EXAMPLE PROBLEMS

EXAMPLE PROBLEMS

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PROBLEM 1.1 

A spacecraft's engine ejects mass at a rate of 30 kg/s with an exhaust velocity
of 3,100 m/s.  The pressure at the nozzle exit is 5 kPa and the exit area is
0.7 m2.  What is the thrust of the engine in a vacuum?


SOLUTION,

   Given:  q = 30 kg/s
           Ve = 3,100 m/s
           Ae = 0.7 m2
           Pe = 5 kPa = 5,000 N/m2
           Pa = 0
    
   Equation (1.6),

      F = q × Ve + (Pe - Pa) × Ae
      F = 30 × 3,100 + (5,000 - 0) × 0.7
      F = 96,500 N




PROBLEM 1.2 

The spacecraft in problem 1.1 has an initial mass of 30,000 kg.  What is the
change in velocity if the spacecraft burns its engine for one minute?

SOLUTION,

   Given:  M = 30,000 kg
           q = 30 kg/s
           Ve = 3,100 m/s
           t = 60 s

   Equation (1.16),
 
      V = Ve × LN[ M / (M - qt) ]
      V = 3,100 × LN[ 30,000 / (30,000 - (30 × 60)) ]
      V = 192 m/s  




PROBLEM 1.3 

A spacecraft's dry mass is 75,000 kg and the effective exhaust gas velocity
of its main engine is 3,100 m/s.  How much propellant must be carried if the
propulsion system is to produce a total v of 700 m/s?


SOLUTION,

   Given:  Mf = 75,000 kg
           C = 3,100 m/s
           V = 700 m/s

   Equation (1.20),
 
      Mo = Mf × e(DV / C)
      Mo = 75,000 × e(700 / 3,100)
      Mo = 94,000 kg

   Propellant mass,

      Mp = Mo - Mf
      Mp = 94,000 - 75,000
      Mp = 19,000 kg




PROBLEM 1.4 

A 5,000 kg spacecraft is in Earth orbit traveling at a velocity of 7,790 m/s.
Its engine is burned to accelerate it to a velocity of 12,000 m/s placing it
on an escape trajectory.  The engine expels mass at a rate of 10 kg/s and an
effective velocity of 3,000 m/s.  Calculate the duration of the burn.


SOLUTION,

   Given:  M = 5,000 kg
           q = 10 kg/s
           C = 3,000 m/s
           V = 12,000 - 7,790 = 4,210 m/s

   Equation (1.21),

      t = M / q × [ 1 - 1 / e(DV / C) ]
      t = 5,000 / 10 × [ 1 - 1 / e(4,210 / 3,000) ]
      t = 377 s




PROBLEM 1.5 

A rocket engine burning liquid oxygen and kerosene operates at a mixture ratio
of 2.26 and a combustion chamber pressure of 50 atmospheres.  If the nozzle is
expanded to operate at sea level, calculate the exhaust gas velocity relative
to the rocket.


SOLUTION,

   Given:  O/F = 2.26
           Pc = 50 atm
           Pe = Pa = 1 atm

   From LOX/Kerosene Charts we estimate,

      Tc = 3,470 K
      M = 21.40           
      k = 1.221

   Equation (1.22),

      Ve = SQRT[ (2 × k / (k - 1)) × (R* × Tc / M) × (1 - (Pe / Pc)(k-1)/k) ]
      Ve = SQRT[ (2 × 1.221 / (1.221 - 1)) × (8,314.46 × 3,470 / 21.40) × (1 - (1 / 50)(1.221-1)/1.221) ]
      Ve = 2,749 m/s





PROBLEM 1.6 

A rocket engine produces a thrust of 1,000 kN at sea level with a propellant
flow rate of 400 kg/s.  Calculate the specific impulse.


SOLUTION,

   Given:  F = 1,000,000 N
           q = 400 kg/s

   Equation (1.23),

      Isp = F / (q × g)
      Isp = 1,000,000 / (400 × 9.80665)
      Isp = 255 s (sea level)




PROBLEM 1.7 

A rocket engine uses the same propellant, mixture ratio, and combustion chamber
pressure as that in problem 1.5.  If the propellant flow rate is 500 kg/s,
calculate the area of the exhaust nozzle throat.


SOLUTION,

   Given:  Pc = 50 × 0.101325 = 5.066 MPa
           Tc = 3,470 K
           M = 21.40
           k = 1.221
           q = 500 kg/s

   Equation (1.27),

      Pt = Pc × [1 + (k - 1) / 2]-k/(k-1)
      Pt = 5.066 × [1 + (1.221 - 1) / 2]-1.221/(1.221-1)
      Pt = 2.839 MPa = 2.839×106 N/m2

   Equation (1.28),

      Tt = Tc / (1 + (k - 1) / 2)
      Tt = 3,470 / (1 + (1.221 - 1) / 2)
      Tt = 3,125 K

   Equation (1.26),

      At = (q / Pt) × SQRT[ (R* × Tt) / (M × k) ]
      At = (500 / 2.839×106) × SQRT[ (8,314.46 × 3,125) / (21.40 × 1.221) ]
      At = 0.1756 m2




PROBLEM 1.8 

The rocket engine in problem 1.7 is optimized to operate at an elevation of 2000
meters.  Calculate the area of the nozzle exit and the section ratio.


SOLUTION,

   Given:  Pc = 5.066 MPa
           At = 0.1756 m2
           k = 1.221

   From Atmosphere Properties,

      Pa = 0.0795 MPa

   Equation (1.29),

      Nm2 = (2 / (k - 1)) × [(Pc / Pa)(k-1)/k - 1]
      Nm2 = (2 / (1.221 - 1)) × [(5.066 / 0.0795)(1.221-1)/1.221 - 1]
      Nm2 = 10.15
      Nm = (10.15)1/2 = 3.185

   Equation (1.30),

      Ae = (At / Nm) × [(1 + (k - 1) / 2 × Nm2)/((k + 1) / 2)](k+1)/(2(k-1))
      Ae = (0.1756 / 3.185) × [(1 + (1.221 - 1) / 2 × 10.15)/((1.221 + 1) / 2)](1.221+1)/(2(1.221-1))
      Ae = 1.426 m2

   Section Ratio,

      Ae / At = 1.426 / 0.1756 = 8.12





PROBLEM 1.9

For the rocket engine in problem 1.7, calculate the volume and dimensions of a
possible combustion chamber.  The convergent cone half-angle is 20 degrees.


