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.

Launch loop capsule drag and heating is acceptable above 80 km altituded. 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 Return 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 :

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:

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 book cites 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:

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 energy is E = 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, CD = 2.0 and density at 80, 100, and 120 km:

altitude km

80

100

120

density kg/m3

1.85e-5

5.60e-7

2.22e-8

\dot Q_c ~~ kW

3500

610

120

\dot Q_r ~~ kW

120

0.62

0.005

exponent

t_{eff}

Q_c ~~ MJ

330

57

11

3

94

Q_r ~~ KJ

2100

110

0.09

20

18

Q_{total}~~ MJ

330

57

11

heat fraction

1.1e-3

1.9e-4

3.7e-5

drag loss MJ

14000

440

17

drag fraction

4.8e-2

1.4e-3

5.7e-5

heat/drag

0.023

0.13

0.65

HeatDrag.ods

Vehicle

Mass kg

Diameter

Length

Loop

5000

2 m

tbd

Apollo CM

5900

3.9 m

3.2 m

Shuttle

68600

300 m² surface est.

Dragon

4200+3310

3.7 m

6.1 m

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 velocity. Apollo ballistic parameter 322 kg/m² ± 40%, launch loop 1590 kg/m². A "pill shaped" launch loop capsule may not reenter safely; wings or cone-shaped capsule is probably required for manned entry.