U.S. Conventional Prompt Global Strike: Issues for 2008 and Beyond (2008)

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G The Why and How of Boost-Glide Systems Given the prominence of the boost-glide technology in some of the options under consideration in this report, it is useful to include an appendix explaining semiquantitatively what the technology can and cannot accomplish, its relation to the fractional orbit bombardment systems (FOBSs) technology discussed during the 1960s and 1970s, and some of the technical challenges involved. Another issue is the extent to which such vehicles can be expected to defeat “garden-variety” and advanced air defenses. A boost-glide vehicle (BGV), or “lifting body” without propulsion, can be used to extend the range of a ballistic-missile payload beyond the purely ballistic range. It can also be used for out-of-plane or “dogleg” maneuvers to avoid over- flight of certain areas or to allow the dropping of initial rocket stages into the sea or into another body of water not under the ballistic path. The space shuttle on reentry is an example of a hypersonic lifting body. First consider the BGV for relatively short-range systems—up to a few thousand kilometers in range—in the approximation of a flat Earth. An important simplification arises from the fact that the atmosphere is shallow; the air density falls by a factor e = 2.72 for each 8 km increase in altitude. As is the case with a normal glider, the aerodynamic support of the vehicle against gravity (the “lift” L) is accompanied by “drag due to lift,” as characterized by the lift-to-drag ratio (L/D); for a clean subsonic glider aircraft this may be as much as 40, but for a hypersonic lifting body an L/D = 2.2 is an achievement. In the numerical examples, it is assumed that L/D = 2.2. With the glider aircraft or for a powered vehicle that has run out of fuel and that is gliding for as long a distance as possible, the drag, D, extracts energy from the vehicle—from its store of kinetic energy MV2/2 and potential energy MgH; 206 

APPENDIX G 207 here M, V, g, and H are the vehicle mass in kilograms, the velocity in meters per second, the gravitational acceleration 0.0098 kilometers per second, and the flight altitude in kilometers. Every pilot has first in mind the glide range from altitude, which by the same token (equating the loss of potential and kinetic energy to the drag times the distance) is exactly L/D multiplied by (H + (MV2/2g)). An airliner with L/D about 20 can glide 20 times the initial altitude, which is quite significant—about 200 km from an initial altitude of 10 km. Added to this is the contribution of initial kinetic energy, corresponding to an additional altitude of 4.5 km at Mach 1, about 300 m/s. For the private pilot, the kinetic energy term is not large, since small aircraft may travel at 0.3 Mach, so that the equivalent height is only about 0.5 km, and glide range comes mostly from altitude. The relative importance of speed and altitude is very different for hypersonic speed, since Mach 20 would equate to (20)2 × 4.5 km, or 1,800 km altitude. Consider the use of glide for range extension of the maximum-range (“mini- mum energy”) ballistic trajectory. A pure ballistic trajectory to intercontinental range has the reentry vehicle (RV) reentering the atmosphere near the target at a grazing angle typically on the order of –22º and slowing abruptly in the atmosphere according to the M/Cd A of the RV, with M the RV mass, Cd the drag coefficient, and A the base area of the RV. This ballistic reentry wastes the kinetic energy of the RV at the time of reentry, whose velocity is the minimum required to achieve the desired range in the first place. This is the best that could be done if Earth had no atmosphere. But it does, and in principle the RV could be designed as a lifting body for the hypersonic regime, and if the thermal insult could be man- aged it could transition in the upper regions of the atmosphere to near-horizontal flight, and then use lift and change of altitude, air density, and change of angle of attack to support the RV weight for a substantial range extension beyond the purely ballistic trajectory. This approach was validated decades ago by flights of the Mk-500 “Evader” RV. Successful implementation of boost-glide technology could yield additional benefits for the prompt global strike mission by means of the ability to maneuver and thus to aid in avoiding undesired overflight of various countries. The launch would be similar to that for a minimum-energy trajectory—that is, maximum range for a given missile—typically with a high apogee and the transition on ballistic reentry to either level or phugoid (porpoise-like) flight—in which the RV bounces in and out of the atmosphere several times and supports its weight by aerodynamic lift only a relatively small fraction of the time, say 10 percent. Supporters of the BGV often argue that this phugoid flight provides range exten- sion at little cost, because for much of this flight—between bounces—the drag is almost zero. It is important to recognize that there is “no free lunch” in phugoid flight, because the lift averaged over this portion of the flight is precisely the weight of the vehicle, and so the time-average lift (and drag) are the same as if the RV were flying at steady altitude and speed in order to maintain the same average aerody- 

