Okay, so...

I'm not going to divulge how, why or what's going to happen after, but I want to drop the moon onto Equestria. Earth-impact. I would like to keep this upcoming story scientifically accurate as much as possible, so I would like a little insight as to what would happen if this were to occur on Earth. If there are any budding astrophysicists out there, I could use your help.

-For reference, say for example the moon lands on North America. How would Afro-Eurasia fare on the opposite side of the planet?

Preliminary google searches have given me conflicting answers as to whether or not all life on the planet would be wiped out. Obviously, in order to have an existing story, complete extinction must not happen. If such is the case, would the outcome if the following are considered?

-Equestria's moon is controlled by magic, so it probably doesn't have the same kind of gravitational orbit around the planet - perhaps none at all. Given the magical nature of the celestial bodies, I plan to have the moon drop straight down like an apple onto a table.
-In a few scenes, I believe, the moon appears quite large, indicating that it's perhaps closer to Equestria's earth than our moon is to us.
-If the outcome is still everyone ded, could the impact be lessened if the moon were smaller than ours?

If anyone has any insight, please let me know. I greatly appreciate all the help I can get! Ciao!

Dunno if you saw this in your searches, but it might help.

5069356
It depends on a variety of factors, most importantly impact speed and size. Something the size of Earth's moon striking the Earth would effectively destroy the entire crust of the planet; it probably isn't survivable, at least on the surface, even on the far side of the planet, and there would be numerous secondary impacts. Even at slow speeds the gravitational attraction between the two bodies would cause too much damage; the Moon is just too large relative to the Earth to peacefully accrete on one side of the planet (which may have happened on the far side of the Moon with a smaller body).

A 60 mile wide body would probably be sufficient to destroy all surface life, assuming standard asteroid velocity (something along the lines of what took out the dinosaurs). The slower something travels, the larger it could be.

A slow enough impact might not cause what most people would think of as an impact crater, but would instead result in accretion, where the colliding body basically fell apart and spread out over a large portion of the side of the planet it "struck". This would be less bad, but would still be pretty devastating; if it fell into an ocean, for instance, you're talking about 100% displacement of several miles of seawater, times the area. Something with a 60 mile radius would have a volume of 15079 cubic miles of rock, which could displace that much seawater. Note that this is small relative to the Earth (332,500,000 cubic miles of stuff) but 15,079 cubic miles of rock would fill up a five mile deep ocean to a radius of 70 miles (slightly greater than the radius of the initial body). This... would be bad, as that water has to go somewhere. We're talking huge coastal tsunamis.

5069356
Um.
Well, to put things into perspective, if you dropped the Moon on the Earth at a 90 degree angle, and it hit the earth at the lowest possible speed of something free-falling into Earth (11km/s), no matter where you stood on the planet, you would officially be in the crater it caused.

I have absolutely no idea what the thermal effects would be, going that slow. Traditionally you wouldn't get a huge fireball from that. But then again, the impact against the atmosphere is 10^15 Megatons of energy. So... um... yeah. Let's assume that magically doesn't create a wall of fire to end all walls of fire.

Now, you ran to safety across an ocean. Say, in China, compared to the impact on the east coast of the US. If you stood 10000km away, you would be hit by magnitude 14.7 earthquakes in about half an hour. The worst recorded earthquake in the world was a 9.5. The area of this one is 160,000x larger. And 63 MILLION times more energetic. Every standing structure would be reduced to rubble. The rubble would bounce. Then likely sink. The earth would probably be a roiling liquid under your feet.

But, of course, that's really child's play. Cities are destroyed by an overpressure of 5 PSI, as a rule of thumb. Basically even heavy concrete structures are gone at 20PSI. You would be hit by a wave of pressure at 19500PSI a few hours after impact. And 21100mph winds.

Every living thing on the planet would be liquified.

But really, more than likely the whole planet would be a fireball anyway, and that might be for the best.

5069356

-For reference, say for example the moon lands on North America. How would Afro-Eurasia fare on the opposite side of the planet?

Well, you're missing a few things that we need to give you a rough estimate.
Here,
Mass of the moon, Velocity of the impact into the planet.

