Sub.:    Range  vs.   difficulty   for   ballistic   missile

       Date:    March 27, 1998
       From:    R.L.  Garwin    (914) 945-2555   
                26-234   Yorktown Heights, NY
                IBM Fellow Emeritus
                Email: [email protected]
                       [email protected] 
                       (Jean A. Hernandez, Secretary)

       The  question frequently arises as to the difficulty or ease
       of extending the range  of  a  ballistic  missile  from  the
       common  Scud B of perhaps 300 km range to a range of 9000 km
       or 10,000 km.

       One way to answer this question is simply to  point  to  the
       various  missiles that exist in the inventory of the various
       nations, to show that ICBM  range  is  indeed  possible  and
       conventional.    But  this  conceals  the  very  substantial
       difference between the technology of a short-range ballistic
       missile or medium-range ballistic missile  and  that  of  an

       Another approach is to go back to first principles-- namely,
       the  rocket  equation--  to see what is involved actually in
       reaching such range for a ballistic  missile.    The  reader
       should  bear  with us while we explain some simple concepts,
       probably familiar to just about everybody.

       A "ballistic missile" is a vehicle or object that  continues
       on  its  path  under  the  force of the Earth's gravity.  It
       continues  because  of  Newton's  laws  of  motion.     More
       specifically,  it  exchanges  its  kinetic  energy of radial
       motion for potential energy as it rises to apogee  and  then
       recovers  kinetic  energy as it loses potential energy as it
       comes closer to the center of the earth.  That applies to  a
       rock just as well as it does to an ICBM.

       But  how  does  the  missile "go ballistic"?   The artillery
       projectile follows a ballistic trajectory after having  been
       accelerated  by  high  pressure  gas  in a gun barrel to its
       final velocity.   But  a  bazooka  or  other  shoulder-fired
       rocket  does  not need a heavy gun barrel.  It does not push
       on the gun barrel to be accelerated in a short time; rather,
       it pushes on the exhaust gases of the combustion  of  rocket

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            Of  course, since the manufacturer and owner of the gun
            barrel probably believes that it is more important (and
            certainly costs more) than  the  individual  projectile
            that  it  shoots,  this  is  usually presented from the
            point  of  view  of  the  gun  barrel's  pushing on the

       A given rocket stage operates ordinarily at constant chamber
       pressure.  Hot gas is produced by the combustion  of  liquid
       or  solid  fuel,  exchanging  chemical  energy  for  thermal
       energy.  The hot gas creates some chamber  pressure  in  the
       combustion  chamber,  which  is  connected  to  space by the
       throat of a rocket nozzle.  If there were just  an  aperture
       in  the  combustion  chamber, one would still have a rocket,
       but the gas  would  be  emitted  from  that  aperture  in  a
       hemisphere,  rather  than  in  a  directed  stream.   So the
       momentum associated with the thermal energy would  be  lower
       than if all the gas went in the same direction.

       Furthermore,  even  if  the  gas  were all to go in the same
       direction, but were to remain hot, the gas stream would have
       less momentum than if the gas were cooled by  expansion  and
       still kept largely uni-directional.

       In  fact, the function of the rocket nozzle is to expand the
       combustion  gas  in  such  a  fashion  that  it  cools   and
       constitutes  a  largely uni0directional jet.  At the throat,
       one has hot gas with the molecules going in every direction.
       As the gas expands  in  the  rocket  nozzle,  it  repeatedly
       pushes  on  the  material  of  the  nozzle  (except  in  the
       direction of the exit circle),  and  as  the  rocket  nozzle
       diameter  increases  with  distance from the throat, the gas
       expands and cools itself.   So in the  rocket  jet  that  is
       ejected  into  space,  the  gas  is  quite  cold, but moving
       extremely rapidly.  The best that can be done  in  obtaining
       exhaust  velocity is to convert all of the thermal energy of
       the fuel into kinetic energy of the exhaust (leaving nothing
       left over for internal thermal energy of the exhaust plume).
       For rocket exhaust into space (above the  atmosphere),  that
       condition  is  closely  approached.    Certainly billions of
       dollars have gone into wringing the last bit of  performance
       out  of  rocket nozzles.   Eventually one gets to a point at
       which the size and mass of additional nozzle  outweighs  the
       small  benefit  that can be obtained by reducing the already
       small residual thermal energy in the exhaust plume.