208 U.S. Conventional Prompt Global Strike namic lift. The average lift must be equal to the weight of the vehicle: W = gM; the average drag is thus the weight divided by (L/D). On the assumption of constant L/D, it turns out that there are simple closed- form formulas not only for the glide portion of flight but also for the velocity and kinetic-energy loss in the transition from ballistic flight to glide. FLAT EARTH 1. On a flat Earth, the maximum ballistic range is achieved with a constant launch angle of 45º to the horizontal; the purely ballistic range is Rb = V2/g, as is readily derived. Here V is the initial speed of the projectile, T the time of bal- listic flight, g the acceleration of gravity (9.8 m/s2), and Rb the resulting ballistic range: Vh = Vv = V/√2; T = 2Vv/g; Rb = Vh × T = V2/g. Vh and Vv are the horizontal and vertical components of V. 2. Let s be the horizontal path length at a given time. It is shown that the glide range Rg with constant L/D is half the ballistic range multiplied by L/D, and so the glide range equals the ballistic range for L/D = 2: dV/ds = dV/dt × dt/ds = (g/(L/D)) × (1/V); VdV = (g/(L/D))ds; Integrating, V2/2 = gRg/(L/D), so Rg = (L/D) × V2/2g = (L/D) × Rb /2 (Eq. G-1) The glide range can thus significantly extend the ballistic range, but not to the full extent implied here. The launch velocity (and reentry velocity) cannot be taken as the initial glide speed because of the loss of speed due to lift-induced drag in the maneuver to glide flight. 3. Although it is probably not optimum, the transition between ballistic reentry and horizontal glide can be assumed to be made with a high-g (g' >> g) trajectory of constant radius, r, and with the assumption of constant L/D the logarithmic fraction of velocity decrement is just the angle of the arc, θ (in this case 45º) divided by (L/D): dV/dθ = dt/dθ × dV/dt = (r/V) × dV/dt = (r/V)(D/M) = (r/V) g'/(L/D), with g' = V2/r the centripetal acceleration. So dV/dθ = V/(L/D), from which dV/V = dθ/(L/D). Integrating, ∆ln V = (D/L)∆θ, so that 

APPENDIX G 209 (Vf /Vi) = exp((D/L) × ∆θ) = exp –((1/2.2) × (π/4)), (Eq. G-2) where Vi is the vehicle initial speed at the beginning of the transition, and Vf is the speed at the end of the transition and the beginning of the glide portion of flight. For ∆θ = 45º (= π/4) and (D/L) = 1/2.2, the velocity falls to 0.6998 of its initial value, and the square of the velocity to 0.4897. The glide portion of the trajectory, Rb, is thus reduced to 49 percent of what it would have been had the projectile been fired horizontally in the first place into glide mode at V = Vi and at constant L/D = 2.2. The ballistic-trajectory range extension is thus 53.9 percent of the ballistic range, rather than the 110 percent if the vehicle could have negotiated the –45º pull-up without velocity loss on a flat Earth or at short range, as would be the case in principle if a long tether could have been used to supply the centripetal force for the maneuver. At intercontinental ballistic missile (ICBM) range on a round Earth with a reentry dip angle of –22.5º, the logarithmic loss of velocity is only half as large as at –45º (the velocity emerging from the maneuver is 0.8365 of the reentry speed, and its square 0.700). CALCULATION OF TRAJECTORY FOR A ROUND, NONROTATING EARTH As indicated in Figure G-1, much of the range benefit from boost-glide in general and phugoid flight in particular is only available on the round Earth and with near-orbital initial speed of the RV. For speeds in the upper atmosphere com- parable to the orbital velocity in low Earth orbit (LEO), almost no aerodynamic lift is necessary, so the glide range can be astonishing—say, on the order of 13,000 nautical miles (nmi) in some cases. Since Earth’s circumference is 360 degrees of arc and each arc minute is 1 nmi, the circumference of the world is 21,600 nmi (precisely 40,000 km, by the definition of the meter). For simplicity, one can calculate numerically the time-reversed trajectory, starting at zero speed over the target with an (unphysical) angle of attack that is assumed to support the weight of the vehicle (gM) at these very low speeds. Since the weight of the vehicle, W, is constant (it does not have fuel or propulsion), the drag is also constant at W/(L/D), and thus the horizontal acceleration of the vehicle is constant and known: D/M = g/(L/D). The calculation is done in an Excel spreadsheet, providing at each step the new velocity and the integral of velocity thus far, corresponding to the range from the assumed target. A plot of velocity versus range, for example, Figure G-2, shows that cen- trifugal force provides 64 percent of the overall lift of the vehicle with a range of 7,168 km (3,870 nmi) to go, and a time-to-go of 1,950 s. Supplementing the curve of velocity versus range of Figure G-2 are Figure G-3, “Range versus time 