SOLUTION,

   Given:  At = 0.1756 m2 = 1,756 cm2
           Dt = 2 × (1,756/)1/2 = 47.3 cm
            = 20o

   From Table 1,

           L* = 102-127 cm for LOX/RP-1, let's use 110 cm

   Equation (1.33),

           Vc = At × L*
           Vc = 1,756 × 110 = 193,160 cm3

   From Figure 1.7,

           Lc = 66 cm (second-order approximation)

   Equation (1.35),

           Dc = SQRT[(Dt3 + 24/ × tan  × Vc) / (Dc + 6 × tan  × Lc)]  
           Dc = SQRT[(47.33 + 24/ × tan(20) × 193,160) / (Dc + 6 × tan(20) × 66)]  
           Dc = 56.6 cm (four interations)




PROBLEM 1.10 

A solid rocket motor burns along the face of a central cylindrical channel 10
meters long and 1 meter in diameter.  The propellant has a burn rate coefficient
of 5.5, a pressure exponent of 0.4, and a density of 1.70 g/ml.  Calculate the
burn rate and the product generation rate when the chamber pressure is 5.0 MPa.


SOLUTION,

   Given:  a = 5.5
           n = 0.4
           Pc = 5.0 MPa
           p = 1.70 g/ml
           Ab =  × 1 × 10 = 31.416 m2

   Equation (1.36),

      r = a × Pcn
      r = 5.5 × 5.00.4 = 10.47 mm/s

   Equation (1.37),

      q = p × Ab × r
      q = 1.70 × 31.416 × 10.47 = 559 kg/s




PROBLEM 1.11 

Calculate the ideal density of a solid rocket propellant consisting of 68%
ammonium perchlorate, 18% aluminum, and 14% HTPB by mass.


SOLUTION,

   Given:  wAP = 0.68
           wAl = 0.18
           wHTPB = 0.14

   From Properties of Rocket Propellants we have,

      AP = 1.95 g/ml
      Al = 2.70 g/ml
      HTPB = ≈0.93 g/ml

   Equation (1.38),
      
      p = 1 / i (w / )i
      p = 1 / [(0.68 / 1.95) + (0.18 / 2.70) + (0.14 / 0.93)]
      p = 1.767





PROBLEM 1.12 

A two-stage rocket has the following masses:  1st-stage propellant mass 120,000
kg, 1st-stage dry mass 9,000 kg, 2nd-stage propellant mass 30,000 kg, 2nd-stage
dry mass 3,000 kg, and payload mass 3,000 kg.  The specific impulses of the
1st and 2nd stages are 260 s and 320 s respectively.  Calculate the rocket's
total V.


SOLUTION,

   Given:  Mo1 = 120,000 + 9,000 + 30,000 + 3,000 + 3,000 = 165,000 kg
           Mf1 = 9,000 + 30,000 + 3,000 + 3,000 = 45,000 kg
           Isp1 = 260 s
           Mo2 = 30,000 + 3,000 + 3,000 = 36,000 kg
           Mf2 = 3,000 + 3,000 = 6,000 kg
           Isp2 = 320 s

   Equation (1.24),

      C1 = Isp1g
      C1 = 260 × 9.80665 = 2,550 m/s

      C2 = Isp2g 
      C2 = 320 × 9.80665 = 3,138 m/s

   Equation (1.39),

      V1 = C1 × LN[ Mo1 / Mf1 ]
      V1 = 2,550 × LN[ 165,000 / 45,000 ]
      V1 = 3,313 m/s  

      V2 = C2 × LN[ Mo2 / Mf2 ]
      V2 = 3,138 × LN[ 36,000 / 6,000 ]
      V2 = 5,623 m/s

   Equation (1.40),

      VTotal = V1 + V2       
      VTotal = 3,313 + 5,623
      VTotal = 8,936 m/s




PROBLEM 4.1 

Calculate the velocity of an artificial satellite orbiting the Earth in a
circular orbit at an altitude of 200 km above the Earth's surface.


SOLUTION,

   From Basics Constants,

      Radius of Earth = 6,378.14 km 
      GM of Earth = 3.986005×1014 m3/s2  
   
   Given:  r = (6,378.14 + 200) × 1,000 = 6,578,140 m

   Equation (4.6),

      v = SQRT[ GM / r ]
      v = SQRT[ 3.986005×1014 / 6,578,140 ]
      v = 7,784 m/s





PROBLEM 4.2 

Calculate the period of revolution for the satellite in problem 4.1.


SOLUTION,

   Given:  r = 6,578,140 m

   Equation (4.9),

      P2 = 4 × 2 × r3 / GM

      P = SQRT[ 4 × 2 × r3 / GM ]
      P = SQRT[ 4 × 2 × 6,578,1403 / 3.986005×1014 ]
      P = 5,310 s




PROBLEM 4.3 

Calculate the radius of orbit for a Earth satellite in a geosynchronous orbit,
where the Earth's rotational period is 86,164.1 seconds.


SOLUTION,

   Given:  P = 86,164.1 s

   Equation (4.9),

      P2 = 4 × 2 × r3 / GM

      r = [ P2 × GM / (4 × 2) ]1/3
      r = [ 86,164.12 × 3.986005×1014 / (4 × 2) ]1/3
      r = 42,164,170 m




PROBLEM 4.4 

An artificial Earth satellite is in an elliptical orbit which brings it to
an altitude of 250 km at perigee and out to an altitude of 500 km at apogee.
Calculate the velocity of the satellite at both perigee and apogee.