Those two are the most basic things I need to even begin doing any math.

5069381
I don't think it would matter much.

You pretty much need to get down to the 1/10th the size of our moon to get to the point where you MIGHT be able to survive in the world's most hardened and seismically isolated bunker deep underground on the opposite side of the planet if you were very lucky. Even with the moon going the slowest possible speed it could go. At that point you're only dealing with magnitude 12.4 quakes. The surface is still unsurvivable due to the blast wave.

5069383

Even with the moon going the slowest possible speed it could go.

Exceeeedingly slooooow death at the rate of 1 nanometer per second.

Less bad than

Very very quick death at normal gravitational speeds.

You pretty much need to get down to the 1/10th the size of our moon to get to the point

Hence why I asked for the mass of it, just in case according to headcanon or something, the moon is significantly smaller.

5069356 It would wipe out everything.

5069356 If you just gently set the moon on the Earth, I think you'd still destroy all life on the Earth. Dropping the moon onto the Earth would probably result in a new molten mass orbiting between Venus and Mars.

We've seen Equestria's moon rise behind hills so, if it isn't as big as Earth's moon, it's still mountain-sized or more. We don't know how big those hills are... Or how big the horizon is... Or the free-fall speed on Equestria...

5069380

I have absolutely no idea what the thermal effects would be, going that slow. Traditionally you wouldn't get a huge fireball from that. But then again, the impact against the atmosphere is 10^15 Megatons of energy. So... um... yeah. Let's assume that magically doesn't create a wall of fire to end all walls of fire.

I don't have time to try a more detailed calculation right now, but consider the fact that air molecules (or ions, rather. The energies involved are more than enough to break up any molecules and probably disassociate electrons) near the surface of the moon will need to be moving at speeds of at least 11km/s. This corresponds to a temperature of order 10^5 K -- far hotter than, say, the surface of the sun. If the moon's speed does not change significantly significantly as it passes through the atmosphere, that means that the entire volume of air in the path of the moon will be heated to approximately this temperature.

5069373
5069380

Thanks for the very in-depth answers!

Titanium, the moon would not be landing in an ocean.

Am I correct in understanding that Luminary's given speed of 11 km/s is not constant and would accelerate before it hit the Earth? If so, what if the speed were instead a constant 11 km/s? What if it were a slower constant speed, such as 5 km/s or even 1? (I know this might not be scientifically possible, but we're in a realm of magic remember)

Earthquakes are not a problem within the canon of my story, but could you explain the 'pressure' a little more? Is that similar to the shockwave from an explosion?

5069366 In this simulation, it shows a bunch of volcanoes erupting. Is this something that would happen?

5069825
The speed of an infalling object towards Earth would only be appreciably slowed by the atmosphere at fairly low altitudes. At 11 km/s, you're going through that portion of the atmosphere in a matter of seconds, so it wouldn't make an appreciable difference (and in any case, that kinetic energy is still going somewhere, namely into the atmosphere, so it isn't really much of an improvement regardless).

The lower an object is when released, the slower it will be travelling. An object that started off just at the edge of space (100 km) that fell straight down would be falling considerably slower than 11 km/s by the time it hit the ground. Even assuming the planet had no atmosphere, you'd be looking at...

EDIT: Fixed a typo in my initial calculation. I accidentally punched in 10,000 instead of 100,000 initially, which lead to incorrect results.

100000 = 9.8 * t^2

t = 101 seconds

v = a t = 9.8 * 101 = 989.8 m/s (or about 1 km/s).

It would actually be travelling somewhat slower than this because there is, in fact, an atmosphere, but as it turns out, it makes little difference on this scale.

Assuming a density of 2.65 g/cm^3, you'd be looking at something like:

2 * (2.65 g/cc * 4/3 pi r^3) * 9.8/ (p * pi * r^2 * 0.47) = 147.3 * r/p

Where r is the radius of the infalling object and p is the density of air at the altitude you're interested in.

Air density at sea level is 1.225 kg per cubic meter at 15C, with it falling off with height.

Most likely terminal velocity wouldn't be especially relevant, given the sheer scale of the object.

