# What is the smallest yet lethal meteorite which can reach the surface of Earth? [closed]

Based on naturally occurring materials in space, what is the smallest possible size a 'natural' meteorite can be that can pierce the Earths atmosphere and kill a human on the ground?

Also, could said meteorite cause the human to burst into flames upon impact?

• This question is about real world physics, and isn't about world building. A candidate for physics SE? – Binary Worrier Dec 5 '17 at 10:17
• – JDługosz Dec 5 '17 at 10:35
• Yes it is very possible. Tens of thousands of meteorites between 10 grams and 1 kilogram fall on the surface of the Earth each year. – AlexP Dec 5 '17 at 10:54
• I'm afraid that physics SEs are more science-inclined which is not always needed in WB. If L.Dutch needs only approximate size of potential killer-meteorite, no need to go physicists just yet :). If it were me to decide, I'd keep this question as appropriate for WB. Besides, I anticipate some good detective story about a murder that was discarded as an unfortunate accident, huh. – user2851843 Dec 5 '17 at 11:00
• @user2851843, you got my plot the other way around :) they look for a killer until they realize sky is the killer – L.Dutch - Reinstate Monica Dec 5 '17 at 11:01

This answer is an approximation. Assuming the lethal meteorite strikes its victim with the equivalent kinetic energy of a bullet. A bullet will have a velocity of about 300 m/s, while meteors can have velocities up to 50 km/s.

Taking the upper bound for the velocity of meteorites of 50 km/s, this means a 'meteor bullet' will have 27,777.78 times the kinetic energy per unit mass.

Now taking this value it is possible to estimate the mass of the lethal bolide. of course, there is a range of masses for bullets.

Depending on the gun, the mass of a bullet usually ranges between 0.02 kilograms and 0.04 kilograms. The mass of a bullet depends on the caliber and type of gun used.

There is this caveat on the mass of bullets.

A bullet can be no more than 0.02 to 0.4 kilograms. Otherwise its ability to glide through the air would be halted. It wouldn't be the right size to pierce through anyone's skin or cut through a block of wood without the correct mass.

However, this mass limit might be overcome by the meteorite's velocity and its attendant kinetic energy. Also, it is probable that the meteorite's velocity will be reduced by its passage through the atmosphere.

This suggests a possible minute meteoric missile which can have the equivalent power of a bullet of a given mass and carrying its equivalent kinetic energy. This is complicated by the range of masses of bullets.

This answer is for guidance, but should provide sufficient information to devise a suitable lethal 'meteor bullet'.

According to this lovely page one of the smallest reported meteorites is 340-gm (12 ounces) piece of rock that penetrated roof of a house. It is still quite big, though, about a size of a tennis ball as far as my imagination goes. There is no record as to whether it was lethal or not, but if this thing is capable of going through the roof, there is no doubt that a direct hit would kill a human.

meteorite shower; boy hit on head by 3.6-g fragment after it hit tree first

Again, no lethality information provided, but it's still worth mentioning, I believe. Besides, this shot was slowed down by the tree, so the original impact force would've been much greater otherwise.

So, without hard science, my educated guess is that a piece of a space rock with 5-10 gramms weight can effectively kill almost any human, 100+ gm meteorite can do it even through the roof/car/etc, but that won't appear as a gun shot, I'm afraid.

Edit: all assumptions are made at the moment of the considered impact. As stated in the comments, the original size of meteorites is lot bigger before they burn away. Since we're talking about a detective story, I assume we're more interested in its size upon impact.

• 5-10 grams on impact. You can bet your life it was way bigger when it entered the atmosphere. – Rekesoft Dec 5 '17 at 11:58
• Thank you for the note, updated my answer to add clarity. – user2851843 Dec 5 '17 at 12:02
• @Rekesoft Wouldn't "way bigger" depend on the composition? A meteorite mainly made of ice would deteriorate differently than one made mainly of unobtanium or vibranium? I would think dense rock/metal/alloys would remain mainly intact as it falls - even through earths dense atmosphere... – WernerCD Dec 5 '17 at 13:30
• @WernerCD Surely. See my own answer down there. – Rekesoft Dec 5 '17 at 15:11

A dust/ice/compact earths/heterogenous sediments meteorite will burn out and dispel away in the atmosphere even if it's several tonnes big 99 out of 100 times. You must have some kind of meteorite which is bond together by something stronger than its (nearly nonexistant) gravity or the lousy amalgamations than water or other liquids can provide. You need metallic links.