SOLUTION,

   Given:  Rp = (6,378.14 + 250) × 1,000 = 6,628,140 m
           Ra = (6,378.14 + 500) × 1,000 = 6,878,140 m

   Equations (4.16) and (4.17),

      Vp = SQRT[ 2 × GM × Ra / (Rp × (Ra + Rp)) ]
      Vp = SQRT[ 2 × 3.986005×1014 × 6,878,140 / (6,628,140 × (6,878,140 + 6,628,140)) ]
      Vp = 7,826 m/s

      Va = SQRT[ 2 × GM × Rp / (Ra × (Ra + Rp)) ]
      Va = SQRT[ 2 × 3.986005×1014 × 6,628,140 / (6,878,140 × (6,878,140 + 6,628,140)) ]
      Va = 7,542 m/s




PROBLEM 4.5 

A satellite in Earth orbit passes through its perigee point at an altitude of
200 km above the Earth's surface and at a velocity of 7,850 m/s.  Calculate the 
apogee altitude of the satellite.


SOLUTION,

   Given:  Rp = (6,378.14 + 200) × 1,000 = 6,578,140 m
           Vp = 7,850 m/s

   Equation (4.18),

      Ra = Rp / [2 × GM / (Rp × Vp2) - 1]
      Ra = 6,578,140 / [2 × 3.986005×1014 / (6,578,140 × 7,8502) - 1]
      Ra = 6,805,140 m 

      Altitude @ apogee = 6,805,140 / 1,000 - 6,378.14 = 427.0 km




PROBLEM 4.6 

Calculate the eccentricity of the orbit for the satellite in problem 4.5.


SOLUTION,

   Given:  Rp = 6,578,140 m
           Vp = 7,850 m/s
           
   Equation (4.20),

      e = Rp × Vp2 / GM - 1
      e = 6,578,140 × 7,8502 / 3.986005×1014 - 1
      e = 0.01696




PROBLEM 4.7 

A satellite in Earth orbit has a semi-major axis of 6,700 km and an eccentricity
of 0.01.  Calculate the satellite's altitude at both perigee and apogee.


SOLUTION,

   Given:  a = 6,700 km
           e = 0.01

   Equation (4.21) and (4.22),

      Rp = a × (1 - e)
      Rp = 6,700 × (1 - .01)
      Rp = 6,633 km

      Altitude @ perigee = 6,633 - 6,378.14 = 254.9 km

      Ra = a × (1 + e)
      Ra = 6,700 × (1 + .01)
      Ra = 6,767 km

      Altitude @ apogee = 6,767 - 6,378.14 = 388.9 km




PROBLEM 4.8 

A satellite is launched into Earth orbit where its launch vehicle burns out at
an altitude of 250 km.  At burnout the satellite's velocity is 7,900 m/s with the
zenith angle equal to 89 degrees.  Calculate the satellite's altitude at perigee
and apogee.


SOLUTION,

   Given:  r1 = (6,378.14 + 250) × 1,000 = 6,628,140 m
           v1 = 7,900 m/s
            = 89o 
            
   Equation (4.26),

      (Rp / r1)1,2 = ( -C ± SQRT[ C2 - 4 × (1 - C) × -sin2  ]) / (2 × (1 - C))

      where  C = 2 × GM / (r1 × v12)
             C = 2 × 3.986005×1014 / (6,628,140 × 7,9002)
             C = 1.927179

      (Rp / r1)1,2 = ( -1.927179 ± SQRT[ 1.9271792 - 4 × -0.927179 × -sin2(89) ]) / (2 × -0.927179)
      (Rp / r1)1,2 = 0.996019 and 1.082521

   Perigee Radius, Rp = Rp1 = r1 × (Rp / r1)1

      Rp = 6,628,140 × 0.996019
      Rp = 6,601,750 m

      Altitude @ perigee = 6,601,750 / 1,000 - 6,378.14 = 223.6 km 

   Apogee Radius, Ra = Rp2 = r1 × (Rp / r1)2

      Ra = 6,628,140 × 1.082521
      Ra = 7,175,100 m

      Altitude @ agogee = 7,175,100 / 1,000 - 6,378.14 = 797.0 km
 



PROBLEM 4.9 

Calculate the eccentricity of the orbit for the satellite in problem 4.8.


SOLUTION,

   Given:  r1 = 6,628,140 m
           v1 = 7,900 m/s
            = 89o

   Equation (4.27),

      e = SQRT[ (r1 × v12 / GM - 1)2 × sin2  + cos2  ]
      e = SQRT[ (6,628,140 × 7,9002 / 3.986005×1014 - 1)2 × sin2(89) + cos2(89) ]
      e = 0.0416170




PROBLEM 4.10 

Calculate the angle  from perigee point to launch point for the satellite
in problem 4.8.


SOLUTION,

   Given:  r1 = 6,628,140 m
           v1 = 7,900 m/s
            = 89o

   Equation (4.28),

      tan  = (r1 × v12 / GM) × sin  × cos  / [(r1 × v12 / GM) × sin2  - 1]
      tan  = (6,628,140 × 7,9002 / 3.986005×1014) × sin(89) × cos(89) 
        / [(6,628,140 × 7,9002 / 3.986005×1014) × sin2(89) - 1]
      tan  = 0.48329

       = arctan(0.48329) 
       = 25.794o




PROBLEM 4.11

Calculate the semi-major axis of the orbit for the satellite in problem 4.8.


SOLUTION,

   Given:  r1 = 6,628,140 m
           v1 = 7,900 m/s

   Equation (4.32),

      a = 1 / ( 2 / r1 - v12 / GM )
      a = 1 / ( 2 / 6,628,140 - 7,9002 / 3.986005×1014) )
      a = 6,888,430 m




PROBLEM 4.12

For the satellite in problem 4.8, burnout occurs 2000-10-20, 15:00 UT.  The
geocentric coordinates at burnout are 32o N latitude, 60o W longitude, and the
azimuth heading is 86o.  Calculate the orbit's inclination, argument of perigee,
and longitude of ascending node.


SOLUTION,

   Given:   = 86o
            = 32o
           2 = -60o

   From problem 4.10,

       = 25.794o

   Equation (4.33),

      cos(i) = cos() × sin()
      cos(i) = cos(32) × sin(86)
      i = 32.223o

   Equations (4.34) and (4.36),

      tan() = tan() / cos()
      tan() = tan(32) / cos(86)
       = 83.630o

       =  - 
       = 83.630 - 25.794
       = 57.836o

   Equations (4.35) and (4.37),

      tan() = sin() × tan()
      tan() = sin(32) × tan(86)
       = 82.483o

      1 = 2 - 
      1 = -60 - 82.483 
      1 = -142.483o

       = Sidereal time at -142.483 longitude, 2000-10-20, 15:00 UT 
       = 7h 27' 34" = 111.892o





PROBLEM 4.13

A satellite is in an orbit with a semi-major axis of 7,500 km and an eccentricity
of 0.1.  Calculate the time it takes to move from a position 30 degrees past
perigee to 90 degrees past perigee.

