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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:

https://discover.hubpages.com/education/Earthquakes-richter-scale-and-what-that-means-in-relation-to-dynamite

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

Here's another back-of-the-envelope way of looking at the energy in one of those tungsten rods. The energy comes from converting chemical energy (from rocket fuel) into potential energy (by being at a certain height above the earth) and kinetic energy (going at a certain speed). A Saturn V rocket has about 1,000 tons of fuel. If 100% of that were converted into potential and kinetic energy, at 100% efficiency, of a single tungsten rod, then the rod could potentially have the energy of one kiloton of fuel, perhaps equivalent to one kiloton of TNT.

There are huge efficiency losses, however. First, using rocket fuel to create kinetic energy is not 100% efficient -- I'm guessing 50%. That takes us down to half a kiloton energy equivalent. Second, one rod weighs about 15,000 lbs or 7.5 tons. The cargo capacity of the Saturn V is about 130,000 lbs, or about seventeen tungsten rods. So each of the rods can have at most 1/17 of the energy from the rocket fuel. Now we're down to about 30 tons of rocket fuel equivalent energy per tungsten rod. Of course a huge amount of energy was wasted lifting the Saturn V and its fuel into space, though most of it not all the way into space. I'll take the energy total down to 10 tons of rocket fuel per tungsten rod, combining the kinetic energy of orbital speed and the potential energy of being 100 miles above the earth's surface. Neglecting energy losses from re-entry, this is the energy available to create chaos when the rod hits the earth.

This 10 tons of energy is focused onto a 1-foot diameter area. This is the equivalent cross-section of about 2,000 .22 caliber bullets. A .22 cartridge has about 1/10 ounce of powder in it. With our tungsten rod, we have the equivalent of 10 tons behind 2000 bullets or about 10 pounds of powder/rocket fuel behind each of the bullets, compared to the usual 1/10 ounce. That is, each bullet-equivalent area of the tungsten rod has about 1600 times the energy of an actual bullet, and the tungsten rod has the area of 2000 bullets, so the tungsten rod has the energy of about 3 million bullets coming at you.

Depending on the depth of the target, I imagine the tungsten rod has a pointed titanium tip so the energy is focused on a much smaller cross-section than 1 foot across.

That's going to hurt.