What would happen if the Earth struck a tiny but immovable object

Question

We are living on this big lump of rock called the Earth, drifting ever so continuously through space.

Now imagine a tiny but immovable object in space, a single atom of 'unexplanium', locked in space-time (relative to our universe or whatever you need). It will not move, bulge or transform; it's just there having indestructible mass. Assume it has the properties of a hydrogen atom (1 proton + 1 electron) for all physical and chemical purposes, except that it is endowed with infinite inertia.

What would happen if our Earth hit that one immovable atom? For simplicity, we may assume that the Earth's movement is linear and that the atom will hit the center of the Earth. I imagine something like a truck hitting a stationary object. If it's super narrow it'll slice the half, if it's broad it'll smash into it.

Would we even know it happened? Or would the amount of energy rip a giant tunnel through the planet? If a single atom is too small for something significant to happen, what would happen if it's the size of a marble or something bigger?

---

I've found this XKCD, which is somewhat related and might be an interesting read. It also described super fast and tiny particles:

"If a meteor made out of diamond and 100 feet in diameter was traveling at the speed of light and hit the earth, what would happen to it?”

Answer 1

As @Thucydides mentioned, researchers posited a similar magnitude of impact and thought they had found evidence for it, though they later retracted their findings they did calculate the potential effects.

The team, from the Southern Methodist University in Texas, analysed more than a million earthquake reports, looking for the tell-tale signal of strangelets hitting Earth.

While their very high speed gives strangelets a huge amount of energy their tiny size suggests that any effects might be extremely localised, and there is unlikely to be a blast big enough to have widespread effects on the surface.

The scientists looked for events producing two sharp signals, one as it entered Earth, the other as it emerged again. They found two such events, both in 1993. The first was on the morning of October 22. Seismometers in Turkey and Bolivia recorded a violent event in Antarctica that packed the punch of several thousand tons of TNT. The disturbance then ripped through Earth on a route that ended with it exiting through the floor of the Indian Ocean off Sri Lanka just 26 seconds later - implying a speed of 900,000 mph.

The speed of 900,000mph is important here. Since you're determining that the particles are absolutely stationary we need to know how fast the Earth is moving relative to them.

What is the speed of earth relative to absolute space?

In practice we approximate the comoving frame as the frame at rest with respect to the microwave background radiation. Thus we measure the speed of Earth relative to the comoving coordinates assuming, that the background radiation has no "natural" dipole anisotropy.

The speed of Earth wrt the local comoving frame measured this way in of the order of 500 km/s. And, of course, it varies as the Earth orbits the Sun. If we subtract the velocity of Earth in orbit around the sun, velocity of the sun relative to galactic center and velocity of the Milky Way with respect to the centre of the local group of galaxies, we find, that the local group is moving and slightly above 600 km/s relative to the comoving frame.

900,000mph is ~400km/s, you're looking for an impact appropriate to 600km/s, but the orders of magnitude have the same approximation as the incident in the Telegraph article

several thousand tons of TNT

Answer 2

Conclusion:

In the case of the atom, you wouldn't see anything without special instruments. This is because the object is so small that very few atoms will interact with it, so it imparts almost no energy to any matter that it encounters.

In the case of the marble, there would be a bright flash streaking through the sky, comparable to a large meteor, but much faster. Upon impact, the top layers of soil would be vaporized, resulting in a blast comparable in intensity to a very large conventional explosion (tens of tons of TNT). This would coincide with a small localized earthquake and a ground shockwave comparable to an underground explosion, due to the object passing through the rock at greater depths.

At the exit site, almost identical effects would occur, except there would be a large explosion followed by a flash streaking upward into the sky.

The exit event would happen $34$ seconds after the original impact, assuming the object comes down vertically and is moving at $370 \text{km} / \text{s}$ (see below about these assumptions).

