1
$\begingroup$

Let's assume that the Earth has a body of equivalent mass directed at it that is made up purely of antimatter. This antimatter is held back in a containment bubble I created that moves it towards the Earth without interference from the Earth's atmosphere or particles moving around it. The bubble withstands reentry until it hits the surface of the Earth, when the antimatter collides with the matter. When the two collided, how much energy would be released? What would happen to the solar system, in particular, the sun? Would there be debris left over or not? Would the moon be destroyed?

I need a source of lots of energy to be harvested so my race can continue to scope out planets for use all across the galaxy (and this will destroy my enemy's planet as well, muahahaha.)

The raw amount of Joules is useful but can this be described in terms of energy outputs by comparison to a star? A supernova? A galaxy? How dangerous is it? Would it affect the sun or the moon? Ultimately the scale of the energy release is important to understand whether arranging this fate for my enemy would be strategically useful or just plain dangerous, since my home planet is in the same system.

$\endgroup$
  • 10
    $\begingroup$ This is really just a physics problem that requires googling a few things. $\endgroup$ – Cort Ammon - Reinstate Monica May 2 '18 at 22:03
  • 7
    $\begingroup$ @JBH I'm voting to close this as off-topic, because it is about physics and not worldbuilding. There is no known mechanism to make such a planet happen, and the OP has provided none, so this is basically just suspending disbelief and plugging a planet's mass into $E = mc^2$. $\endgroup$ – kingledion May 2 '18 at 23:43
  • 3
    $\begingroup$ @JBH I would say that the justifications for closure here are that a) this is a physics question, plain and simple (I believe the answer can be easily solved using the mass-energy equivalence equation); and b) it doesn't have anything to do with worldbuilding. It neither helps conceiving of the physical nature of a fictional universe nor helps in investigating the cultures, histories, mythologies, languages, technologies, & scientific application of any of the peoples of that fictional universe. (.i., what makes the fictional universe different from the primary universe we inhabit.) $\endgroup$ – elemtilas May 2 '18 at 23:44
  • 4
    $\begingroup$ @elemtilas, we answer physics questions on this site all the time.. The definition of "worldbulding" is in the tour and help center. Until changed or updated, those (and only those) are the rules. "Worldbuilding" is specifically defined as things smaller than a village and larger than a multiverse. Scope is therefore not justification for closure. Show me in the help center where the OP violated the rules, justifying closure. $\endgroup$ – JBH May 2 '18 at 23:58
  • 10
    $\begingroup$ This question is perfectly on-topic for world building. Suggesting to the poster that they may get a better answer on physics.se is perfectly fine. But that is the user's choice. If the user flags it to be moved to physics we will migrate otherwise we keep it here. $\endgroup$ – James May 4 '18 at 17:42
4
$\begingroup$

How much energy is released?

Let's talk about energy.

Imagine, first, two point particles - one of matter, one of antimatter - each with the mass of the Earth. The Earth orbits the Sun at around 30 km/s; if the particles collide on, moving in opposite directions, then their relative velocity will be 60 km/s. Let's use some special relativistic formulae to figure this out. In the reference frame of one of the planets, its speed is 30 km/s1. Therefore, the sum of its kinetic and mass energies is $$E=\frac{m_{\oplus}c^2}{\sqrt{1-\frac{v^2}{c^2}}}\approx5\times10^{41}\text{Joules}$$ where $v$ is the velocity and $c$ is the speed of light. Thus, the total energy of the two planets, in this frame, is $\sim10^{42}$ Joules, as others have said.

However, there are other types of energy we might need to take into account - and in particular, we have to think about gravity. First, consider gravitational potential energy. The gravitational potential energy of one planet due to the Sun is $$U_{\odot}=\frac{GM_{\odot}m_{\oplus}}{r}\approx5\times10^{33}\text{ Joules}$$ where $r$ is the distance to the Sun, and so the total potential energy due to the Sun is $\sim10^{34}$ Joules. The potential energy of each planet due to the other is assumed to be close to 0, as they begin far away from each other, and any increase in potential energy corresponds to a corresponding decrease in kinetic energy.

We also have to consider gravitational binding energy. For a sphere of uniform density - and we'll say that each body has roughly uniform density - the binding energy is $$U_{\oplus}=\frac{3Gm_{\oplus}^2}{5R}\approx2.2\times10^{32}\text{ Joules}$$ where $R$ is the radius of Earth. Therefore, the total binding energy of the system is $\sim5\times10^{32}$ Joules.

