# How far would an antimatter asteroid made inside our solar system? [closed]

Imagine right now an antimatter asteroid the size of Hawaii of extraterrestrial origin enters our solar system and is on a collision course with one of the inner planet, Mercury at a fantastic speed of 100km/s. I wonder how far would it made assuming Murphy's law holds true(excluding crash landing on object 100 times more massive than itself) ? But you know antimatter would completely and utterly annihilate ordinary matter to release huge amount of energy, term and condition apply.

• actually, this asteroid would detonate the moment it come into being, so this scenario is not anything I would regard as possible, if not unlikely Commented Jun 21, 2020 at 6:57
• It wouldn't in a real vacuum. It would only annihilate if it came into contact with positive matter. Commented Jun 21, 2020 at 9:18
• Related question. Commented Jun 21, 2020 at 17:00

Perhaps it would get to Mercury, although I'm a little doubtful. 0.1% (as per Charisturcear's answer) of an object the size of Hawaii is a huge amount. Double that mass (when it comes into contact with the anti matter) then convert that into energy and the object would be slowly ripped apart on the approach and also decelerated. The nearer it got the larger and more defuse the cloud of fragments would get and the more matter particles it would encounter making things even worse.

In addition closer into the sun in the proximity of the Earth particle density is much higher. Probably more like 5,000,000 particles per cubic metre. The asteroid would slow and expand into a catastrophic maelstrom of radiation that would be powerful enough to sterilize planet Earth and possibly also remove planetary atmospheres in the inner solar system. The deceleration effect would be difficult to predict and if it survived at all it would probably miss Mercury.

Well ignoring the inability of existence of such an object, as 'Murphy's law holds true' the asteroid will pass the Kuiper-Belt and all other objects of our solar system without contact so we can skip them for our considerations. We also ignore the possibility of shatering by all the anihilation energy it gets when getting in contact with particles.

Hawaii has a surface of $$10,500 km^2$$. Lets assume the asteroid is roughly cubic, he would have a length of $$100km$$ in every direction. For unknown asteroids its usual to take an average-density of $$2g/cm^2$$ or an average $$1.5*10^{23} particles/cm^2$$. Particle density in outer space is $$10^6 particles/m^2$$ or $$10^2 particles/cm^2$$. For simplicity we will assume al particles to be identical so one particle negates one anti-particle. By that our asteroid will lose $$1cm$$ of its length for every $$10^{21}cm$$ it moves. The asteroid is about $$10^7 cm$$ in length so it roughly gets $$10^{28}cm=10^{23}km=6,7*10^{14}AU$$ into the solar system. As the solar system is less than 2,000 AU in diameter, rhe asteroid will hit Mercury with more than 99.9% of its initial mass, the energy of this extinction event should destroy mercury.

As said, these equations are all very simplified and ignore special cases and side effects like anihilation energy and all of that. In our universe, such an eventy is not possible in any way. The asteroid would never come into existence, even if it did the first few particle contacts would rip him apart (proton-antiproton-anihilation can produce more then 2GeV easily) and send the fragments in every direction, causing havoc in half the solar system and so on.

• particle density is better expressed in units volume, and the number of intercepted particles is then surface time velocity times particle density per volume
– L.Dutch
Commented Jun 21, 2020 at 8:27