SOLUTION,

   Given:  a = 7,500 × 1,000 = 7,500,000 m
           e = 0.1
           tO = 0
           O = 30 deg × /180 = 0.52360 radians 
            = 90 deg × /180 = 1.57080 radians
 
   Equation (4.40),

      cos E = (e + cos ) / (1 + e cos )

      Eo = arccos[(0.1 + cos(0.52360)) / (1 + 0.1 × cos(0.52360))]
      Eo = 0.47557 radians

      E = arccos[(0.1 + cos(1.57080)) / (1 + 0.1 × cos(1.57080))]
      E = 1.47063 radians

   Equation (4.41),

      M = E - e × sin E

      Mo = 0.47557 - 0.1 × sin(0.47557)
      Mo = 0.42978 radians

      M = 1.47063 - 0.1 × sin(1.47063)
      M = 1.37113 radians

   Equation (4.39),

      n = SQRT[ GM / a3 ]
      n = SQRT[ 3.986005×1014 / 7,500,0003 ]
      n = 0.00097202 rad/s

   Equation (4.38),

      M - Mo = n × (t - tO)

      t = tO + (M - Mo) / n
      t = 0 + (1.37113 - 0.42978) / 0.00097202
      t = 968.4 s




PROBLEM 4.14

The satellite in problem 4.13 has a true anomaly of 90 degrees.  What will be the
satellite's position, i.e. it's true anomaly, 20 minutes later?

SOLUTION,

   Given:  a = 7,500,000 m
           e = 0.1
           tO = 0
           t = 20 × 60 = 1,200 s
           O = 90 × /180 = 1.57080 rad

   From problem 4.13,

      Mo = 1.37113 rad
      n = 0.00097202 rad/s

   Equation (4.38),

      M - Mo = n × (t - tO)

      M = Mo + n × (t - tO)
      M = 1.37113 + 0.00097202 × (1,200 - 0)
      M = 2.53755


 METHOD #1, Low Accuracy:

   Equation (4.42),

       ~ M + 2 × e × sin M + 1.25 × e2 × sin 2M
       ~ 2.53755 + 2 × 0.1 × sin(2.53755) + 1.25 × 0.12 × sin(2 × 2.53755)      
       ~ 2.63946 = 151.2 degrees


 METHOD #2, High Accuracy:

   Equation (4.41),

      M = E - e × sin E
      2.53755 = E - 0.1 × sin E

      By iteration, E = 2.58996 radians

   Equation (4.40),

      cos E = (e + cos ) / (1 + e cos )

      Rearranging variables gives,

      cos  = (cos E - e) / (1 - e cos E)

       = arccos[(cos(2.58996) - 0.1) / (1 - 0.1 × cos(2.58996)]
       = 2.64034 = 151.3 degrees




PROBLEM 4.15

For the satellite in problems 4.13 and 4.14, calculate the length of its position
vector, its flight-path angle, and its velocity when the satellite's true anomaly
is 225 degrees.

SOLUTION,

   Given:  a = 7,500,000 m
           e = 0.1
            = 225 degrees

   Equations (4.43) and (4.44),

      r = a × (1 - e2) / (1 + e × cos ) 
      r = 7,500,000 × (1 - 0.12) / (1 + 0.1 × cos(225))
      r = 7,989,977 m

       = arctan[ e × sin  / (1 + e × cos )]
       = arctan[ 0.1 × sin(225) / (1 + 0.1 × cos(225))]
       = -4.351 degrees 

   Equation (4.45),

      v = SQRT[ GM × (2 / r - 1 / a)]
      v = SQRT[ 3.986005×1014 × (2 / 7,989,977 - 1 / 7,500,000)]
      v = 6,828 m/s




PROBLEM 4.16

Calculate the perturbations in longitude of the ascending node and argument of 
perigee caused by the Moon and Sun for the International Space Station orbiting
at an altitude of 400 km, an inclination of 51.6 degrees, and with an orbital 
period of 92.6 minutes.

SOLUTION,

   Given:  i = 51.6 degrees
           n = 1436 / 92.6 = 15.5 revolutions/day

   Equations (4.46) through (4.49),

      Moon = -0.00338 × cos(i) / n 
      Moon = -0.00338 × cos(51.6) / 15.5 
      Moon = -0.000135 deg/day

      Sun = -0.00154 × cos(i) / n
      Sun = -0.00154 × cos(51.6) / 15.5 
      Sun = -0.0000617 deg/day

      Moon = 0.00169 × (4 - 5 × sin2 i) / n
      Moon = 0.00169 × (4 - 5 × sin2 51.6) / 15.5 
      Moon = 0.000101 deg/day

      Sun = 0.00077 × (4 - 5 × sin2 i) / n
      Sun = 0.00077 × (4 - 5 × sin2 51.6) / 15.5 
      Sun = 0.000046 deg/day




PROBLEM 4.17

A satellite is in an orbit with a semi-major axis of 7,500 km, an inclination 
of 28.5 degrees, and an eccentricity of 0.1.  Calculate the J2 perturbations in 
longitude of the ascending node and argument of perigee.

SOLUTION,

   Given:  a = 7,500 km
           i = 28.5 degrees
           e = 0.1

   Equations (4.50) and (4.51),

      J2 = -2.06474×1014 × a-7/2 × (cos i) × (1 - e2)-2
      J2 = -2.06474×1014 × (7,500)-7/2 × (cos 28.5) × (1 - (0.1)2)-2
      J2 = -5.067 deg/day

      J2 = 1.03237×1014 × a-7/2 × (4 - 5 × sin2 i) × (1 - e2)-2
      J2 = 1.03237×1014 × (7,500)-7/2 × (4 - 5 × sin2 28.5) × (1 - (0.1)2)-2
      J2 = 8.250 deg/day




PROBLEM 4.18 

A satellite is in a circular Earth orbit at an altitude of 400 km.  The satellite 
has a cylindrical shape 2 m in diameter by 4 m long and has a mass of 1,000 kg.
The satellite is traveling with its long axis perpendicular to the velocity
vector and it's drag coefficient is 2.67.  Calculate the perturbations due to
atmospheric drag and estimate the satellite's lifetime.