Detailed explanation and calculations:

I. Hydrogen Atom

Let's first do a quick back-of-the-envelope estimate for how much energy would be imparted to the atmosphere.

1. Object's speed relative to the Earth.

First of all, what does "at rest" mean? Let's suppose the object is at rest with respect to the cosmic microwave background (CMB). The rationale is that if it originated in the primordial Universe, then it would have originally been at rest with respect to the CMB, and if it has infinite inertia, then nothing can change its motion, so it will remain at rest with respect to the CMB. The Earth moves at about $370 \text{km} / \text{s}$ with respect to the CMB, so that's how fast the object would hit.

2. Interaction with the atmosphere.

When the object passes through the atmosphere, then any atoms it comes into contact with will bounce off at approximately this speed, on average. Let's first calculate how much mass of air it will encounter. A column of Earth's atmosphere (from the surface to space) has a mass of about $10\mathpunct{,}000$ kilograms per square meter of surface. The object is about the size of a hydrogen atom, with a radius of $10^{-10}$ meters, so if it's a sphere, it will take out a cylinder of atmosphere with an area of $\pi R^2 \approx 3\mathrm{x}10^{-20} \text{m}^2$. Multiplying this by $10\mathpunct{,}000$ kilograms gives us $3\mathrm{x}10^{-16} \text{kg}$ of air that directly interacts with the object.

After this object passes, these air molecules will be moving around at about $370 \text{km} / \text{s}$. Using the expression $E = \frac{MV^2}{2}$ for kinetic energy, we get $$E = \frac{(10^{-16} \text{kg})(370\mathpunct{,}000 \text{m} / \text{s})^2}{2} \approx 7 \mathrm{x}10^{-6} \text{J}$$ energy imparted, in Joules. This is an absolutely minuscule amount of energy; for comparison, a normal household 10W halogen bulb emits 10 Joules of light per second.

3. Effects on rock would also be negligible, so there is no need to calculate them.

The amount of energy deposited when the object strikes the surface will be similarly negligible.

In the above calculation, the energy imparted to an object is proportional to its column mass (mass per surface area).

Consider the first three meters of soil that the object passes. A layer of rock 3 meters deep has about the same column mass as the atmosphere. Therefore, the object would also deposit about $7 \mathrm{x}10^{-6} \text{J}$ in the first three meters of rock that it penetrates. Again, this is practically undetectable. It will, of course, continue depositing these tiny amounts of energy as it passes through the Earth.

II. Marble

In the case of the marble, the calculation is almost the same as above, except the object's radius is now more like $0.01$ meters instead of $10^{-10}$.

So its radius is a factor of $10^{8}$ bigger, the area of the column of atmosphere it takes out is $10^{16}$ times bigger (as it's proportional to the radius squared) and the amount of energy deposited is also $10^{16}$ times bigger.

The marble would therefore deposit $7 \mathrm{x}10^{-6} \text{J} * 10^{16} = 7 \mathrm{x}10^{10} \text{J}$ of energy in the atmosphere, and a similar amount in the first three meters of rock.

Since we're looking at explosive-like effects, let's convert this to tons of TNT. One ton of TNT releases $4.2 \mathrm{x}10^{9} \text{J}$. So, our object would deposit $$(7 \mathrm{x}10^{10} \text{J})(\frac{1 \text{ ton}}{4.2 \mathrm{x}10^{9} \text{J}}) \approx 17 \text{ tons}$$ of TNT in the atmosphere and in the first three meters of soil.

The effects from the atmosphere would be comparable to a sizable meteor fireball. On the ground, the energy from the first few meters will reach the surface, producing an explosion comparable to a few tens of tons of TNT (a very large bomb).

Anything deeper than a few tens of meters would produce a relatively short-range surface rumble, like a small earthquake, but relatively little energy will reach the surface.

The same effects would happen at the exit site but in reverse order. The diameter of the earth is about $12700 \text{km}$, and the object is moving at $370 \text{km} / \text{s}$, so if the object comes down vertically, the exit event would happen $12700 / 370 \approx 34$ seconds after the first impact.