Note that the kinetic energy of this object would be:

Ek = 1/2 m v^2

Given that a 60 km radius meteor will destroy all life on the planet at 11 km/s, we can then calculate out out much energy that will take:

E = 1/2 (4/3 pi 6000000^3 * 2.65) 11000^2

E = 1.45 x 10^29

So we can assume anything which releases this much energy would kill all life on the planet.

You could make this into an equation, then:

1.45 x 10^29 = 1/2 * 4/3 * pi * r^3 * 2.65 * v^2

As you can see from this, velocity must decrease much faster than radius increases due to power scaling.

So our hypothetical 989.8 m/s object, to destroy all life, would need a radius of:

1.45 x 10^29 / (1/2 * 4/3 * pi * 2.65 * 989.8^2) = r^3

r = 29,876,028 cm = 298.8 km

Note that an object of this size at 100 km in altitude would pretty much blot out the sky, with an angular diameter of 143 degrees.

Their moon is obviously smaller than this, so if it is perched on the edge of space, there's a good chance that it freefalling would be survivable. If you could calculate out the angular diameter of their moon, you could figure out how big it was and how much damage it would do.

5069380

Now, you ran to safety across an ocean. Say, in China, compared to the impact on the east coast of the US. If you stood 10000km away, you would be hit by magnitude 14.7 earthquakes in about half an hour. The worst recorded earthquake in the world was a 9.5. The area of this one is 160,000x larger. And 63 MILLION times more energetic. Every standing structure would be reduced to rubble. The rubble would bounce. Then likely sink. The earth would probably be a roiling liquid under your feet.

Note that an earthquake caused by an impact like this wouldn't be exactly the same as an Earthquake caused by a fault. Most really large earthquakes end up with such large magnitudes not only because of their moment-to-moment strength but also because they last longer. An earthquake caused by an impact like this would not be quite the same as a hypothetical 14.7 earthquake caused by a fault moving. I would expect it would be much worse.

5069825

Earthquakes are not a problem within the canon of my story, but could you explain the 'pressure' a little more? Is that similar to the shockwave from an explosion?

That's all a shockwave from an explosion is - air pressure.

In this simulation, it shows a bunch of volcanoes erupting. Is this something that would happen?

I think those are secondary impacts from debris thrown up from the initial impact, rather than volcanoes. Something which hits with enough force will launch a lot of rock into unstable low-earth orbit, which will then fall back down and hit elsewhere. Rather like throwing some object into the sand, and seeing the sand puff up and then fall back down. Basically, planetary shrapnel.

But yes, it could trigger volcanic activity. In fact, there's a theory that it would set off a lot of volcanic activity on the exact opposite side of the planet from where it struck, because the shockwaves would all converge there, creating a secondary pressure center which would want to drive rock up and out. So being on the exact opposite side of the planet might be worse than being slightly displaced from that.

5069825
5069996
Comparing a picture of the real moon, which has an angular diameter of half a degree:

To a picture of Equestria's moon in similar conditions:

Suggests their moon has an angular diameter of about 2.5 degrees.

This would suggest a radius of only about 2.2 km if it was sitting on the edge of space. Note that this would be impossible if their moon was only held together by gravity, as it would be well within the Roche limit, but if it was a fairly solid object, like a big solid chunk of rock, it would be feasible.

At this diameter, the energy would be:

E = 1/2 * 4/3 * pi * 220000^3 * 2.65 * 989.8^2 = 5.8 x 10^22

This sounds like a lot of energy, but is, as you can see, about seven orders of magnitude less energy than what is necessary to extinguish all life on the planet.

The Chicxulub impactor released about 420 ZJ of energy, which is 4.2x10^23, so our impact about an order of magnitude less energy than that. The Mount Tambora eruption of 1815 released 1.4 x 10^20 joules of energy, so our impact would be about two orders of magnitude more than that.

So how bad would that be?

You'd probably be looking at a crater about 550 m deep, with a diameter of about 8 km. This would be a local catastrophe, and would probably start fires for many miles nearby, and destroy everything nearby. It would kick up a bunch of crap into the atmosphere and likely lower global temperatures for several years by a fair bit.

