# Earth Exploding

If the Earth all of sudden was instantaneously transmuted into pure light, what would happen?

• How much energy would be released?
• How much damage would it do to surrounding space?
• Would it destroy space-time?
• Would it vaporize our solar system?
• Would it destroy the sun or even nearby stars?
• If all planets had life on them, how close would the closest planets be that still had multi-cellular life?
• depend what you mean by a ball of light. Without any combustible, it's going to fade quickly. – Vincent Nov 14 '14 at 21:08
• The total mass of the light would be equal to that of Earth's. Initially the density of the light(weight wise) would be equivalent to Earth. – TheNumberOne Nov 14 '14 at 21:16
• Not sure if you can 'transmute' mass to light, mass to energy yes...are you looking for the effects of a mass the size of earth instantaneously turned to energy? 10^41 is generally regarded as the energy-mass of the earth as per wiki here en.wikipedia.org/wiki/Orders_of_magnitude_(energy) while a supernova is 10^44. So about 1/1000 the energy released in a supernova. – Twelfth Nov 14 '14 at 21:34
• Apologies for the edit. I did it so I could undo my downvote, and to organize the question a bit better. Feel free to roll it back if you don't like it. – HDE 226868 Nov 14 '14 at 22:57

Such explosion is of course practically impossible in reality, but from the point of view of modern physics this would not break any absolute conservation laws:

1. Earth is almost electrically neutral like everything such big. If you take into account the relatively small charge, you would need to leave some electrons for negative charge and positrons or protons for positive charge, but this would not make much difference.
2. Baryon number (B) and lepton number (L) conservation would forbid transformation of normal matter into photons, but this conservation laws are not absolute. Sphalerons in the Standard Model, if not grand unification theory, let change them, conserving only B-L. So you would need antineutrino (B-L=0-(-1)=1) for every neutron (B-L=1-0=1). Electrically neutral pair of proton and electron has B-L=0.
3. B-L also probably is not absolutely conserved. For example if neutrinos have Majorana mass, neutrinos are at the same time antineutrinos, so L=1 can transform into L=-1.

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:

• $$\pi = 3.14159265$$
• $$c = 299792458 \,\mathrm{\frac{m}{s}}$$ (exactly)
• $$h = 6.62606957 × 10^{-34} \,\mathrm{J s}$$
• $$\hbar = \frac{h}{2 \pi} = 1.054571726 × 10^{-34} \,\mathrm{J \cdot s}$$
• $$k_B = 1.3806488 × 10^{-23} \,\mathrm{\frac{J}{K}}$$
• $$1 \,\mathrm{M}e\mathrm{V} = 1.60217657 × 10^{-13} \,\mathrm{J}$$
• $$G = 6.67384 × 10^{-11} \,\mathrm{\frac{N \cdot m^2}{kg^2}}$$
• $$m_E = 5.97219 × 10^{24}\,\mathrm{kg}$$
• $$r_E = 6.3844 × 10^{6}\,\mathrm{m}$$
• Wow. Really nice analysis. – HDE 226868 Nov 15 '14 at 22:46
• This is one of the best first posts I've ever seen. +1, and welcome to Worldbuilding! – Shokhet Nov 16 '14 at 0:10
• IOW it is almost possible to convert E=mc.c, after we add some exotic particles (which might or might be not allowed by original question - not obvious). But you are right, of course nobody else would notice. – Peter M. - stands for Monica Dec 17 '14 at 22:25
• I have a problem with the planetary destruction numbers. While it might deliver more energy than the gravitational binding energy it's going to show up as heating the visible face to incredible temperatures. How much of that energy actually goes into disrupting the planet vs how much simply blasts/radiates into space? I would expect the efficiency of this process to be very low, not the 100% that this math assumes. – Loren Pechtel Nov 7 '19 at 0:06

This cannot happen. Here's why:

• Conservation of electric charge would be violated: The Earth is made of protons, neutrons, and electrons. Electrons are charge; protons and neutrons contain particles with charge. If you turned the Earth into a bunch of chargeless photons, all this would be gone. The only way out would be to say that the Earth previously contained an even number of protons and electrons, but this is highly unlikely.
• Conservation of baryon and lepton number: Electrons are leptons, which are a certain class of elementary particle. Destroying the Earth would mean destroying electrons, which would violate conservation of lepton number. Protons and neutrons are baryons, a class of composite particles made of quarks. Destroying the Earth would mean destroying protons and neutrons, violating conservation of baryon number.

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

• Total theoretical mass to energy of earth comes out to about 10^41 Joules (listed on wiki). Interesting answer to a seemingly silly question +1 – Twelfth Nov 15 '14 at 0:14
• @Twelfth Thanks. I've been tempted to use the Earth's momentum and mass to figure out the wavelength of the photons, but I might wait for tomorrow to do that. – HDE 226868 Nov 15 '14 at 0:17
• Your last edit just kind of ends here. – kingledion May 6 '17 at 17:18

## You'd likely do some damage

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.

• energy mass of earth is 10^41...mass in your equation should be closer to 10^25, no? – Twelfth Nov 14 '14 at 21:35
• Correct I actually accidentally put in the answer for the calculation. – Jake Nov 17 '14 at 12:39