So, if you have an almost pure ball of iron, you can be relatively sure that it will go all the way down to the Earth surface without melting (too much). Even then, its mass and aerodynamic coefficient will play a significant role in its letality. However, everything bigger that a prune should be able to hit the soil at speeds near the 500 m/s or more, so more than enough to kill someone.

An object which would do damage comparable to a bullet, would have lost its speed by the time it landed on earth.

That is why nobody ever died from a falling bullet, that was shot straight up into the air. It loses its momentum.

I would like to link to this post on the Space Exploration SE:

https://space.stackexchange.com/questions/12774/what-impact-would-a-pea-size-meteor-have-when-it-hits-the-ground

The accepted answer has all the information.

• Are you sure nobody ever died from a falling bullet? en.wikipedia.org/wiki/Celebratory_gunfire – Erik Dec 5 '17 at 13:34
• You are right there seems to be some information on that. However it does state that it's more likely when bullets are fired at an angle, which makes the bullet maintain its trajectory. Bullets fired straight up into the air tend to start to tuble and lose more momentum. – Whacko Dec 5 '17 at 13:42
• at expected terminal velocity of 100 - 200 m/s, a falling meteorite can do damage comparable to a bullet. But as it is not bullet shaped it most likely will need to be much heavier and will not be confused by authorities to a bullet. More likely they think it was a shotgun loaded with stone pebbles. – Angelo Fuchs Dec 5 '17 at 13:55

Where it hits, how fast it is going, and its composition also make a difference.

I am not sure how fast such an object would be going near the surface of the earth (after the atmosphere reduced its speed). Speed makes a difference but not in the way you might expect. As war guns were improved it was oddly noticed that bullet wounds to the head became less lethal. Faster bullets seemed to travel through the brain without damaging it as much. Of course bullets are mostly heterogeneous, a rocky meteorite fragment might break up in the head causing more damage.

An object less than one gram weight (by the time it reaches the surface) could kill someone if it went through a critical part of the heart, spine, or brain

Probably not.

Terminal velocity is inversely proportional to the size of the object. A very small (bullet-sized) meteorite would have a terminal velocity under 100mph, while a typical bullet travels at 1700mph. It would have to be very unusual circumstances (e.g. your character was looking skyward and took the meteor right in the eye and into the brain), so under normal circumstances I don't see how this could be fatal and look like a gunshot.

• Meteorites can start with 161 thousands mph, what exactly makes you think they will have enough time to slow down to terminal velocity? – Mołot Dec 5 '17 at 13:30
• The fact that most slow down to terminal velocity? – Keith Morrison Dec 5 '17 at 21:52
• @Molot The only way they would not slow down to terminal velocity is if they were much too large to be applicable to OP's question. In the first link I gave, the author does the math to show that a pea-sized meteorite would be slowed from 30km/s to terminal velocity in less than a second once it hits sea-level air pressure. Of course it would absolutely not survive the trip to sea level, nor the shock of this deceleration if it somehow did, so the whole thing is moot. If you've ever seen a shooting star, you've seen this in action. – Haydentech Dec 5 '17 at 22:28
• But they would need to start too large to burn down to applicable size, wouldn't they? So can you show a math for the bullet size at the end of the trip? If you can, then this will be a good answer. Now it looks like you use example that's unrelated to the question, as your last comment apparently admitted – Mołot Dec 5 '17 at 22:40
• Meteorites don't burn down nice and neatly, smaller and smaller. The shock of the atmosphere and the differential heating (front side vs back side) typically blows it into a million pieces. Could one of those bullet size pieces survive to the ground? Of course, but it would arrive there at about 70mph as mentioned above. Leaving aside the explosion heard for hundreds of miles, and chunks raining down all over the place, as we saw recently in Russia. So as to OP's question, exactly no one is going to mistake that for a single gunshot, and if so, it almost certainly would be non-fatal. – Haydentech Dec 6 '17 at 16:16

Based on this (the first hit when searching for skull mechanical strength), the average human skull has an ultimate compressive strength of $10.7 \times 10^3$psi = $7.4 \times 10^7$ Pa. Also Strength analysis of human skull on high speed impact

Let's say an average bullet has a 9 mm diameter,
Cross sectional area of the bullet = $\pi \times {(9 \times 10^{-3})^2} \over 4$ = $0.000063617 m^2$.
Force required for bullet to pierce skull ($F$), therefore, = $7.4 \times 10^7$ Pa$\times 0.000063617 m^2=4708$N.
Using $F=mg$, ignoring terminal velocity effects, $m$=$4708 \over 9.81$ = 500 kg.
This is the upper bound for a sphere at rest; we would have to integrate over the height of the atmosphere for $g$