We can see that both $U_{\odot}$ and $U_{\oplus}$ are much less than $E$, and so we can say that the total energy released is $\sim10^{42}$ Joules, released in a very short amount of time (although I don't know just how quickly). We've shown that the other types of energy are negligible.

How much is this, really?

Let's talk about other events that release a lot of energy.

  • Superflares: These are, essentially, just really powerful stellar flares, observed on a number of (mostly Sun-like) main sequence stars. An early search discovered energies of up to $\sim10^{31}$ Joules, although that number could be increased.
  • Helium flashes: Helium flashes occur in red giants. Helium in the core of the star eventually becomes degenerate. At this point, the core stops shrinking, but temperatures continue to rise, until the pressure and temperature reach levels where sudden fusion of helium takes place, releasing $\sim5\times10^{41}$ Joules of energy - about the same amount of energy released by your collision.
  • Type Ia supernovae: A Type Ia supernova releases $\sim3\times10^{44}$ Joules of energy. This energy is roughly the same for all Type Ia supernovae, since they (mostly) arise from progenitors of the same mass, $1.4M_{\odot}$. An example is SN 1572, also known as Tycho's Supernova. That said, a supernovae with double degenerate profenitors (i.e. two white dwarfs) could release more energy; a possible example is the supernovae responsible for SNR 0509-67.5.
  • Kilonovae: A kilonova - resulting from the collision of two neutron stars - should release a fairly large amount of energy. The kilonova that was detected by LIGO and Virgo in August of 2017 radiated perhaps $5\times10^{45}$ Joules, according to some models, although kilonovae could be less energetic by a factor of perhaps $10$.
  • Type II supernovae: A Type II supernova releases more energy than a Type Ia supernova, on the order of $\sim10^{46}$ Joules. Some of this energy is emitted as photons, but most of it is released as neutrinos, which are harder to detect. Type II supernovae (and Type Ia supernovae) slowly fade over time, and remain luminous for weeks or months or years, but the majority of the energy is released very quickly, in a matter of seconds. An example is the famous SN 1987A, which, by the way, led to an historic detection of neutrinos from the event.
  • Gamma-ray bursts: These events have a number of possible prognitors, including Wolf-Rayet stars, kilonovae, and highly luminous supernoave. Many release more energy than a typical Type II supernova, but others are less powerful (see GRB 101225A), a long gamma-ray burst that released a little under $\sim10^{44}$ Joules of energy.

I think, then, that the closest comparable event would be a helium flash, although helium flashes are typically not visible outside the host star.

A possible failed annihilation

After doing some more thinking, I considered that the collision might only directly result in a partial annihilation of both planets - stopped, interestingly enough, by the radiation emitted from the initial collision. When an electron and a positron annihilate, gamma rays are produced: $$e^-+e^+\to\gamma+\gamma$$ each photon has an energy of about $0.511\text{ MeV}$. If there are $\sim10^{50}$ atoms in the world, then there are at least $\sim10^{50}$ electrons (we assume that there is at least one proton for atom, and that most atoms are electrically neutral). If each photon has a momentum $p=E/c$, then each collision produces two photons of momentum $2.73\times10^{-22}\text{kg m s}^{-1}$. If each planet is traveling at 30 km/s,2 it has a momentum $\sim10^{29}\text{kg m s}^{-1}$. Therefore, if perhaps $\sim5\times10^{49}$ electrons are annihilated in the earliest phase of the collision - roughly one half - and one photon from each annihilation travels towards one of the planets,3 then $\sim10^{28}\text{kg m s}^{-1}$ worth of momentum would oppose each moving planet. We might have underestimated the number of electrons in Earth, and if we're off by an order of magnitude, the annihilation of half of the planets could blow away (and blow apart) the remnants.

We also haven't taken into account photons produced by proton-antiproton and neutron-antineutron annihilations, which should produce even more photons of higher energies (as the masses of protons and neutrons are much, much greater than the mass of an electron). In short, I think the collision would pan out something like this:

  1. The planets initially collide. Annihilation begins in the atmosphere and crust.
  2. Photons produced by the annihilation travel in various directions, many applying radiation pressure to the planets and slowing them down.
  3. The planets continue moving together albeit at a lower speed. The increased radiation begins to seriously push against them.
  4. Parts of the remnants are either broken apart or expelled outwards by the streams of photons.