SOLUTION,

   Given:  a = (6,378.14 + 400) × 1,000 = 6,778,140 m
           A = 2 × 4 = 8 m2
           m = 1,000 kg
           CD = 2.67

   From Atmosphere Properties,

       = 2.62×10-12 kg/m3
      H = 58.2 km

   Equation (4.6),

      V = SQRT[ GM / a ]
      V = SQRT[ 3.986005×1014 / 6,778,140 ]
      V = 7,669 m/s

   Equations (4.53) through (4.55),

      arev = (-2 ×  × CD × A ×  × a2) / m
      arev = (-2 ×  × 2.67 × 8 × 2.62×10-12 × 6,778,1402) / 1,000
      arev = -16.2 m 

      Prev = (-6 × 2 × CD × A ×  × a2) / (m × V)
      Prev = (-6 × 2 × 2.67 × 8 × 2.62×10-12 × 6,778,1402) / (1,000 × 7,669)
      Prev = -0.0199 s 

      Vrev = ( × CD × A ×  × a × V) / m
      Vrev = ( × 2.67 × 8 × 2.62×10-12 × 6,778,140 × 7,669) / 1,000
      Vrev = 0.00914 m/s 

   Equation (4.56),

      L ~ -H / arev 
      L ~ -(58.2 × 1,000) / -16.2
      L ~ 3,600 revolutions





PROBLEM 4.19 

A spacecraft is in a circular parking orbit with an altitude of 200 km.
Calculate the velocity change required to perform a Hohmann transfer to a
circular orbit at geosynchronous altitude.

SOLUTION,

   Given:  rA = (6,378.14 + 200) × 1,000 = 6,578,140 m

   From problem 4.3,

      rB = 42,164,170 m

   Equations (4.58) through (4.65),

      atx = (rA + rB) / 2
      atx = (6,578,140 + 42,164,170) / 2
      atx = 24,371,155 m

      ViA = SQRT[ GM / rA ]
      ViA = SQRT[ 3.986005×1014 / 6,578,140 ]
      ViA = 7,784 m/s

      VfB = SQRT[ GM / rB ]
      VfB = SQRT[ 3.986005×1014 / 42,164,170 ]
      VfB = 3,075 m/s

      VtxA = SQRT[ GM × (2 / rA - 1 / atx)]
      VtxA = SQRT[ 3.986005×1014 × (2 / 6,578,140 - 1 / 24,371,155)]
      VtxA = 10,239 m/s

      VtxB = SQRT[ GM × (2 / rB - 1 / atx)]
      VtxB = SQRT[ 3.986005×1014 × (2 / 42,164,170 - 1 / 24,371,155)]
      VtxB = 1,597 m/s 

      VA = VtxA - ViA
      VA = 10,239 - 7,784
      VA = 2,455 m/s

      VB = VfB - VtxB
      VB = 3,075 - 1,597
      VB = 1,478 m/s

      VT = VA + VB
      VT = 2,455 + 1,478
      VT = 3,933 m/s




PROBLEM 4.20 

A satellite is in a circular parking orbit with an altitude of 200 km.  Using
a one-tangent burn, it is to be transferred to geosynchronous altitude using a
transfer ellipse with a semi-major axis of 30,000 km.  Calculate the total
required velocity change and the time required to complete the transfer. 


SOLUTION,

   Given:  rA = (6,378.14 + 200) × 1,000 = 6,578,140 m
           rB = 42,164,170 m
           atx = 30,000 × 1,000 = 30,000,000 m

   Equations (4.66) through (4.68),

      e = 1 - rA / atx
      e = 1 - 6,578,140 / 30,000,000
      e = 0.780729

       = arccos[(atx × (1 - e2) / rB - 1) / e ]  
       = arccos[(30,000,000 × (1 - 0.7807292) / 42,164,170 - 1) / 0.780729 ]
       = 157.670 degrees 

       = arctan[ e × sin  / (1 + e × cos )]
       = arctan[ 0.780729 × sin(157.670) / (1 + 0.780729 × cos(157.670))]
       = 46.876 degrees

   Equations (4.59) through (4.63),

      ViA = SQRT[ GM / rA ]
      ViA = SQRT[ 3.986005×1014 / 6,578,140 ]
      ViA = 7,784 m/s

      VfB = SQRT[ GM / rB ]
      VfB = SQRT[ 3.986005×1014 / 42,164,170 ]
      VfB = 3,075 m/s

      VtxA = SQRT[ GM × (2 / rA - 1 / atx)]
      VtxA = SQRT[ 3.986005×1014 × (2 / 6,578,140 - 1 / 30,000,000)]
      VtxA = 10,388 m/s

      VtxB = SQRT[ GM × (2 / rB - 1 / atx)]
      VtxB = SQRT[ 3.986005×1014 × (2 / 42,164,170 - 1 / 30,000,000)]
      VtxB = 2,371 m/s 

      VA = VtxA - ViA
      VA = 10,388 - 7,784
      VA = 2,604 m/s

   Equation (4.69),

      VB = SQRT[ VtxB2 + VfB2 - 2 × VtxB × VfB × cos  ]
      VB = SQRT[ 2,3712 + 3,0752 - 2 × 2,371 × 3,075 × cos(46.876)]
      VB = 2,260 m/s

   Equation (4.65),

      VT = VA + VB
      VT = 2,604 + 2,260
      VT = 4,864 m/s

   Equations (4.70) and (4.71),

      E = arccos[(e + cos ) / (1 + e cos )]
      E = arccos[(0.780729 + cos(157.670)) / (1 + 0.780729 × cos(157.670))]
      E = 2.11688 radians 

      TOF = (E - e × sin E) × SQRT[ atx3 / GM ]
      TOF = (2.11688 - 0.780729 × sin(2.11688)) × SQRT[ 30,000,0003 / 3.986005×1014 ] 
      TOF = 11,931 s = 3.314 hours




PROBLEM 4.21

Calculate the velocity change required to transfer a satellite from a circular
600 km orbit with an inclination of 28 degrees to an orbit of equal size with
an inclination of 20 degrees.

