irradiation would be the biggest effect (destruction of surface organisms on one side of a planet and ozone layer).
Values of used constants:
This cannot happen. Here's why:
Would it vaporize our solar system?
No. Destroying the solar system would violate the above laws.
Would it destroy space-time?
No. Absolutely not. Why would it "destroy" space-time? There's no possible mechanism.
Would it destroy the sun or even nearby stars?
I highly doubt it. The photons would spread out in a spherical shell, so the vast majority of them would not come near the Sun. And by the time they reached the nearest stars (4 light-years away - the Alpha Centauri system), they would not be very dense.
How much energy would be released?
Use $E^2=p^2c^2+m^2c^4$, or $E=\sqrt{p^2c^2+m^2c^4}$. Note, though, that the Earth also has kinetic energy, which would have to be accounted for, as well as gravitational potential energy.
If all planets had life on them how close would the closest planets be that still had multi-cellular life?
Ummm. . . Not sure I understand this one.
Okay, let's do some calculations.
Mass of Earth ($M_E$): $5.97219 \times 10^{24} \text{ kg}$
Speed of Earth: The Earth completes one orbit ($2 \pi r = 2 \pi (150,000,000 \text {km}) = 300,000,000 \pi \text { km}$) in $31557600 \text { seconds}$ (using 1 years as 365.25 days. This means it has an angular velocity $\omega$ of $1.991021278 \times 10^{-7} \text { radians/second}$. Using the formula for tangential velocity, $v=r \omega$, we find $v=29865.31916 \text { m/s}$, if the Sun if the point of reference. This leads us to the kinetic energy by $KE=\frac{1}{2}mv^2=2.663409478 \times 10^{33} \text{ Joules}$.
The Earth is moving, so we have to use $$E=\sqrt{p^2c^2+m^2c^4}$$ Defining $p$ as $mv=1.783613604 \times 10^{29} \text{ kg} \text{ m/s}$, We find $$E=5.374971 \times 10^{41} \text{ Joules}$$ Add this energy to the kinetic energy to find that the total energy released is $$5.374971027 \times 10^{41} \text{ Joules}$$ Now we use
Assume you could somehow convert the entire mass of the earth into energy in a moment, say a single second.
The total energy output of the sun is 3.8 x 10^26 J per second.
The total energy mass of the earth is 5.4 x 10^41 J
The fireball would be 15 orders of magnitude brighter than the sun, an unbelievably powerful event, almost as powerful as a supernova, probably powerful enough to destroy our sun.
The damage caused would depend on the type of radiation released. Visible light would look pretty but neighbouring solar systems would likely not be damaged. A gamma ray burst might irradiate one hemisphere of any planets orbiting nearby stars.
The flash would be briefly visible in the night sky throughout the galaxy.
Warning: shooting off the top of my head since I am about to go home.
The amount of energy would be given by E=mc^2 which would be 5.97219 × 10^24 kg × (299792458 m/s)^2 = 5.36753 × 10^41 J. Which is a lot of energy.
However, the damage done would be based on the type of energy. You say pure light. If by light you mean electromagnetic radiation, it would depend on how much of that energy is in a frequency that would damage anything.
If it's gamma radiation, for example, it might destroy life on other planets based on their atmospheres and whether there is a magnetosphere to deflect that radiation.
I doubt it would destroy the sun or any nearby stars though. The amount of radiation is space is enormous and your average supernova would probably release more energy than our planet in this situation. I would venture to guess that unless an object was relatively close, like in our solar system, the effects would be minimal.
|
asked |
|
|
viewed |
532 times |
|
active |