In other words, the planets might be blown apart by radiation pressure. We see a related problem in massive stars, where radiation pressure limits the upper mass a star could have. At masses greater than a certain value, radiation pressure in the interior would be so strong it would blow the star apart.


1 I'm considering 30 km/s to be its initial kinetic energy - its "velocity at infinity".
2 They might be traveling faster than this, as gravity accelerates them toward each other. Of course, my value of 30 km/s for the initial velocities could be an underestimate or overestimate.
3 In reality, I suppose the emission would be isotropic, and the photons wouldn't have to hit the planets, or even hit head-on. However, note that we've severely underestimated the number of photons produced so far, by ignoring protons and neutrons.

$\endgroup$
  • $\begingroup$ This is just so, so much better than the other two answers. $\endgroup$ – FoxElemental May 16 '18 at 15:58
  • $\begingroup$ "Imagine, first, two point particles - one of matter, one of antimatter - each with the mass of the Earth": what's the point? Earth is not a particle. In particular, as you have shown, the gravitational binding energy of Earth is one billion times smaller than the annihilation energy; which means that very very quickly the gravitational binding energy will be overwhelmed and Earth will explode in a cloud of debris, which will then proceed to transform into superheated gas. About half of this gas cloud will be moving away from the other planet. The same thing will happen to the antimatter. $\endgroup$ – AlexP May 16 '18 at 18:32
  • $\begingroup$ @AlexP That was just a first approximation, which I extended afterwards to account for that. I'm planning to add some stuff on what the radiation from the first contact might to do the explosion (e.g. could it drive the remaining bits of the planets apart?). $\endgroup$ – HDE 226868 May 16 '18 at 18:35
4
$\begingroup$

The Earth weighs $5.972\cdot10^{27}$ grams. One gram of antimatter equals $42.96 \cdot 4.184\cdot10^{12}$ Joules ($1.7974464 \cdot 10^{14}$ Joules.) $1.7974464 \cdot 10^{14} \cdot 5.972\cdot10^{27}$ equals $1.07343499\cdot10^{42}$. That means that $1.07343499\cdot10^{42}$ Joules are the equivalent output. To put that into perspective, the amount of energy the Virgo Supercluster (including the Milky Way) puts out every second is $1\cdot10^{42}$. Nearly equivalent to one Earth-mass of antimatter colliding with matter.

So quite a lot. It is also equivalent to $2.7 \cdot 10^{32}$ tons of TNT. Compare this to a lower number: $4.184 \cdot 10^{15}$ Joules (equivalent to 1 megaton of TNT or 67 Hiroshima bombs) or a slightly higher number: $7.0 \cdot 10^{27}$ (2 Exatons TNT or the max output of the Death Star's superlaser). The amount of energy output is stunning. You could get good fuel, as long as you weren't vaporized.

$\endgroup$
  • 2
    $\begingroup$ Wouldn't annihilation release the energy of both matter and anti-matter colliding? So double that? A bigger issue is that the weaponized planet won't have anywhere near enough kinetic energy to actually fully interact with the target. $\endgroup$ – Ville Niemi May 2 '18 at 22:07
  • 1
    $\begingroup$ @VilleNiemi: I based my calculations on the power contained in the matter/antimatter explosion. As for the kinetic energy problem--I'll leave that to whomever uses this schematic for their idea. $\endgroup$ – FoxElemental May 2 '18 at 22:12
  • $\begingroup$ That sounds like a smart decision ;) $\endgroup$ – Ville Niemi May 2 '18 at 22:19
  • 4
    $\begingroup$ The problem is probably not nearly so simple as applying E=MC^2. I would guess that the first tenuous contact will release enough energy to vaporize the planets, and sending the vapor expanding outwards, much like a type 1a supernova. $\endgroup$ – jamesqf May 3 '18 at 4:22
2
$\begingroup$

The relevant equation is Einstein's famous

$E=mc^2$

In plain English, this says that the energy equals the mass that you annihilate, times the speed of light squared. In our case, we're taking the mass of earth plus the mass of the antimatter planet.

I went to wolframalpha.com, and typed in this

2 * (mass of earth) * c^2

And the answer is $1.0735089\times10^{42}$ J (joules)

Wolfram Alpha gave the following helpful equivalency: $( \frac{1}{93} )\times$ energy released from a supernova $(\approx 1\times 10^{44} J$ )

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.