       For exhaust into the atmosphere in  the  boost  phase  of  a
       rocket,  such  large  expansion  ratios  are  not available,
       because the atmospheric density and pressure is finite.   So
       first-stage   rocket  nozzles  lack  the  long  skirt  of  a
       deep-space nozzle.

       Often if one's goal is synthesis the best  or  only  way  to
       proceed  is  through  analysis of various specific examples.

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       For instance, it is a simple matter to follow the trajectory
       of a  specific  choice  of  rocket  design,  launched  at  a
       specific  angle  to  the  horizontal;  one can determine the
       entire  course  of  propulsion,  burnout,  and  the range to
       impact.  One would, of course, like to  know  the  elevation
       angle  corresponding  to  maximum  range,  which is a simple
       problem for a flat Earth  (the  result  is  45 deg,  without
       taking  into  account gravity loss or atmospheric drag), but
       even  that  could  be  obtained  through  multiple  computer
       trials.   Even when a complex synthesis technique-- based on
       experience, complex computer models, and the like-- purports
       to provide a solution to the problem, it is a good  idea  to
       check  it  by simple analysis and by elementary step-by-step
       following of the trajectory.

       Quantitatively one can have an  excellent  approximation  to
       rocket performance by the assumption that all of the thermal
       energy of the fuel emerges as uni-directional kinetic energy
       of  the  exhaust.    Then Newton's third law states that for
       every action there is an equal and opposite reaction.   This
       means  that  in a coordinate frame centered on the rocket at
       any given time, the momentum  associated  with  the  exhaust
       emitted  in  a  very  short  interval  (say, one second), is
       countered by the momentum given to the remaining rocket plus
       un-exhausted fuel.  So one has an equation relating the two,
       and specifically giving  the  rate  of  increase  of  rocket
       momentum  as  determined  by the rate m (in grams/second) at
       which mass is exhausted, together with the  velocity  Ve  of
       the exhausted mass.

       Eq. 1:   mVe = M dV/dt

       In  this  Equation,  m is the mass ejected per second in the
       rocket plume; Ve is the exhaust velocity of the  jet;  M  is
       the  remaining  mass of the rocket (decreasing with time); V
       is the instantaneous velocity of the rocket, and  t  is  the

       This equation has a simple solution which can be written as:

       Eq. 2:   Mf/Mo = e**-(Vg/Ve)

       Here  Mf is the final mass of the rocket stage-- the payload
       plus the dry weight of the rocket; M0 is  the  gross  launch
       overall  weight (GLOW); Vg is the velocity gain overall; and
       Ve is again the rocket exhaust speed.  The reader  is  asked
       to   pardon  the  rather  gross  appearance  of  the  simple
       mathematics; expedience may be  worth  more  than  elegance.
       "**"  stands  for  exponentiation;  "r x s"  or "r*s" is the
       product of r and s, as is "r s".

       Note that this simple  Eq. 2  has  its  ultimate  simplicity
       because  it ignores the time at which the combustion occurs,
       and this means that for rockets operating in space  and  not

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       captive  in  the  Earth's gravitational field, a tiny rocket
       nozzle achieving a certain exhaust velocity  and  ultimately
       expelling  the  same  total  mass  will  ultimately give the
       residual  rocket mass the same speed as would a large rocket
       nozzle.  And the smaller nozzle and combustion  chamber  and
       pumps  have  less  parasitic  mass  and  allow more mass for
       payload or a higher velocity of that payload.  So deep-space
       systems have low-thrust rockets and long burn times.

       Obviously one cannot push this  too  far  in  the  ballistic
       missile  application,  because once the rocket engine is lit
       for the missile on the ground, the missile  must  produce  a
       thrust greater than the Earth's gravity, or else the missile
       will  continue  to  sit  on  the ground, despite the furious
       expenditure of fuel.

