The thought popped into my head a while about about what would happen if you dropped Spire from space. Imagine the carnage! Imagine the fire! Without further ado:
The volume of a cone is one-third of the area of the base multiplied by the height:
0.3 * π * r^2 * h
The Spire is 121.2 metres in height and 3 metres in diametre at the base:
0.3 * 3.141 * 1.5 * 1.5 * 121.2 ≈ 257 m^3
The spire is mostly hollow inside:
It is constructed from eight hollow tubes of stainless steel and features a tuned mass damper[…]
The Spire’s declared mass is 126 metric tonnes. The kinetic energy delivered by an impact is:
0.5 * m * v^2
Energy is measured in Joules released.
That’s simple. The more mass or velocity, the more kinetic energy is released on impact. There’s more of increase in kinetic energy from an increase in velocity than an increase in mass. I said “dropped from space”, but dropped from space can mean a great many things.
I have read the fantastic Space Weapons Earth Wars by the RAND Corporation more than once, and I am aware of considerations that will affect the impact:
- The Earth’s gravity.
- The Spire’s initial orbit inclination, altitude, and velocity.
- Ablation as the Spire travels through the atmosphere.
For sake of this post ending before my beer does, the Spire is dropped perfectly straight through a vacuum-that is to say it travels directly perpendicular to the ground through a perfect vacuum.
The International Space Station has an average orbital velocity of 7.66 km/s.
0.5 * 126 * 7660 * 7660 ≈ 3696562800 J
Round it to
3.696 x 10^10 Joules. A quick look at this chart finds that it is a little under four-fifths the energy of America’s air-dropped GBU-43/B Massive Ordnance Air Blast bomb. Here is an explosion for reference.
If the Spire impacted at
8910 m/s then it would release an amount comparable to a MOAB blast.
0.5 * 126 * 8910 * 8910 ≈ 5001450300 J
I do not have a single source to site, but several websites anecdotally relate that the blast radius of the MOAB is ~150 metres. Secondary effects like broken windows and launched shrapnel would project far beyond this. To reiterate: Everything within 150 metres of the bomb would be obliterated. There would still be lethal effects out to as far as 1.5 kilometres.
I measured distances of 150 m and 1500 m in Google Maps, and filled in the circles in the Gimp. Everyone in the red circle would die, everyone in the yellow circle would be decreasingly affected by a MOAB explosion as you moved out from ground zero. Most of the energy of a kinetic impactor is directed straight into the ground (see pages 139-141 of Space Weapons Earth Wars).
Hilariously-speaking, everyone in red circle would die horribly.
Realistically-speaking, most of the energy goes into the ground, and only a fraction of the painted radius is actually affected. Consider the red circle the limits of effect. I’m out of beer, however, so the math ends here.
Either way, the damage proceeds downward and outward from the point of impact. In contrast, the damage for an ordinary explosive detonated at the same point would proceed outward in all directions.
If the projectile is long and rodlike, on the other hand, the steady-state phase dominates the effects. The crater is more cylindrical and its depth is proportional to the square root of the ratio of projectile density over target density. If the kinetic-energy weapon must penetrate shielding, e.g., a ship’s hull or a bunker, the depth to be penetrated determines the minimum projectile length, depending on the density of the shielding material. For example, a 1-m-long tungsten hypervelocity penetrator should be able to penetrate about 1.5 m of steel, almost 3 m of clay or stone, and 1 m of uranium. What penetrates through that depth (or less) of target will be a very hot mixture of target and penetrator material and any remaining penetrator length. The damage is done almost entirely in the direction of the impact, as with a shaped charge explosive, except for damage caused by secondary fires or explosions ignited by the impact.
Bob Preston et al., Space Weapons Earth Wars (RAND Corporation, 2003), 141.
The direction in which damage is done after a kinetic impact depends on:
- The shape of the impactor.
- The velocity of the impactor.
- The hardness of the impacted surface.
All of the energy from the Spire would be directed into the ground. The big difference in impact comes from whether the Spire is travelling faster than the speed of sound in concrete. It is-a velocity of 8910 m/s versus a speed of sound of ~3600 m/s. In the comments below Paul linked NUKEMAP, an interactive tool for the armchair nuclear strategist. The problem is that unless you use a nuclear shaped charge, the energy in a nuclear explosion is released omnidirectionally. That is to say it radiates equally in all directions.
In a kinetic impactor, all of the energy is initially delivered to a point on the ground equal in area to the surface area of the weapon.
3.141 * 1.5 * 1.5 = 7.06725 m^2
5.014 * 10^10 J of energy is delivered to an area of ground about seven metres square. There would be hypersonic shock wave through the ground-expect a lot of broken legs, angles, and spines as that energy transfers itself from the ground into bystanders. There would be a supersonic aerial shockwave, albeit at a fraction of the energy of the ground impact. Windows turned to deadly aerosol, cars knocked over, and people killed by debris nearby.
You would hear the impact. You would feel it in the ground and the air. You would be knocked off your feet a few blocks away. Windows would be blown out for up to a kilometre around. That’s in a perpendicular impact. If the Spire impacted at an angle, then the energy would be transferred at a tangent to the ground, and destruction in the vicinity would be correspondingly greater.
Worst part? It’s hypersonic. The sound of the impactor’s passage is several seconds behind the impactor. If it came in at an angle in a built-up urban area, you wouldn’t see or hear it until it already hit.
Watch the skies!
EDIT 2: How many would be hurt?
I cannot find good numbers for the density of people in Dublin city centre. The population there is transient, in that while relatively few people live within, say, a kilometre of O’Connell Street, many thousands of people will be on the street on a given day. To extract from the 2011 census data:
|Area (Sq km)||318|
|Population (including Suburbs or Environs)(Number)||1,110,627|
|Population Density (persons per sq km) (Number)||3,498.2|
|Population within Legally Defined Boundary (Number)||527,612|
|Population of Suburbs or Environs (Number)||583,015|
|Population Change Since Previous Census (%)||6.2|
I’ll arbitrarily triple the density by assuming Dublin city centre on a Saturday afternoon. There would be upwards of almost eleven thousand people in the area affected by the impact.