Answer 3

With an impactor the size of a hydrogen atom it's probable that we wouldn't even notice overly much. There might be a bright flash as it hits the atmosphere, it's even more probable there'd be one as it hit the ground, hard enough to break some molecular bonds. That flash is going to go all the way through too but we're not going to see it below ground level. It's almost impossible, certainly at such low relative velocities, to get an individual atom to actually hit such an object, even one with no charge, but because of it's exotic nature there might be some atomic fission events. Taken altogether I would expect very little energy to be imparted to the Earth, and very little if any noticeable damage.

Answer 4

At first I thought this would be a planet-ending event, but upon further consideration, it wouldn't be that big of a deal.

First off, nothing can really be immovable, and nothing is really stationary in an absolute sense as others have mentioned. However, if we assume that the earth is going around a few hundred thousand miles per hour relative to the cosmic background, that's a number we can work with: let's call it 400,000 mph. That's not a relativistic number, so nuclear reactions are unlikely.

Neutrinos have a small cross section and routinely pass through the earth unimpeded. However, a hydrogen atom has a much larger area; this is the reason hydrogen can be stored in a pressure vessel, because it bounces off the walls of the container.

Now hydrogen has a diameter of about 50 picometers, so at the very least the particle would scrape out a 50 picometer-wide tunnel all the way through the earth. The mass of that tunnel would be about $6\times10^{-10}\text{ kg}$.  If all of that mass were converted to pure energy, which is a worst-case scenario ($\text{E}=\text{mc}^2$), it would be about $5.6\times10^{7}\text{ Joules}$, which is about the amount of energy released by burning a couple liters of gasoline.

So the good news is we'd all survive.

Answer 5

As we now know - everything is relative.

A fixed particle in space might as well be a moving particle in space, relative to us, when we are moving.

This being the case, the atom in question would simply be like many other atoms already bombarding the Earth - individually the effect would likely be negligible, collectively a different story.

The only difference in your case is your comment 'infinite inertia' - I presume this atom doesn't interact like normal atoms do then when in proximity or colliding with another atom. Still, one atom is only one atom, it may not directly interact with others at all on its brief journey through the Earth, or if so only imperceptibly.

Answer 6

Nothing will happen.

Remember, there are (comparatively, when you think about the ratio of macroscopic distances to the sizes of macroscopic objects), huge gaps between atoms, which is why hydrogen will gradually leak out of any container. In much the same way, the fixed hydrogen atom will mostly push other atoms out of the way. In the atmosphere, we might see a parabolic arc of lightning as the the atoms gets knocked away at the speed of the earth's rotation. When it hits the ground, the fixed atom will simply force its way through, displacing atoms where it comes in contact with the nucleus, until it comes out on the other side. Then we might see another parabolic lightning arc going the other way.

Answer 7

As the asker of the question has already learned, it was asked using some phraseology which is difficult to mesh with contemporary physics. As such, I am attempting to answer as if “immovable” is describing:

• an elastic collision in which the relative motion between a third body which does not collide and one participant in the collision is unchanged.

While the Earth is moving in relation to this “immovable atom” — IA — which you propose, each atom of the Earth has a velocity vector with respect to the IA. The distance between the IA center of mass and the Earth's center of mass is decreasing. When any atom considered to be a part of the Earth collides with this IA, the result of that collision does not cause any change of the IA's relative velocity vector with respect to the velocity vector of any other atom.
In simpler language, the incident atom which is not the IA will bounce back with exactly the same speed but with a deflected direction. Any other atom will continue to see the IA as having the same velocity vector as it did prior to the collision.