On Earth, we get an impact this bad roughly once every 200,000 years or so.

5069753

We've seen Equestria's moon rise behind hills so, if it isn't as big as Earth's moon, it's still mountain-sized or more.

There's no way in hell it's merely mountain sized. Not unless the moon is close enough to Equestria that it wouldn't take you 8 hours and 35 minutes of travelling at 36,373 mph to get there.

5069356
Pretty much all the data on Equestria's moon is up for grabs so just decide what you want the result of the impact to be for your story and then finagle its size and velocity to get it to come out that way. The thing might not even be a rocky astronomical body in the traditional sense—It could be a kind of magical hologram for all we know, or made of some heavenly material as light as aerogel.

5070118

There's no way in hell it's merely mountain sized. Not unless the moon is close enough to Equestria that it wouldn't take you 8 hours and 35 minutes of travelling at 36,373 mph to get there.

I agree completely -- mountain-sized is just the lower bound for its size. It could be a lot closer than Earth's moon, of course. We could take the speed of Nightmare Moon's trip to the moon when banished as indicating that it's just above the atmosphere... But that is probably reading a little too much into the animation. And then it'd appear a lot bigger or smaller as it or the viewer traveled around Equestria.

5070098 I am discounting gravity in most cases when it comes to the moon because it is completely controlled by magic. In essence, though the Equestrian moon could theoretically be in orbit perhaps, it is not. It is dragged across space by a mare's horn. While not confirmed in any circumstance, many stories and headcanons hyposit that Luna created the moon, so having it be a big hunk of solid rock is also feasible. A big magic hunk of solid rock.

Because of the moon's magical nature, I can also play with its mass in order to get the desired effects.

Question; If OUR moon was not there and then suddenly POOFED into existence out of nowhere (along its normal revolution) what would it do? Would it fall into an orbit, or crash into the Earth?

So how bad would that be? [...]

Looking at what you've described here, ironically, might not be enough for what I have planned. I refer back to my question about the impact site being in North America (say the middle, like South Dakota or something). What sort of effects happen on the complete opposite end of the Earth? What about distant but nearby South America? I assume this would still cause earthquakes and pressure, but not enough to kill everything.

2.2km for a radius is probably too small for what I'm aiming for. I like what you've described for me there as it sort of provides a lower bound. I want more devastation than that, but the upper bound is destroying everything which I also don't want.

5070961

Question; If OUR moon was not there and then suddenly POOFED into existence out of nowhere (along its normal revolution) what would it do? Would it fall into an orbit, or crash into the Earth?

Depends on its relative motion.

Orbits are, fundamentally, when an object is going so fast sideways that it falls around the object it is orbiting. This is why spacecraft have to go so fast to orbit the Earth - they have to be flying so fast sideways that they fall at the ground and miss.

A stable circular orbit is when the amount you fell downwards plus the amount you flew sideways meant that the total distance between you and the center of mass of the object you are orbiting didn't change.

So if it came into creation with zero motion? No momentum in any direction?

When I said normal revolution, I just meant that it would 'poof' into appearance somewhere along the invisible circle that the moon orbits around, but it wouldn't be traveling along this orbit.

5071016 if it had no motion relative to the planet and gravity was a thing, it would fall into the planet and kill everything. If gravity weren't a thing, it would sit stationary in the sky.

5071052 Cool, that's what I thought.

5071052

If gravity weren't a thing, it would sit stationary in the sky.

5071180
Actually, dependant on Equestria's internal planetary rotation, it'll still seem like it's moving across the sky for them, but it's actually the planet's own spinning.

5071285 I thought about specifying that, but figured it wasn't really relevant. It would be sitting still in the sky relative to the surrounding stars though.

5069996
That is some delicious calculation.

Though I would wager that 100km is rather low. Not just because it's sitting in the Roche limit, and would never have formed there (Can be hand waved by maaaaagic). But just because the thing would be experiencing atmospheric drag at all times. (Not much, less the 1/100000th normal. But that's not insubstantial when you're several KM across) If left alone for a while, it would fall into the atmosphere. I guess Celestia learned fast.