SOLUTION,

   Given:  r = (6,378.14 + 600) × 1,000 = 6,978,140 m
            = 28 - 20 = 8 degrees

   Equation (4.6),

      Vi = SQRT[ GM / r ]
      Vi = SQRT[ 3.986005×1014 / 6,978,140 ]
      Vi = 7,558 m/s

   Equation (4.73),

      V = 2 × Vi × sin(/2)
      V = 2 × 7,558 × sin(8/2)
      V = 1,054 m/s




PROBLEM 4.22 

A satellite is in a parking orbit with an altitude of 200 km and an inclination
of 28 degrees.  Calculate the total velocity change required to transfer the
satellite to a zero-inclination geosynchronous orbit using a Hohmann transfer
with a combined plane change at apogee.

   Given:  rA = (6,378.14 + 200) × 1,000 = 6,578,140 m
           rB = 42,164,170 m
            = 28 degrees

   From problem 4.19,

      VfB = 3,075 m/s
      VtxB = 1,597 m/s 
      VA = 2,455 m/s

   Equation (4.74),

      VB = SQRT[ VtxB2 + VfB2 - 2 × VtxB × VfB × cos  ]
      VB = SQRT[ 1,5972 + 3,0752 - 2 × 1,597 × 3,075 × cos(28)]
      VB = 1,826 m/s

   Equation (4.65),

      VT = VA + VB
      VT = 2,455 + 1,826
      VT = 4,281 m/s




PROBLEM 4.23

A spacecraft is in an orbit with an inclination of 30 degrees and the longitude
of the ascending node is 75 degrees.  Calculate the angle change required to
change the inclination to 32 degrees and the longitude of the ascending node to
80 degrees.  

SOLUTION,

   Given:  ii = 30 degrees
           i = 75 degrees
           if = 32 degrees
           f = 80 degrees

   Equation (4.75),

      a1 = sin(ii)cos(i) = sin(30)cos(75) = 0.129410

      a2 = sin(ii)sin(i) = sin(30)sin(75) = 0.482963
 
      a3 = cos(ii) = cos(30) = 0.866025

      b1 = sin(if)cos(f) = sin(32)cos(80) = 0.0920195

      b2 = sin(if)sin(f) = sin(32)sin(80) = 0.521869
 
      b3 = cos(if) = cos(32) = 0.848048

       = arccos(a1 × b1 + a2 × b2 + a3 × b3)
       = arccos(0.129410 × 0.0920195 + 0.482963 × 0.521869 + 0.866025 × 0.848048)
       = 3.259 degrees




PROBLEM 4.24

Calculate the latitude and longitude of the intersection nodes between the
initial and final orbits for the spacecraft in problem 4.23.  

SOLUTION,

   From problem 4.21,

           a1 = 0.129410
           a2 = 0.482963
           a3 = 0.866025
           b1 = 0.0920195
           b2 = 0.521869
           b3 = 0.848048

   Equations (4.76) and (4.77),

      c1 = a2 × b3 - a3 × b2 = 0.482963 × 0.848048 - 0.866025 × 0.521869 = -0.0423757

      c2 = a3 × b1 - a1 × b3 = 0.866025 × 0.0920195 - 0.129410 × 0.848048 = -0.0300543
 
      c3 = a1 × b2 - a2 × b1 = 0.129410 × 0.521869 - 0.482963 × 0.0920195 = 0.0230928

      lat1 = arctan(c3 / (c12 + c22)1/2)
      lat1 = arctan(0.0230928 / (-0.04237572 + -0.03005432)1/2)
      lat1 = 23.965 degrees

      long1 = arctan(c2 / c1) + 90
      long1 = arctan(-0.0300543 / -0.0423757) + 90
      long1 = 125.346 degrees

      lat2 = -23.965 degrees
      long2 = 125.346 + 180 = 305.346 degrees




PROBLEM 4.25

Calculate the escape velocity of a spacecraft launched from an Earth orbit with
an altitude of 200 km.

SOLUTION,

   Given:  r = (6,378.14 + 200) × 1,000 = 6,578,140 m 

   Equation (4.78),

      Vesc = SQRT[ 2 × GM / r ]
      Vesc = SQRT[ 2 × 3.986005×1014 / 6,578,140 ]
      Vesc = 11,009 m/s




PROBLEM 4.26

A space probe is approaching Mars on a hyperbolic flyby trajectory.  When at
a distance of 100,000 km, its velocity relative to Mars is 5,140.0 m/s and
its flight path angle is -85.300 degrees.  Calculate the probe's eccentricity,
semi-major axis, turning angle, angle , true anomaly, impact parameter,
periapsis radius, and parameter p.

SOLUTION,

   From Basics Constants,

      GM of Mars = 4.282831×1013 m3/s2

   Given:  r = 100,000 × 1,000 = 100,000,000 m
           v = 5,140.0 m/s
            = -85.300o

   Equations (4.30) and (4.32),

      e = SQRT[ (r × v2 / GM - 1)2 × cos2  + sin2  ]
      e = SQRT[ (100,000,000 × 5,1402 / 4.282831×1013 - 1)2 × cos2(-85.3) + sin2(-85.3) ]
      e = 5.0715

      a = 1 / ( 2 / r - v2 / GM )
      a = 1 / ( 2 / 100,000,000 - 5,1402 / 4.282831×1013 )
      a = -1,675,400 m

   Equations (4.80) through (4.85),

      sin(/2) = 1 / e
       = 2 × arcsin( 1 / 5.0715 )
       = 22.744o

      cos  = -1 / e
       = arccos( -1 / 5.0715 )
       = 101.37o

       = arccos[ (a × (1 - e2) - r) / (e × r) ]
       = arccos[ (-1,675,400 × (1 - 5.07152) - 100,000,000) / (5.0715 × 100,000,000) ]
       = -96.633o

      b = -a / tan(/2)
      b = 1,675.4 / tan(22.744/2)
      b = 8,330.0 km

      ro = a × (1 - e)
      ro = -1,675.4 × (1 - 5.0715)
      ro = 6,821.4 km

      p = a × (1 - e2)
      p = -1,675.4 × (1 - 5.07152)
      p = 41,416 km





PROBLEM 4.27

The space probe in problem 4.26 has moved to a true anomaly of 75 degrees.
Calculate the radius vector, flight path angle, and velocity.

