Damage of a high mass object falling at terminal velocity

[Anthony]

Originally Posted by Agemegos

Check your impactor density. It's off three orders of magnitude.

Haha. Oops, that does make a difference.

So, apparently that calculator doesn't understand seriously degenerate cases. I decided to simulate a 1 meter chunk of white dwarf matter (density 100 tons/cc = 1e+11 kg/m^3), and it failed to pass through the crust and into the mantle...

[Agemegos]

Yeah. I got results that were obviously wrong when I asked it to calculate a low-speed impact with an object the size of the Moon. e.g. a permanent crater wider than the Earth and 1,600 km deep, but no ejecta landing at the antipode.

[Agemegos]

Let's look at the back of our envelopes.

"640 tons" probably means 640 short tons, because no-one uses Imperial tons any more. That's 1 280 000 pounds, which is 581 000 kg to three significant figures. "Metal" could be anything from potassium (862 kg/m³) to osmium (22 590 kg/m³), but let's take it easy and assume iron (7 874 kg/m³). Calculator: 73.7 m³.

Let's model a "rod" as a cylinder with length eight times its diameter, to make the maths easy.

V = πr² L
= πD²/4 × 8D
= 2πD³

D = ∛(V/2π) = 2.27 m.

The rod is going to have the cross-sectional area of, confront the air like, and receive the ram pressure of an object with a diameter of 2.27 m.

But the rod will have 12 times the volume of a spherical object that diameter, which is to say 12 times the mass at given density. We can represent that by giving it 12 times the density. So we model our 640-ton metal rod as an object with a diameter of 2.27 metres and a density of 94 488 kg/m³.

"Dropped" from orbit it will have low-Earth circular orbit velocity of about 7.9 km/s. Supposing that it is de-orbited using nothing more than a railgun with a muzzle velocity of 3 km/s, it will impact the atmosphere at an angle of arctan (3/7.9) = 21° or less, or with diminished velocity.

Impacting at 5°: the rod lands intact at 4.16 km/s and makes a crater 94 m wide and 20m deep in hard rock, having shattered the bedrock for another 9 m below that. Buildings will be knocked down by the air blast at 400m (quarter of a mile), but there will be no fireball nor significant seismic effects.

Impacting at 20°: the rod lands intact at 6.71 km/s and makes a crater 183 m wide and 39m deep in hard rock, having shattered the bedrock for another 18 m below that. Buildings will be knocked down by the air blast at 550m, but there will be no fireball nor significant seismic effects.

[Anthony]

I would note that dropping things from orbit is one of those things that sounds cooler than it actually is; the destructive power is much lower than a comparably sized nuclear warhead, and because the impactor is mostly blind and deaf until it drops below around 3 km/s due to forming a plasma sheath, accuracy is terrible.

[Agemegos]

Originally Posted by Anthony

I would note that dropping things from orbit is one of those things that sounds cooler than it actually is; the destructive power is much lower than a comparably sized nuclear warhead, and because the impactor is mostly blind and deaf until it drops below around 3 km/s due to forming a plasma sheath, accuracy is terrible.

Also, if you want to hit a target in less than about 45 minutes or that is significantly closer that "the opposite side of the Earth", "drop" means "throw, very hard".

[Agemegos]

Originally Posted by Agemegos

Let's look at the back of our envelopes.

"640 tons" probably means 640 short tons, because no-one uses Imperial tons any more. That's 1 280 000 pounds, which is 581 000 kg to three significant figures. "Metal" could be anything from potassium (862 kg/m³) to osmium (22 590 kg/m³), but let's take it easy and assume iron (7 874 kg/m³). Calculator: 73.7 m³.

Let's model a "rod" as a cylinder with length eight times its diameter, to make the maths easy.

V = πr² L
= πD²/4 × 8D
= 2πD³

D = ∛(V/2π) = 2.27 m.

The coefficient of drag for a long cylinder is C = 0.82, this cylinder has a frontal area of A = 4.05 m² and a mass of m = 581 000 kg. The density of air at surface temperature and pressure is about ρ = 1.2 kg/m³, and the acceleration due to gravity near Earth's surface is g = 9.81 m/s².

v(terminal) = √ {(2 m g)/(ρ A C)}
= 1 691 m/s

That is, the terminal velocity of such a rod near Earth's surface is about 1.69 km/s. But the velocity at which it hits the surface when "dropped" from orbit is about 4.16 – 6.71 km/sec. So when the rod is dropped from orbit hits the surface before it can slow to terminal velocity.

[Kfireblade]

Originally Posted by Anthony

I would note that dropping things from orbit is one of those things that sounds cooler than it actually is; the destructive power is much lower than a comparably sized nuclear warhead, and because the impactor is mostly blind and deaf until it drops below around 3 km/s due to forming a plasma sheath, accuracy is terrible.

Well, this is kinda more or less a what if thing. I have a mage with a asinine amount of energy to work with who could get to and from orbit and I noticed just how much material I can create, shape, and turn to say iron with certain earth spells. Now of of course actually dropping said rod with even the slightest degree of accuracy would be a another story.

[Agemegos]

Originally Posted by Kfireblade

Well, this is kinda more or less a what if thing. I have a mage with a asinine amount of energy to work with who could get to and from orbit and I noticed just how much material I can create, shape, and turn to say iron with certain earth spells. Now of of course actually dropping said rod with even the slightest degree of accuracy would be a another story.

Do you have a way to speed that material up to orbital velocity? Because if not you're dropping it from space, but not from orbit. That's a very different kettle of fish.

[Kfireblade]

Originally Posted by Agemegos

Do you have a way to speed that material up to orbital velocity? Because if not you're dropping it from space, but not from orbit. That's a very different kettle of fish.

Not without doing a lot of other work. I straight up admit I don't know much about the physics involved, if you let an object fall from high orbit gravity would in fact pull it down and it would accelerate to terminal velocity right? I mean, things in orbit have to keep a certain speed to stay in orbit right? so if something was just created and had no velocity it would get pulled in by the planets gravity yes?

[Anthony]

Originally Posted by Kfireblade

Not without doing a lot of other work. I straight up admit I don't know much about the physics involved, if you let an object fall from high orbit gravity would in fact pull it down and it would accelerate to terminal velocity right?

'Orbit' is not an altitude. If you're actually in orbit, and you 'drop' something, it will just drift slowly away from you, and never hit the ground, because if you can fall and hit the ground, you aren't in orbit (by definition).

However, dropping something from the edge of space is possible; the object you describe will reach terminal velocity if dropped from more than about 80 miles up.