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AstralFire
2013-08-10, 04:29 PM
A primarily concrete mass of 21 gigagrams spread out over a volume of 91,000 cubic meters impacts dense soil. As a rough estimate, accounting for terminal velocity and air resistance (assume the shape is a cube and the density is uniform), what would the damage be when the mass is dropped: 25 meters? 75 meters? 225 meters? 675 meters?

By damage, I'd like to know the rough size of the impact crater, dirt spewed into the air, how far the kinetic shockwave could be felt through the ground to the average person.

If I need to specify more information, let me know. If this is far too much work for you to do and would rather give me a more ballpark estimate, go for it. About all I remember from mechanical physics is acceleration due to gravity. If you're curious, this is some wild ballparking to begin with on the size of a fictional building I am writing, and since I've already taken some pains to make the astronomy not make science cry (maybe sniffle), I might as well make my physics teacher not think I was a complete failure.

rs2excelsior
2013-08-10, 06:15 PM
Some back-of-the-envelope calculations:

Volume-> 91000m^3
Length of one side -> 45m
Surface area of one side -> 2023m^2

According to Wikipedia (http://en.wikipedia.org/wiki/Terminal_velocity), the formula for terminal velocity is v=sqrt(2*m*g/d*A*C) where
v=terminal velocity
m=mass
g=acceleration due to gravity
d=density of medium (which, again according to Wikipedia (http://en.wikipedia.org/wiki/Density_of_air), is about 1.225kg/m^3 at sea level)
A=area of bottom side
C=drag coefficient (For a cube, once more thanks to Wikipedia (http://en.wikipedia.org/wiki/Drag_coefficient), is 1.05)

Using 21*10^6 kg for mass instead of 21*10^9 g (same thing, just easier to work with in SI), that gives a terminal velocity of 398m/s (which honestly seems a little high to me).

I don't know how to work out the size or depth of an impact crater. I seem to remember a crater's depth is about 1/4 its diameter, but take that with a whole pile of salt. I can, however, figure the energy it has.

The kinetic energy formula is KE=0.5*m*v^2 where
KE=kinetic energy
m=mass
v=terminal velocity (in this case)
Which, by my calculations, results in 1.663*10^12 Joules of energy.

How much is this, you ask? Well, I referred to the ever-useful Boom Table (http://www.projectrho.com/public_html/rocket/spacegunconvent.php) (scroll down a bit). According to that, a Tomahawk cruise missile yields an energy of 1.9*10^9 J, and the Hiroshima nuclear bomb 6.3*10^13 J. So, this mass falling is about 1000 cruise missiles, or 1/60th of a (small) nuclear bomb. And this much kinetic energy probably acts enough like an explosion that the difference isn't really noticeable.

DISCLAIMER: I did this on a windows calculator and a sheet of paper. I very well could have made a mistake in the process. At least double-check my math (that's why I listed the equations). Still, this should give a rough ballpark estimate.

Hope this helps!

AstralFire
2013-08-10, 06:17 PM
Ooh. Big help. Thank you. I just want to check, what's the ballpark around when it hits terminal velocity? How high does it have to fall from?

rs2excelsior
2013-08-10, 06:24 PM
v=a*t

x=0.5*a*t^2

v=velocity
a=acceleration (in this case due to gravity, 9.81m/s^2)
t=time
x=distance traveled

So it would take 40.6 seconds to reach terminal velocity, and would fall a distance of 8074m. So depending on the context, it probably won't be at terminal velocity.

Now, these equations don't take air resistance into account, so it'd actually take longer. It'd probably fall about the same distance. Effectively, acceleration would be a bit reduced.

Eldan
2013-08-10, 06:25 PM
Hm. Accelerating to freefall takes about 12 seconds, I seem to remember. With g=9.81 m/s^s...

So d=1/2 a*t^2, it needs to fall about 700 meters.

rs2excelsior
2013-08-10, 06:28 PM
Technically it's in freefall as soon as it begins to fall. And the time depends on the specifications of the object falling. This has a higher terminal velocity, which will take longer to reach.

AstralFire
2013-08-10, 06:46 PM
Alright, I got my answers from the provided equations and stuff. Thanks very much! My rough suspicions were correct, but it's good to know that there's some basic physics behind it.