SOLUTION,

   Given:  a = -1,675,400 m
           e = 5.0715
            = 75o

   Equations (4.43) through (4.45),

      r = a × (1 - e2) / (1 + e × cos ) 
      r = -1,675,400 × (1 - 5.07152) / (1 + 5.0715 × cos(75))
      r = 17,909,000 m

       = arctan[ e × sin  / (1 + e × cos )]
       = arctan[ 5.0715 × sin(75) / (1 + 5.0715 × cos(75))]
       = 64.729o 

      v = SQRT[ GM × (2 / r - 1 / a)]
      v = SQRT[ 4.282831×1013 × (2 / 17,909,000 - 1 / -1,675,400)]
      v = 5,508.7 m/s




PROBLEM 4.28

A spacecraft is launched from Earth on a hyperbolic trajectory with a semi-major
axis of -36,000 km and an eccentricity of 1.1823. How long does it take to move
from a true anomaly of 15 degrees to a true anomaly of 120 degrees?

SOLUTION,

   Given:  a = -36,000 × 1,000 = -36,000,000 m
           e = 1.1823
           O = 15o
            = 120o

   Equation (4.87),

      cosh F = (e + cos ) / (1 + e cos )

      Fo = arccosh[(1.1823 + cos(15)) / (1 + 1.1823 × cos(15))]
      Fo = 0.07614

      F = arccosh[(1.1823 + cos(120)) / (1 + 1.1823 × cos(120))]
      F = 1.10023

   Equation (4.86),

      t - tO = SQRT[(-a)3 / GM ] × [(e × sinh F - F) - (e × sinh Fo - Fo)]
      t - tO = SQRT[(36,000,000)3 / 3.986005×1014 ] × [(1.1823 × sinh(1.10023) - 1.10023) 
               - (1.1823 × sinh(0.07614) - 0.07614)]
      t - tO = 5,035 s = 1.399 hours




PROBLEM 4.29

A spacecraft launched from Earth has a burnout velocity of 11,500 m/s at an
altitude of 200 km.  What is the hyperbolic excess velocity?

SOLUTION,

   Given:  Vbo = 11,500 m/s

   From problem 4.25,

      Vesc = 11,009 m/s

   Equation (4.88),

      V2 = Vbo2 - Vesc2
      V = SQRT[ 11,5002 - 11,0092 ]
      V = 3,325 m/s




PROBLEM 4.30

Calculate the radius of Earth's sphere of influence.

SOLUTION,

   From Basics Constants,

      Dsp = 149,597,870 km
      MP = 5.9737×1024 kg
      MS = 1.9891×1030 kg

   Equation (4.89),

      REarth = Dsp × (MP / MS)0.4
      REarth = 149,597,870 × (5.9737×1024 / 1.9891×1030)0.4
      REarth = 925,000 km





PROBLEM 5.1 

Using a one-tangent burn, calculate the change in true anomaly and the
time-of-flight for a transfer from Earth to Mars.  The radius vector of Earth at
departure is 1.000 AU and that of Mars at arrival is 1.524 AU.  The semi-major
axis of the transfer orbit is 1.300 AU.

SOLUTION,

   Given:  rA = 1.000 AU
           rB = 1.524 AU
           atx = 1.300 AU × 149.597870×109 m/AU = 194.48×109 m

   From Basics Constants,

      GM of Sun = 1.327124×1020 m3/s2

   Equations (4.66) and (4.67),

      e = 1 - rA / atx
      e = 1 - 1.0 / 1.3
      e = 0.230769
 
       = arccos[(atx × (1 - e2) / rB - 1) / e ]  
       = arccos[(1.3 × (1 - 0.2307692) / 1.524 - 1) / 0.230769 ]
       = 146.488 degrees 

   Equations (4.70) and (4.71),

      E = arccos[(e + cos ) / (1 + e cos )]
      E = arccos[(0.230769 + cos(146.488)) / (1 + 0.230769 × cos(146.488))]
      E = 2.41383 radians 

      TOF = (E - e × sin E) × SQRT[ atx3 / GM ]
      TOF = (2.41383 - 0.230769 × sin(2.41383)) × SQRT[ (194.48×109)3 / 1.327124×1020 ] 
      TOF = 16,827,800 s = 194.77 days





PROBLEM 5.2 

For the transfer orbit in problem 5.1, calculate the departure phase angle, given
that the angular velocity of Mars is 0.5240 degrees/day.

SOLUTION,

   Given:  2-1 = 146.488o
           t2-t1 = 194.77 days
           t = 0.5240o/day

   Equation (5.1),

       = (2-1) - t × (t2-t1)
       = 146.488 - 0.5240 × 194.77
       = 44.43o




PROBLEM 5.3 

A flight to Mars is launched on 2020-7-20, 0:00 UT. The planned time of flight
is 207 days.  Earth's postion vector at departure is 0.473265X - 0.899215Y AU.
Mars' postion vector at intercept is 0.066842X + 1.561256Y + 0.030948Z AU.
Calculate the parameter and semi-major axis of the transfer orbit.

SOLUTION,

   Given:  t = 207 days
           r1 = 0.473265X - 0.899215Y AU
           r2 = 0.066842X + 1.561256Y + 0.030948Z AU
           GM = 1.327124×1020 m3/s2
              = 1.327124×1020 / (149.597870×109)3 = 3.964016×10-14 AU3/s2

   From vector magnitude,

      r1 = SQRT[ 0.4732652 + (-0.899215)2 ]
      r1 = 1.016153 AU

      r2 = SQRT[ 0.0668422 + 1.5612562 + 0.0309482 ]
      r2 = 1.562993 AU

   From vector dot product,

       = arccos[ (0.473265 × 0.066842 - 0.899215 × 1.561256) / (1.016153 × 1.562993) ]       
       = 149.770967o

   Equations (5.9), (5.10) and (5.11),

      k = r1 × r2 × (1 - cos )
      k = 1.016153 × 1.562993 × (1 - cos(149.770967))
      k = 2.960511 AU

