Hypervelocity Drag

Drag on an accelerating launch loop vehicle with a hemispherical nose

I am not an aeronautical engineer and probably misunderstand the sources. In any case, the numbers are approximate, and should be treated skeptically.

The drag and heating is acceptable at all altitudes. However, incoming debris impactors in decaying orbits will be more unpredictable when the drag is higher, and more difficult to shield or dodge. Perhaps most of them may be intercepted a few orbit earlier, reducing flux at launch loop track level.

Based on Trajectory Optimization for an Apollo-type Vehicle under Entry Conditions Encountered During Lunar Returm by John W. Young (famous astronaut) and Robert E. Smith Jr., May 1967, NASA TR-R-258, Langley Research Center.

Equations on Page 5 in Foot-second-slug-BTU :

• 1a convective power: ~ ~ \dot Q_c = 20 \rho^{1/2} \left( V \over 1000 \right)^3 Btu/ft²-s

• 1b radiative power: ~ ~ \dot Q_r = 6.1 \rho^{3/2} \left( V \over { 10 000 } \right)^{20} Btu/ft²-s

• Equations assume an effective nose radius of 1 foot

• Equations from Shock Layer Radiation During Hypervelocity Re-Entry by Robert M. Nerem and George H. Stickford, AIAA Entry Technology Conference, CP-9, American Institute of Aeronautics and Astronautics, Oct. 1964, pp 158-169. (not downloaded or read yet)

Density in slugs/ft³: multiply kg/m³ by 1.9403203e-3

Power in Btu/ft²-s: multiply by 11350.54 to get W/m²

Velocity in ft/s: divide m/s by 0.3048

Metric equations:

• Metric 1a convective power: ~ ~ \dot Q_c = 3.53e-4 \rho^{1/2} ~ V^3 Watts

• Metric 1b radiative power: ~ ~ \dot Q_r = 1.24e-69 \rho^{3/2} ~ V^{20} Watts

These are for a 1 foot diameter nose, and scale by {r_n}^{-1/2} according to equation 4B-4 on page 520 of Part 4B (Entry Heat Transfer) of the SAE Aerospace Applied Thermodynamics Manual. That sites reference 1, A study of the motion and aerodynamic heating of missiles entering the earth's atmosphere at high supersonic speeds, H. Julian Allen and A. J. Eggers, Jr, NACA TN 4047, 1957. If r_n is in meters, scale by 0.552 {r_n}^{-1/2} .

If we scale these for a half-spherical nose, area \pi {r_n}^2 , we get:

• Total nose convective power: ~ ~ \dot Q_c = 6.1e-4 ( {r_n}^3 ~ \rho )^{1/2} ~ V^3 Watts

• Total nose radiative power: ~ ~ \dot Q_r = 2.2e-69 ( {r_n} ~ \rho )^{3/2} ~ V^{20} Watts

Effective time:

Assume constant acceleration for the vehicle, v = a t , to a maximum velocity V = a T .

define t_{eff} = {\Large { T \over { n+1 } } } = { \Large { V \over { a ( n+1) } } }

If the drag power \dot Q = k v^n = k a^n t^n , then the time integrated power:

Q=k a^n{\Large {T^{n+1}\over {n+1}}}=k a^n T^n{\Large {T\over{n+1}}} = k V^n t_{eff} = \dot Q_{max} t_{eff}

There will also be additional exit or climb-out time for the launch loop added to t_{eff} , TBD. This additional time will be proportionally larger for the radiation fraction, but that will remain small, especially in thinner, higher altitude atmosphere.

The drag losses are much higher; most of the lost energy ends up heating the upper atmosphere (where it radiates efficiently into space, not to the ground). The drag power is P = C_D \rho Area V^3 and the drag loss is P = C_D \rho Area V^3 T/4

Examples

For a 1 meter diameter nose, V=11 km/s, a=3*9.8m/s, T=374 s, C_D = 2.0 and density at 80, 100, and 120 km:

Add spreadsheet here.

Validity?

The Young and Smith paper was for Apollo lunar reentry, and the equations may not generalize to lower drag regions. Exit trajectory must still be computed, keeping in mind that Earth's rotation velocity (470 m/s) should be added to the outgoing orbit vector.