       It  is  quite  remarkable  that  large   liquid-fuel   space
       boosters,  for instance, have an excess acceleration of only
       about 0.3 g, so that most  (1.0/1.3)  of  their  thrust  and
       their  initial  fuel  expenditure simply replaces the upward
       thrust of the launch stand.  Ballistic missiles as  weapons,
       however,  typically have a larger initial thrust and have in
       this way less "gravity  loss"  associated  with  their  burn

       In fact, it is a useful approximation to ignore gravity loss
       in the first estimate of ballistic missile performance.

       We  have now demonstrated how to calculate the ratio between
       gross launch overall weight (GLOW) and payload.  That is, we
       know how for a single stage to relate  the  GLOW,  the  fuel
       mass,  the  dry  weight, and the payload.  There is a direct
       trade-off between dry weight (structure, fuel tanks,  rocket
       nozzle,  pumps,  etc.)  on  the one hand, and payload on the
       other.  Indeed, because there is a maximum speed  even  with
       zero  payload,  and  because  some  of  that  dry  weight is
       associated with the initial thrust and the  initial  nozzle,
       there  is  benefit  or  even  necessity  to "staging" rocket

       For a given technology level, the idea is to throw  away  as
       much  of  the  inert  mass  as possible as the fuel is being
       expended, and to end up with a smaller rocket, with unburned
       fuel, as the second "stage".  The trick can be repeated,  so
       that   in  principle  a  given  level  of  technology  could
       accelerate a second stage to a certain velocity, which would
       then burn and separate from the  third  stage,  which  would
       then burn and separate from the fourth stage, etc.

       It  is  of interest to ask what can be done with "continuous
       staging" in which one assumes that not only rocket  fuel  is
       exhausted,  but  at  the  same time the associated weight of
       tankage and motors.   In this  approximation  of  continuous
       staging,  with  a  small  enough  payload Mf, one is able to

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       calculate a sufficient initial mass Mo, no matter  what  the
       exhaust velocity.  Of course, the payload can be vanishingly
       small,  but  whatever  the  required velocity gain, there is
       such a payload.

       For  continuous  staging,  the  payload fraction is given by
       Eq. 2, but with an equivalent exhaust velocity Vee < Ve,  in
       view  of  the  assumption of zero dry weight that goes along
       with Vee.

       In terms of the exhaust velocity, the payload, and the gross
       launch overall weight, one can now calculate  some  examples
       for  a  single  stage.    But  first  we  want to relate the
       velocity to the range.

            There is an old story about the physicist looking for a
            job who with the confidence common  to  physicists  and
            mathematicians  felt  that  he  could revolutionize the
            dairy industry.  So he was hired on speculation and sat
            down at his brand new, shiny, empty desk with a pen and
            paper and began to write, "Consider a  spherical  cow."
            It  is sometimes instructive to begin with a simple, if
            artificial problem, for which the assumptions  and  the
            mode of calculation are more transparent.

       So  we  consider  first  a  flat Earth, which in our common,
       local experience is a reasonable approximation.

       If one sketches the optimum trajectory for a stone to go the
       greatest distance for given initial  speed  (or  a  baseball
       hit)  one  finds that the optimum launch angle is 45 degrees
       from the horizontal, and one can obtain the range simply  by
       noting  the  time  required  for the vertical velocity Vz to
       reverse.  During flight without thrust or drag, the rocket's
       horizontal velocity Vh is constant.  So we have the  set  of

       Voz = Vo/sr2

       (For brevity, I write "sr2" for the square root of two).

       T = 2Voz/g; R = VhT = (2/g)(Vo/sr2)(Vo/sr2) = Vo**2/g

       g = 9.8
       m/s**2  R = 300 km  Vo = sr(gR) = sr(9.8x3x10**5) =1.7146 km/s.

            Vo km/s    R km(Flat Earth)          Mo/Mb Reality
            1.7146        300          2.140     2.69 Scud B
            2.425         600          2.933     3.80 Al Hussein
            3.1305       1000          4.01
            9.900      10,000          80.82

                        (Round Earth)
            7.1        10,000          23.33

                                                             PAGE 6

       Table 1  required  initial  velocity  Vo  for range R in the
       flat, non-rotating Earth approximation.  Mo is  the  initial
       launch weight and Mb is the burn-out mass of the rocket.

