Reply To: General thought on old coal mines.

[vansig]

Quantitatively, what is the expected drag on a craft as it reaches orbital speed?

A 100 tonne craft, (roughly 3 m² area, and 16 m long), flying at ~36 km altitude (pressure=.01 bar) and mach-24 (8 km/s), needs 980 kN lift to remain climbing, but most of the lift is provided by centripetal motion, since 8 km/s is sufficient for orbital velocity.
( v²/R = 8² /(6371 + 36) = 0.0999 km/s² = 9.99 m/s² )

So only forward drag really matters, here. From the formulae at
http://www.grc.nasa.gov/WWW/K-12/airplane/drageq.html
http://www.grc.nasa.gov/WWW/K-12/airplane/dragco.html and
http://www.grc.nasa.gov/WWW/K-12/airplane/eqstat.html
as well as
http://www.lpi.usra.edu/meetings/lpsc2009/pdf/2059.pdf

We get that form drag dominates at high mach numbers. Assuming drag coefficient, Cd ~ 0.92; cross-sectional area, A ~ 3m², and density of air (at p=.01 bar and 244 K), rho = 0.0143 kg/m³,

then drag = Cd x Area x rho x v²/2
= .92 x 3 m² x 0.0143 kg/m³ x (8000 m/s)² /2 = 1263 kN

practical flight will have to exceed this by a good margin, for at least several minutes as these speeds and heights are obtained.
compare this needed performance to a gang of 3x GEM-40 boosters, (at 1000mm diameter and 492.9 kN each, 64 seconds burn time, only).