Certainly. I think it's striking.
https://earthquake.usgs.gov/earthquakes/map/?extent=30.37288,...
The link goes to usgs.gov, which catalogs all seismic activity. You can search and filter and really have an interesting field day there. I pre-selected an area in China that has had earthquakes from 9.5 to 10.5 km. Notice they are ALL exactly 10.0. This is true world-wide. For grins, you can try widening your search depth - you may see that most "quakes" center around 10.0. REAL quakes will be random numbers. "man-made" are 10.0.
If you remove the depth filter, and view all quakes, you'll see a strange eerie preponderance of 10.0 km depth "quakes".
One question, not to answer... but to start to ponder: If there is a weaponized method to send a telephone-pole like piece of tungsten into the earth from space, how would one know to what depth it could penetrate? Who would know? If one doesn't know... how could one make a guess as to its depth?
F=MA
A-acceleration is 32 ft/s^2 or 9.81 m/S^s
M-estimated mass of a tungsten bar
Then calculate V^2=2AD
Where D is the distance from the satellite to the ground and A is again 32 ft/S^2 or 9.81 m/S^2
Then take the force you get and determine the average density of the earth in the area and you’ll be able to determine the deceleration as a matter of friction..apply that to the formula of D=V^2/2A
Where this time V is the terminal velocity you calculated before at impact and A is the deceleration or friction of god earth in the are (rock will have a higher factor then sod, etc)
Assuming:
1) geosynchronous orbit of a satellite is approximately 22,236 miles for a circular orbit or 39,135,360 ft or 35,785,373 meters
2) a 20 ft long, 1 foot diameter rod of tungsten with a mass of 183.4 g/mol or 19.3g/cm^3 which means 15.708ft^3=444800cm^3 which times 19.3g/cm^3=8585640g=8584.64kg
so...
V^2=2AD=702109018 m^2/s^2 or 26,497 m/s (less as there is a terminal velocity due to air friction so I’ll use 25,000 m/s for ease of calculation and a fudge factor.
Now if we assume that 10 miles is correct then that is 16093 m.
So A (our deceleration) would be = V^2/2D=(-25,000 m/s)^2/(2*16093m)=-19,418 m/s^2.
Force at impact = F=MA = 8584.64kg*19418 m/s^2=166,695,763 kg*m/s^2 or 1634 MegaNewton-m or 367,501,250 lbf. which would be equal to .03984tons of TNT....
by comparison a 7.0 earthquake is equal to 32 million tons of TNT per this website:
What this means is that with a 10 mile deceleration you would not get the same level of force as with a shallower and faster deceleration.
For the force to be what we expect to get out of the impact I’m guessing that deceleration occurs at closer to 1-2 miles which would put us closer to a 1/2 ton of TNT which is more like a 5.5-5.7 Richter earthquake (remember that the Richter scale is logarithmic).
Now this is all really fast, back of the napkin - quick conversion stuff - but you get the idea. feel free to check my math and conversions as I did this quick. This is also another good site for calculations (especially deceleration). remember acceleration isn’t impacted by mass....cannonballs and apples fall at same pace only their surface area increase drag on the larger object....
http://www.hazardcontrol.com/factsheets/pdfs/falling-objects-calculations.pdf