I thought I'd take a stab at working out some numbers for the original scenario which is a 20 kg iron mass hitting atmosphere at 0.013 c or about 3.9 million m/s. There's a lot of variables to deal with such as increasing atmospheric pressure, decreasing mass of the projectile as it's outer layers boil off, etc, but I will be making a number of simplifications to bring a measure of sanity to the equations.
For the atmosphere I will be assuming constant pressure out to 100 km. I'll be using a simple average of the pressure giving a value of 1/2 atmospheres (about 50 kPa). We will be ignoring the effect of gravity on the shell and further make the assumption that the shell's trajectory is constant and is perpendicular to the planet's surface.
From a
naval site I got the following formula for drag on a shell: 0.5*r*V
2 * S * Cd
where r is density of air, V is velocity, S is surface area of the shell and Cd is the drag coefficient.
Cd is going to be tricky to deal with as it is either,
1. determined experimentally,
2. calculated using fluid flow equations, or
3. determined numerically using supercompters
I'm going to go with the ever popular option 4 and pull a number out of my ass, say 0.3. (which is a reasonable value for a spherical projectile)
Given that the density of pure iron is 7870 kg/m³, 20 kgs gives us 0.00254 m³ which makes a nice sphere of 0.0847 m radius (~8.5 cm). This gives us a cross sectional area of 0.0225 m
2.
That just leaves the density of air, which at 1/2 atmosphere is 0.602 kg/m
2.
So the drag force = 0.5*(0.602 kg/m
2)*(3.9 million m/s)
2*(0.0225 m
2)*0.3 = 30.1 GN (giga Newtons).
Since energy is Force * distance, over the 100 km, the atmosphere will do negative work on the shell, slowing it down (we are ignoring this slowdown as it greatly complicates the drag equation), and the difference is energy is converted into heat. This energy works out to be (30.1 GN)*(100 km) = 3.01 e-15 Joules. This value is very large and exceeds the kinetic energy of the shell (1.52 e-14 Joules) which can't happen so my assumption of constant velocity is bogus. But we'll say the shell's entire KE gets converted into heat and made a further assumption that all the heat goes into heating and vaporising the iron.
The specific heat of iron is 0.46 (kJ/kg K), the boiling point is 2750.0 °C (3023.15 K) and the heat of vaporization is 349.60 kJ /mol. We'll say the iron's initial temperature is 4 K (background temperature of space, perhaps the shell has been traveling a while) so the energy to raise the 20kg of iron to 3023 K is about 28 MJoules and the energy to vaporize the iron is 125 MJoules, giving us a grand total of 153 Mjoules (1.53 e-8 Joules). This is such a small fraction of the heat going into the iron, even if a large amount of heat is carried away by the atmosphere, there is plenty left over to vaporize the iron. That was a lot of work to come up the answer that everyone else has been giving in the thread so far, but it was a fun exercise and it was interesting to see the orders of magnitudes involved.
One final note, heat transfer and vaporization are not instantaneous, they do take a finite amount of time. I would conjecture that even a small projectile moving fast enough may have enough time to hit the planet's surface before losing the majority of it's mass. Working out what that speed would be is beyond my capabilities.