It wouldn't even have the gravity to be round, on its own. Its escape velocity is something you could probably achieve with a good sneeze.

If you move it out of the atmosphere, it gets larger. And accelerates further when falling. The whole thing turns into a disaster very quickly.

Then again, maybe someone neglecting it is what happens in 5069356 's story! Who knows!

5071014 Okay, I think the step that needs to be taken is to take our object with the angular diameter of 2.5 degrees and radius of 2.2km, and increase either/or the size, mass and distance from the Earth to make a bigger impact.

Maybe I'll give a little better of an idea of what I'm looking for. I want to pretty much completely devastate the impact site's continent (reference; North America). However, I would also like the incident to have a noticeable impact world wide (reference; as far as China). Through atmospheric problems, earthquakes, magma, what have you. Lots will perish, but still a good amount will manage to survive (unless they live on the impact continent)

I suppose the biggest obstacle is that moving it further from the Earth will increase its downward velocity, right?

5072717 It would be pretty simple to tune the size and distance of the moon to get whatever energy you want.

The energy of impact can by found using conservation of energy. The potential energy that the moon starts with at height h is -GmM/(R + h). When it reaches the surface, its potential energy is -GmM/R. The difference (and thus its kinetic energy when it strikes) is

T = GmM[1/R - 1/(R+h)]

where G is Newton's gravitational constant, m is the mass of the moon, M is the mass of the planet, and R is the radius of the planet. Assuming the moon is a sphere of constant density,

m = 4/3 Pi r^3 ρ

where r is the radius of the moon and ρ is its density.

If we require that the moon have an angular diameter of 2.5 degrees, we can use a small-angle approximation to get

r = h * 1.25 * Pi/180

Putting everything together, we have

T =4/3 G M ρ Pi (1.25 * Pi/180)^3 h^3 [1/R - 1/(R+h)]

Now substituting in some numbers:
G = 6.67408*10^-11 m^3 kg^-1 s^-2
M = 5.972*10^24 kg
ρ = 2650 kg/m^3
R = 6.371 * 10^6 m

We get (in SI units)
T =4.59412*10^13 h^3 [1/(6.371 * 10^6) - 1/(6.371 * 10^6+h)]

Now all that's left is to solve for h. The solution is really messy:
Wolfram Alpha

Just replace "T" in the formula in the link with your energy in Joules and Wolfram Alpha will give you the needed height in meters.

For example, to get an impact comparable to that which killed the dinosaurs (4.2*10^23 J), you'd need to drop a 2.5 degree angular diameter moon from a height of 804 km. This moon would have a radius of 17.5 km

It turns out that if h is much less than R, you can use the approximation

G M [1/R - 1/(R+h)] = gh

where g is the gravitational field strength at the surface of the planet, and get the much simpler formula

h = 0.9699 T^(1/4)

So, 5.8 x 10^22J is Titanium's 2.2km object, which causes a 'local catastrophe'. Destruction for miles around, but only atmospheric crap and temperature changes for the rest. How far would earthquakes and pressure from this go? (Again, assume we're striking the area of South Dakota for reference)

Conversely, 4.2*10^23 J killed the dinosaurs. Would someone be able to briefly describe what would happen if this event were repeated today? (same location and everything). I'm presuming this is another end all- or most life scenario, but let me know.

I'm positive that my catastrophic event lies in between those two values somewhere. I just need a little more understanding of the scientific effects of each rather than the mathematical ones.

Again, thank you guys for your continued help. I should have said this before, but you will all most certainly be credited for your help

5074320

What was an extinction-level event for the dinosaurs would not necessarily be one for ponies. Ultimately, it was the dust kicked up by the meteor that killed the dinosaurs by blocking out sunlight and preventing photosynthesis. However, we already know that ponies are capable of surviving in a world without sunlight (the Nightmare Moon alternate future from the season 5 finale).