       It  should be noted that an exhaust velocity Ve = 2.254 km/s
       corresponds to a "specific impulse" (Isp) of 230 sec = Ve/g.
       The specific impulse is a time that characterizes the energy
       per  unit  mass  of  the  propellant.    Equipped  with   an
       appropriate  nozzle, it is equal to the time that an initial
       mass of propellant could  provide  a  thrust  equal  to  the
       Earth's gravitational force on that initial mass.

       From  the  MIT  group,  we have data on the Al Hussein.  The
       empty weight is 1785 kg with a warhead of 300 kg for  a  dry
       weight of 1485 kg.  Similarly the Scud B has an empty weight
       of  2370 kg  and  a  warhead  of  965 kg for a dry weight of
       1405 kg.   The Iraqis did not add  very  much  structure  in
       lengthening the tanks of the Scud B to make the Al Hussein.

       The  Al Hussein is supposed to have a gross GLOW of 6785 kg,
       so  that  Mb/Mo = 1485/6785 = 0.2189.    From  Eq. 2,   this
       corresponds  to  a  velocity  gain of 1.5193 x Ve.   And for
       Ve = 2.254 km/s for an Isp of 230 s (given by the MIT  group
       as  the  Isp for the Scud B and for the Al Hussein), we find
       that the Vo = 2.254 km/s x 1.5193 = 3.4245 km/s.   From  our
       Table  above,  this  corresponds  to a range of 1200 km with
       zero payload.

       If one wants to have a greater range than 1200 km, even with
       zero payload, one must either change the technology or go to
       a rocket with more than one stage.  First we consider adding
       a second stage.   By "technology"  we  mean  here  only  two
       things--  the structural weight of the rocket and the motors
       (the dry weight or structural fraction SF) and the Isp.

       First we consider the use of staging for a given technology,
       taking Isp = 230 s and a structural fraction  of  22%  (SF).
       We assume that the SF is constant for a payload less than or
       equal  to  the  SF,  assuming  that  not  much  structure is
       required to support the payload.

       Then we have:

     c Eq. 3:   V2 = Ve ln M2/(M2*SF +PL)

       wherefore the second stage,  we  have  written  the  burnout
     c weight  explicitly  as (M2*SF + PL), where PL is the payload
     c in kg.  By "ln" we mean the base-e logarithm.

       And for the first stage we have

       Eq. 4:   V1 = Ve ln M1/(M1*SF + M2)

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       The entire mass of the second stage rocket-- fuel and  all--
       is the payload of the first stage.

       Now  we  want to minimize M1 for a given total velocity gain

       (V1+V2)/Ve = ln M1/(M1*SF+M2)  +  ln M2/(M2*SF+PL)

       e**(V1+V2)/Ve = M1 x M2/(M1*SF+M2)x(M2*SF+PL)

       Eq. 5:   e**Vg/Ve = 1/(SF+M2/M1)(SF+PL/M2)

       Let  PL' = PL/(SF*M2)   SR = M2/(SF*M1),   where   we   have
       introduced the "staging ratio" SR.

       Eq. 6:   e**(V1+V2)/Ve = (SF)**-2  /(1+SR)(1+PL') =

       Note that for a given M1

       Vg  is  maximized  for the minimum of Q = (1+SR)(1+PL') with
       respect to M2. So

     c dQ/dM2 = 0 for -PL*SF*M1 = PL*M2 + M2**2(SF+PL/M2) = 0


       -PL*SF*M1 + M2**2 SF = 0

       or M1*PL = M2**2.

       Therefore M2 = sr(M1*PL).  This means that the velocity gain
       is the same for each stage of the  multi-stage  rocket  with
       the same technology for each stage.