210 U.S. Conventional Prompt Global Strike FIGURE G-1 Reentry trajectories for L/D = 2.2. Note that the last 1,000 km or more of the reentry trajectories are identical for Earth and flat Earth. SOURCE: Data for initial condi- tions (7 km/s, –10o grazing) provided by G. Candler, University of Minnesota, personal communication to the committee, September 17, 2007. Figure G.1, bitmapped, uneditable, color FIGURE G-2 Normalized velocity versus range (Columns C versus E of Table G-1) for pure glide. Figure G,2, bitmapped, uneditable, color (?) 

APPENDIX G 211 FIGURE G-3 Range versus time for pure glide (Columns E versus B of Table G-1). Note that the unit of time here is 10 s. Figure G.3, bitmapped, uneditable, b&w FIGURE G-4 Velocity versus time for pure glide (Columns C versus B of Table G-1). Figure G.4, bitmapped, uneditable, b&w 

212 U.S. Conventional Prompt Global Strike for pure glide,” and Figure G-4 (see previous page), “Velocity versus time for pure glide,” for cases in which time is of interest. The spreadsheet, of which excerpts are shown in Table G-1, embodies cal- culation results using the following formulas: beginning at rest, dV/dt = g D/L; at high speeds and constant altitude, aerodynamic lift does not need to compensate g—only g × (1 – (V/Ve)2), where Ve is LEO speed of 7.90 km/s. Using the prime to indicate time derivative, V′ = g (D/L) (1 – (V/Ve)2), or setting v = V′/Ve, we have v′/(1 – v2) = (g/Ve) (D/L) = c, defining c as (g/Ve) (D/L) = 0.000563 for L/D = 2.2. (Eq. G-3) The scaled speed, v, is obtained by accumulating v' (“v-dot” in the spread- sheet), and the range-to-date by accumulating v. The following simple closed-form integration has been obtained: v(t) = sin (–π/2 + 2 atan (exp(ct)))  (Eq. G-4) In Equation G-3, substitute v = sin(ϕ), so that v' = ϕ' × cos(ϕ) = c, and (dϕ × cos(ϕ)/((cos(ϕ))2) = c dt, or dϕ/(cos(ϕ)) = c dt. This integrates to log tan (π/4 + ϕ/2) = ct, and tan (π/4 + ϕ/2) = exp(ct), so that ϕ = –(π/2) + 2 atan (exp(ct)). Substituting v for sin(ϕ), v(t) = sin(ϕ) = sin (–π/2 + 2 atan (exp(ct))). Spreadsheet column D, headed “v-analytic” (Table G-1) represents values of this last formula, for comparison with the column C, “v,” resulting from numeri- cal integration of v-dot. The agreement to about 0.1 percent is good for a simple first-order integration but, more importantly, shows the absence of blunders in either calculation. COMPARISON WITH TRADITIONAL CALCULATION OF PHUGOID GLIDE Figure G-1 shows glide range for a flat Earth of 4,800 km for the assumed initial conditions, and 7,700 km for the round Earth. According to Equation G-1, for a flat Earth the glide range is s = (L/D) × V2/2g, which for V = 6.5 km/s amounts to 4,742 km. Why “6.5 km/s”? Because 7 km/s at –10º grazing angle yields Vo/Vi = e–(D/L) ∆ θ or 0.93, so that a Vi = 7.0 km/s becomes Vo = 7.0 × 0.93 = 6.5 km/s 