Ergo, the IA will continue to pass through the Earth with each successive collision — however, the relative velocity vector between the Earth's center of mass and the center of mass of the IA will not be unchanged, but will be apparently slowed, albeit almost immeasurably. Such will occur because the velocity of the Earth's center of mass is composited from the velocity vectors of each atom which is factored in the arbitrarily defined center of mass …

Anyway. So, we have not yet defined exactly what fields are contributing to these aforementioned collisions. If the IA is uncharged, i.e. not ionic, then the collision is simply a result of the repulsion between electrons — which is, in turn, a consequence of the Pauli Exclusion property which electrons have — when its electron orbitals approach the similar orbitals of other atoms. It is very improbable that nuclear orbitals will ever interact, but at higher velocities such interactions become more probable — i.e. like the velocities you'd see in an atom smasher. And, even then, not very likely.

Of course, all that lends problems to considering exactly how the IA is indeed ‘immovable’, and how that pertains to its constituent particles: its electrons, quarks, and all the other stuff that perhaps fluxes about between in unperceived spaces inside. Gluons and photons and whatnot.

What if the IA were larger — like a pea, or a ball-bearing, or a marble? Well, all that really does — aside from making its ‘immovability’ more difficult to define — is give the IA a larger cross–section for its collisions: i.e., collisions are more likely.
Because it is immovable, and any collision with it is perfectly elastic, its own mass is irrevelant. Only its size is relevant.

What would be the state of that big mass of atoms which we call the Earth when the IA has passed through them and produce no further interactions?
Well, it really depends on what the initial relative velocity was between the two centers of mass. Many other answers here seem to confuse ‘immovable’ with ‘motionless’.
Whatever relative velocity you finally decide to use in your world, know that a larger magnitude will cause more deflection and thus more damage.

---

All in all, the more details you want require a more rigid definition understanding the concept of ‘immovable’. Others have more or less explained why immovability is almost like a physical limit: you can become more or less immovable, but never perfectly immovable.
Unless, of course, you exist outside the laws of physics which are known to contemporary science.

If that is the case, then this answer is really only useful to you if your physical laws operate in any way which allows my explanation to seem reasonable. Obviously.

Answer 8

Immovable (relative to the Earth), and potentially indestructible, marble sized object? "Immovable" could only mean it exists at the exact center of the universe, around which everything else is in relative motion.

It would be surrounded by a cloud of dust and asteroid fragments from previous impacts, would it not? Unless it sprang into being shortly before impact with Earth. Dust zone size depending on how long this marble has existed.

We would enter that zone of space dust, causing a massive and unexpected meteor shower. Then a flare of light as the marble passed through the atmosphere, followed by minimal impact, and a shudder as it plowed through Earth.

This would be a "bullet" that did not flatten or break on impact. The collision would be no big deal, but pressure would surely increase as it bulled its way through. Resulting in confusing seismic readings on the opposite side, then a small eruption as the marble exited. Or, if it entered and exited through oceans, no discernable effect on the surface of the ocean.

Answer 9

The answer is known, and it’s that nothing happens at all. If something is immovable, it means that nothing can move it, which can only mean that nothing we know of will interact with it. You’re describing a particle of dark matter, which the Earth undoubtedly hits all the time, but since it doesn’t interact with the matter of the Earth (except gravitationally, which is too weak and too long-range to be noticeable) nothing happens.

Answer 10

This scenario already exists. Neutrinos can travel through dense matter such as the Earth without interacting with a single atom, leaving no trace of their passage. To observe even just a few of the extremely rare interactions of neutrinos with matter, physicists build detectors with massive amounts of target material and operate them for many years. The detectors record the tracks of the particles that emerge from the rare collisions of neutrinos with atoms of the target material.

I remember hearing that if you wanted to block a neutrino, you would need a sheet of lead as thick as the distance the Earth is from the Sun.

Answer 11

For every action there is an equal and opposite reaction. If the supposed object is immovable, that tells us that the reaction is zero, and therefore we know that the action is zero; if there is no effect on one of the objects that are colliding, then there can be no effect on the other one either.