Depending on how destructive you want the impact to be, 4.2*10^23 J might be just what you're looking for. According to Wikipedia, the impact would have triggered earthquakes, tsunamis, and volcanic eruptions globally. Furthermore, according to this paper linked in the wikipedia article, re-entering debris would have emitted infrared radiation intense enough to ignite global forest fires and kill unsheltered organisms. Essentially, everything within a 6000 km radius would die (for reference, a 6000 km circle around South Dakota would contain all of North America, the northern part of South America, and a small part of Siberia). Outside that radius, heavy cloud cover or shelter would offer a good degree of protection.

Under these conditions, I imagine that ponies in stone buildings, underground, or under cloud cover would have survived. Remember that ponies control the weather, though. Given enough time to prepare, I imagine many ponies could be saved from the heat pulse.

(Incidentally, by my calculations, Titanium's 2.2 km object dropped from 100 km would actually have an energy of 1.1*10^20 J, not 5.8*10^22J)

5074320

Conversely, 4.2*10^23 J killed the dinosaurs. Would someone be able to briefly describe what would happen if this event were repeated today? (same location and everything). I'm presuming this is another end all- or most life scenario, but let me know.

In the same place? For one thing, insurance rates on the Yucatan Peninsula would go way up.

Massive firestorms over much of the globe -- the ejecta from the impact comes down hot all over the place -- even the re-entry heat from all that material will fry a lot of exposed life. Dust and smoke in the atmosphere hurt plant growth everywhere and bring temperatures down. Starvation for a lot of humans and other animals -- photosynthetic organisms are the base of the food chain. That may not last for decades, though. When the clouds finally part, temperatures go up to higher than their pre-impact levels because now there's a lot more CO2 in the atmosphere.

The original impact killed about three quarters of the plant and animal species on Earth at the time. I wouldn't expect extinction of Homo sapiens sapiens. We're probably better at adapting than dinosaurs -- I don't think any species covers as much of the globe, and as many different climates, as we do. And it's possible this isn't the first time we've gone through a global-scale disaster. But I think there'd be a lot fewer humans and other animals afterward.

Man's civilization would be cast in ruins. But one man would burst his bonds and fight for justice. With his companions, Ookla the Mok and Princess Ariel, he pits his strength, his courage, and his fabulous sun sword against the forces of evil.

The last part is especially speculative, though.

5074432
5075415

I understand what you two are saying, and you're right. This does seem like it's just what I'm looking for. Global earthquakes and eruptions was one of the key things I was looking for, along with atmospheric disturbance. The heat wave might be a little powerful, but I think I can make it work.

There's one thing that's not discussed, though. I imagine the Chicxulub impactor was traveling at a significant velocity through space when it impacted the Earth, but (at least in the initial planning for this story) the moon starts above the planet with an acceleration of zero and then falls to the planet.

A friend of mine came up with a possible workaround to this, but I'd like to hear your thoughts first.

Thanks!

5076901

A lower velocity would mean a less violent impact, of course. And you can play with size a lot here. According to Wikipedia, the Chicxulub impactor is thought to have been about 10km in diameter. Equestria's moon might be that size. Our moon is 3,474km in diameter. My feeling is that Equestria's moon is smaller and closer than ours, but it's not based on anything rock-solid. You can pick any of a wide range of sizes as you need for story purposes, though.

You can also play with composition. Rock is pretty dense, and something lighter would have less mass for the same size, so less kinetic energy. Given the setting, it could be magic moon rocks. Or green cheese. It could be full of Luna's socks. It could be denser than ordinary rock if you needed it to be too -- except for the socks one.

There's a good book about this if you have time to read it:

http://www.goodreads.com/book/show/337134.Moonfall

And if you liked that review, this is the link to where you can buy the paperback for 0.01 cents plus shipping, or the kindle version for $7.
http://www.amazon.com/Moonfall-Jack-McDevitt/dp/0061051128

Hope this helps

5069356
I did a simulation and took some screenshots. Impact speed was 11 km/s.
Right Before Impact:

An Few Minutes After Impact:

An Hour After Impact:

Hours Later, Far Side from Impact:

Average Surface Temperature by that point: 4468 Kelvin

I think it's safe to say that sh*t's f*cked no matter where you were. I can do other simulations if you would like.