TABLE G-1 Excerpts of Calculation Results from Spreadsheet NOTE: Columns A, C, and D are normalized to 7.90 km/s. Columns G and I are km/s2 and km/s. 213 

214 U.S. Conventional Prompt Global Strike after the pull-up of 10º to the horizontal. The comparison between the 4,800 km estimated from the curve in Figure G-1 and the average-lift approximation result of 4,742 km is good. From Figure G-1, the round-Earth glide range can be estimated as 7,700 km. One can interpolate for an initial glide speed of 6.5 km/s to find a glide range of about 7,900 km—again in reasonable agreement with the result of the detailed calculation as displayed in Figure G-1. Note that entry into glide flight horizontally at 7 km/s, rather than at –10º, would provide a glide range of 10,740 km; most of the increase arises from the closer approach to orbital speed of 7.90 km/s, with the resultant reduction in needed aerodynamic lift and hence drag.� Some of the proposals for long-range boost-glide vehicles enter the glide phase at angles from the horizontal two to four times smaller than the 10º example used here, and at speeds considerably closer to orbital speed of 7.90 km/s (25,920 ft/s) than the example of 7 km/s used here. They are essentially “fractional orbital bombardment vehicles” with essentially infinite “range extension” and substantial cross-range maneuver capability. INTERPRETATION It is beyond the scope of the appendix to analyze quantitatively the major challenge to the BGV—the long duration of the heat influx into the thermal protection system that shields the structure and internals of the vehicle from the fiery heat of skin friction with Earth’s atmosphere. BGVs of longest range start at essentially orbital speed of 7.9 km/s and thus have more kinetic energy to dis- sipate than do ICBM RVs. The RV, however, traverses the 8 km “scale height” of the atmosphere at an angle to the horizontal of 22º, in a few seconds, while the BGV supports itself aerodynamically for 10,000 km at near-orbital speed for 1,200 s. The heating due to lift is concentrated on the lower surface of the BGV rather than uniformly around the axis of the RV, usually resulting in a very thick layer of ablative material on the lower surface of the BGV. The function of this inner layer is simple insulation rather than ablation, and so the thermal protection At   steeper reentry angles than the 10º example of Figure G-1, the agreement of detailed calcula- tion with the time-averaged lift and drag calculated here is much worse, because the approximation of horizontal flight is increasingly violated with the steeper angles of the phugoid flight pictured in Figure G-1. For instance, flat-Earth glide with 45º pull-up would give 2,693 km range extension, while detailed calculation (G. Candler, University of Minnesota, personal communication to the committee, September 19, 2007) of conventional phugoid flight provides only 1,790 km. Taking into account that the detailed calculation typified in Figure G-1 does not change the horizontal component of bal- listic reentry velocity, this fraction 0.707 would reduce the time-averaged drag range to 1,904 km, for comparison with the detailed calculation of 1,790 km. This shows the merit of nearly horizontal injection at the altitude that will provide L/D = 2.2 at the injection speed, so that long-distance glide can be achieved by a gradual drop of altitude in order to provide steady 1-g lift as the vehicle speed gradually drops. 

APPENDIX G 215 system has a different optimum design than does that of an RV. Indeed, much of the protection system could be in the form of non-ablating material such as the “tiles” on the space shuttle. The intense heating of the BGV during the whole of the glide phase provides a strong infrared signal to defensive systems equipped to detect it or to use it for an infrared homing intercept. A simple terminal maneuver for a ballistic missile will allow it to deny sanc- tuary to structures and locations shielded by a near-vertical bluff. At intermediate range this can require a 45º maneuver that with an L/D = 2.2 would (according the example following Equation G-2) result in a reduction of warhead speed to 0.6998 of the initial speed. If performed at 10 g transverse acceleration (0.098 km/s2), the maneuver could take on the order of 30 s; an alternative would be to have a high-drag RV to greatly reduce speed to, say, Mach 3 (1 km/s), so that a 45º maneuver could be accomplished in a few seconds (slowdown to turn). The simple kinematic considerations of this appendix indicate the value of the engineering design of a variable-geometry RV, and the competition between the longer-term “better” and the earlier and perhaps “good enough.”