       We can now solve for M1 to obtain

       (SF + sr(M1*PL)/M1)(SF + PL/sr(M1*PL)) = e**-Vg/Ve

       (SF + sr(PL/M1))(SF + sr(PL/M1)) = e**-Vg/Ve


       (SF + sr(PL/M1)) = e**-(Vg/2)/Ve

       Now     take     the     Vg = 7.1 km/s;    Vg/2 = 3.55 km/s;
       Ve = 2.254 km/s

       sr(PL/M1) = e**-(Vg/2)/Ve - SF

       Eq. 7:   (M1/PL) = 1/(e**(-Vg/2)Ve - SF)**2 = 1/(0.2070-0.22)**2

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       Unfortunately, this denominator is negative, and there is no
       launch mass that will propel even a two-stage rocket with Al
       Hussein technology to a range of 10,000 km on a round Earth.

       We  should  have noted that the launch velocity requirements
       for a round Earth are less than those for a  flat  Earth  at
       long  range  simply  because  the  Earth falls away from the
       rocket as it travels-- a phenomenon noted by Newton  in  his
       comparison  between  the  speed  of  a falling apple and the
       speed of the falling moon.  For instance, for a range beyond
       20,000 km, there  is  no  additional  velocity  requirement,
       since  the rocket has gone into orbit and can have any range
       with no additional velocity.

       So maybe one wants to go to better technology rather than to
       a third stage.  For instance, we could imagine a  structural
       fraction  SF = 0.15  instead  of  the structural fraction of

       Better structural fraction can be obtained by use of  higher
       strength  materials,  by chemical milling or other tailoring
       of parts, and so on.   Assuming a 30% reduction  in  SF  (to
       0.15),   the   denominator   in  the  previous  equation  is
       D = 0.2070 - 0.15 = 0.0570,  and  the  launch  mass   of   a
       two-stage  rocket  would be given by M1/PL = (1/D)**2 = 308.
       So for a payload of 300 kg , this would be a  launch  weight
       of 90 tons.

       If  this  seems high, and we want a 30 ton launch weight for
       600 kg payload; (M1/PL) = 50, then we need D = 0.1414.    So
       that    SF = 0.2070 - 0.1414 = 0.0656.        This   is   an
       astonishingly low  structural  fraction  compared  with  the

       Another  option  is to move to higher Isp-- for instance 300
       seconds.  Then Ve = 2.94 km/s and e**(-Vg/2Ve) = 0.271.   So
       we  would  have  the desired denominator of D = 0.1414 for a
       structural fraction SF = 0.130, allowing  a  30  ton  launch
       weight for a 600 kg payload.  Or we could use three stages.

       Now  we discuss "gravity loss".  We have noted early on that
       a rocket launched vertically with a  thrust  less  than  its
       weight  will  just  sit  on  the  pad,  exhausting its fuel,
       without any acceleration at all.   Successful  rockets  thus
       have  acceleration  greater  than  1 g,  but  they are still
       penalized by the Earth's gravity during boost.  Because  the
       burn  time  of  a  Scud B  is 70 s, gravity loss during burn
       imposes a requirement on the  rocket  to  exert  a  vertical
       component  of  thrust  that  would  have sufficed to provide
       additional  vertical  velocity  of  70 x 9.8 m/s,  or  about
       0.69 km/s.    In  the  first  line  of  Table 1, the initial
       velocity for 300 km is 1.7146 at 45 degrees elevation.  This
       has a horizontal component of  1.212 km/s,  and  a  vertical
       component  of  1.212 km/s.   Added to the vertical component

                                                             PAGE 9

       requirement to be produced by the rocket  is  the  0.69 km/s
       that  we  have  just  calculated,  for an effective vertical
       velocity gain of 1.802 km/s.   These two components  of  the
       right  triangle  thus correspond to a total velocity gain of
       2.172 km/s, and  with  the  Isp  of  230 s,  to  a  required
       Mo/Mb = 2.621.    This  accounts  for  a  good  deal  of the
       difference between the  instantaneous  burn  calculation  of
       Mo/Mb = 2.140 and the "Reality" of 2.69 for the Scud B.

       This  rather  lengthy  discussion  indicates  the changes of
       technology toward reduced structural fraction  and  improved
       Isp that are mandatory in producing a rocket of ICBM range.