Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

A familiar trope within science-fiction is that of a large relativistic object hitting a planet such as the earth. This is normally an interstellar spacecraft or a kinetic weapon with a mass in the range 103 - 106 kg. But what would actually happen?

The two scenarios seem to be: (a) the object creates a kind of conical tunnel through the earth with most of the material exiting on the far side; (b) the object dumps all of its kinetic energy within a few tens or hundreds of kilometres of the impact point and we have the equivalent of a conventional asteroid impact.

Light transit time through an earth diameter of 12,800 km is just over 40 milliseconds. There’s not much time for lateral effects as the object barrels in. So what would happen if a 1 tonne object hit the earth, square on, at 0.99995c (γ = 100)?

share|improve this question
add comment

3 Answers

up vote 6 down vote accepted

I think there would not be a hard collision of the object with the ground. It would pass through a column of air of comparable mass, so I suspect the energy of the object would be consumed before it hits the ground. This would be a sort of massive cosmic ray event. The interaction energy of atoms would be comparable to Tevatron energy (1/10 of that or so) and what would reach the ground would be a huge pulse of secondary particles that generate an enormous thermal-mechanical shock wave. At 10^3kg the energy would be $E~\simeq~\gamma mc^2$ or about $10^{19}J$ of energy. One ton of explosives is 4.18 mega Joules of energy, so the energy released would be about the equivalent of $2\times 10^7$ megatons of explosives. Clearly you would not want to be anywhere near this. However, I suspect the damage done to the Earth’s surface would not be due to the body actually impacting the surface, but from this huge shock front. So one would have to search out what is known about that sort of mechanics.

share|improve this answer
The KE of the impactor, as you say, is around 2,000 Gigatons of explosive. This is broadly equivalent to a mile-wide asteroid hitting the earth at 30,000 mph (science.howstuffworks.com/nature/natural-disasters/…). An object with the mass of a tonne is the size of a cubic metre of water or smaller so it has a small cross-section for interacting with the atmosphere (and not much time!). So although it's possible that the whole object will be plasma after atmospheric transit, I think the impactor density at the ground will be enormous. A big, deep crater! –  Nigel Seel Jan 20 '11 at 9:11
The problem is complicated. A cubic meter of material will pass through about 10 tons of atmosphere if it reaches the ground. So if we assume the material is denser then the impactor interacts with about as much atmospheric mass as it has. One might compare this body-air interaction to the collision of comparable masses at relativistc velocities in their center of mass. This makes me think the impactor interacts very strongly with the atmosphere. So there might not be that much left of the impactor as a solid object by the time this event reaches the ground. –  Lawrence B. Crowell Jan 20 '11 at 14:05
add comment

I think that this question should be considered also as a materials physics question. There will certainly be a very high energy and a momentum hundreds of times greater than a traditional asteroid. The intention of the question is that the material not break up before collision. What happens at and after collision is probably vaporization. I say probably because a massless neutrino object could travel through the Earth. Otherwise the vaporization will happen quickly (nanoseconds), but not instantly. So can a material exist which will fail to vaporize until it has burrowed through the Earth some distance? There are discoveries of non-standard materials like metamaterials, but these as yet would be unsuitable. So it might be a research project to find such a material - or prove that none could exist (a proof which would require a broader theory of matter than available today as it would need to incude Dark Matter theory as well).

share|improve this answer
For $E~=~\gamma mc^2$ as a rough guide consider the outer 1cm depth of material as it passes through 10km of the stratosphere. The atmospheric pressure there to be about 1/10 sea level this tube is around $10~-~100$kg of gas, and so assume the $10$kg. That is $~\sim~10^{24}$ molecules. Set $E~=~NkT$ deposited from the out 1cm layer or about 1/1000 the mass. $E = 10^{16}J$ or $T~\sim~10^{17}K$. The object has another 10km to go in about 10^{-4} sec. A nuclear bomb at 10^6K thermalizes itself in 10^{-5} sec. This is vaporized in a gas of hyperons and mesons before it hits the ground. –  Lawrence B. Crowell Jan 20 '11 at 17:49
Doesnt relativistic time dilation play a part in this calculation? From the perspective of a ground clock the missile will maybe take 10^{-4} seconds to traverse the 10 km, but the missile clock will register about 10^{-6} seconds. So depending on thermalization timescales it might well penetrate further. –  Roy Simpson Jan 20 '11 at 19:49
add comment

The answer depends on whether it's heavy enough to reach the ground. If it's small enough, it will fall apart in air and the result will be an explosion. To compute the size of the explosion use the relativistic energy formula.

Small meteors are slowed down by the air and reach the earth at their terminal velocity. These provide the samples you can buy on eBay. Meteors big enough to keep their natural speed of around 40,000 meters per second (i.e. 10 to 40 miles per second) vaporize on hitting the earth. This is what would happen to a relativistic meteor that was big enough to reach the earth.

To see why large meteors vaporize, note that the speed of sound in air is about 330 meters per second. This is the velocity for nitrogen molecules which have a molecular weight of 28 at say room temperature or around 300 Kelvin. Iron has an atomic weight of around twice that, so its velocity, at the same temperature, is around 0.7 as much (i.e. kinetic energy is given by $mv^2/2$ and the molecules in a gas at a given temperature all have the same average kinetic energy no matter how heavy they are, assuming they're a gas). So a meteor traveling at 40,000 meters per second has kinetic energy per atom of around 200 times the kinetic energy per atom of room temperature air. Thus the vapor produced is on the order of around 200 x 300K = 60,000 K. (Actual temperatures are a lot lower because the heat is spread around over more than just the meteor, the nearby rocks or whatever is also heated, but this gives the general idea.)

share|improve this answer
This is a great analysis of ordinary meteors, but my question concerns a highly relativistic, heavy object. If the atmosphere is taken to be ~50 km deep, its transit time is just 170 microseconds. Not much time for thermal ablation/diffusion, or mechanical stresses to propagate! A better metaphor would be a proton-proton particle accelerator. –  Nigel Seel Jan 20 '11 at 9:19
(1) Thermal physics applies to relativistic objects as much as they do to non relativistic. (2) 170 microseconds is long enough for relativistic stresses to propagate 50 km, LOL. If the object is 1 meter across, in 166 us the stresses only have to travel at speeds of 0.00002 c to cross the object. This is highly non-relativistic. –  Carl Brannen Jan 21 '11 at 4:33
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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