       = r1 + r2
       = 1.016153 + 1.562993
       = 2.579146 AU

      m = r1 × r2 × (1 + cos )
      m = 1.016153 × 1.562993 × (1 + cos(149.770967))
      m = 0.215969 AU

   Equations (5.18) and (5.19),

      pi = k / ( + SQRT(2 × m))
      pi = 2.960511 / (2.579146 + SQRT(2 × 0.215969))
      pi = 0.914764 AU

      pii = k / ( - SQRT(2 × m))
      pii = 2.960511 / (2.579146 - SQRT(2 × 0.215969))
      pii = 1.540388 AU

      Since  < , 0.914764 < p < 

   Equation (5.12),

      Select trial value, p = 1.2 AU

      a = m × k × p / [(2 × m - 2) × p2 + 2 × k ×  × p - k2]
      a = 0.215969 × 2.960511 × 1.2 
          / [(2 × 0.215969 - 2.5791462) × 1.22 + 2 × 2.960511 × 2.579146 × 1.2 - 2.9605112]
      a = 1.270478 AU

   Equations (5.5), (5.6) and (5.7),

      f = 1 - r2 / p × (1 - cos )
      f = 1 - 1.562993 / 1.2 × (1 - cos(149.770967))
      f = -1.427875

      g = r1 × r2 × sin  / SQRT[ GM × p ]
      g = 1.016153 × 1.562993 × sin(149.770967) / SQRT[ 3.964016×10-14 × 1.2 ]
      g = 3,666,240

       = SQRT[ GM / p ] × tan(/2) × [(1 - cos ) / p - 1/r1 - 1/r2 ]
       = SQRT[ 3.964016×10-14 / 1.2 ] × tan(149.770967/2) 
          × [(1 - cos(149.770967)) / 1.2 - 1/1.016153 - 1/1.562993 ]
       = -4.747601×10-8

   Equation (5.13),

      E = arccos[ 1 - r1 / a × (1 - f) ]
      E = arccos[ 1 - 1.016153 / 1.270478 × (1 + 1.427875) ]
      E = 2.798925 radians

   Equation (5.16),

      t = g + SQRT[ a3 /  GM ] × (E - sin E)
      t = 3,666,240 + SQRT[ 1.2704783 /  3.964016×10-14 ] × (2.798925 - sin(2.798925))
      t = 21,380,951 s = 247.4647 days

   Select new trial value of p and repeat above steps,

      p = 1.300000 AU,  a = 1.443005 AU,   t = 178.9588 days

    Equation (5.20),

      pn+1 = pn + (t - tn) × (pn - pn-1) / (tn - tn-1)
      pn+1 = 1.3 + (207 - 178.9588) × (1.3 - 1.2) / (178.9588 - 247.4647)
      pn+1 = 1.259067 AU

   Recalculate using new value of p,

      p = 1.259067 AU,  a = 1.336197 AU,   t = 201.5624 days

   Perform additional iterations,

      p = 1.249221 AU,  a = 1.318624 AU,   t = 207.9408 days
      p = 1.250673 AU,  a = 1.321039 AU,   t = 206.9733 days
      p = 1.250633 AU,  a = 1.320971 AU,   t = 206.9999 days <-- close enough





PROBLEM 5.4 

For the Mars transfer orbit in Problem 5.3, calculate the departure and intecept
velocity vectors.

SOLUTION,

   Given:  r1 = 0.473265X - 0.899215Y AU
           r2 = 0.066842X + 1.561256Y + 0.030948Z AU
           r1 = 1.016153 AU
           r2 = 1.562993 AU
           p = 1.250633 AU
           a = 1.320971 AU
            = 149.770967o

   Equations (5.5), (5.6) and (5.7),

      f = 1 - r2 / p × (1 - cos )
      f = 1 - 1.562993 / 1.250633 × (1 - cos(149.770967))
      f = -1.329580

      g = r1 × r2 × sin  / SQRT[ GM × p ]
      g = 1.016153 × 1.562993 × sin(149.770967) / SQRT[ 3.964016×10-14 × 1.250633 ]
      g = 3,591,258

       = SQRT[ GM / p ] × tan(/2) × [(1 - cos ) / p - 1/r1 - 1/r2 ]
       = SQRT[ 3.964016×10-14 / 1.250633 ] × tan(149.770967/2) 
          × [(1 - cos(149.770967)) / 1.250633 - 1/1.016153 - 1/1.562993 ]
       = -8.795872×10-8

       = 1 - r1 / p × (1 - cos )
       = 1 - 1.016153 / 1.250633 × (1 - cos(149.770967))
       = -0.514536

   Equation (5.3),

      v1 = (r2 - f × r1)  / g

      v1 = [(0.066842 + 1.329580 × 0.473265) / 3,591,258] X
           + [(1.561256 + 1.329580 × -0.899215) / 3,591,258] Y
           + [(0.030948 + 1.329580 × 0) / 3,591,258] Z

      v1 = 0.000000193828X + 0.000000101824Y + 0.00000000861759Z AU/s × 149.597870×109
      v1 = 28996.2X + 15232.7Y + 1289.2Z m/s

   Equation (5.4),

      v2 =  × r1 +  × v1

      v2 = [-8.795872×10-8 × 0.473265 - 0.514536 × 0.000000193828] X
           + [-8.795872×10-8 × -0.899215 - 0.514536 × 0.000000101824] Y
           + [-8.795872×10-8 × 0 - 0.514536 × 0.00000000861759] Z

      v2 = -0.000000141359X + 0.0000000267017Y - 0.00000000443406Z AU/s × 149.597870×109
      v2 = -21147.0X + 3994.5Y - 663.3Z m/s




PROBLEM 5.5 

For the Mars transfer orbit in Problems 5.3 and 5.4, calculate the orbital elements.

SOLUTION,

   Problem can be solved using either r1 & v1 or r2 & v2 – we will use r1 & v1.

   Given:  r1 = (0.473265X - 0.899215Y AU) × 149.597870×109 m/AU
              = 7.079944×1010X - 1.345206×1011Y m
           r1 = 1.016153 × 149.597870×109 = 1.520144×1011 m
           GM = 1.327124×1020 m3/s2

   From problem 5.4,

      v1 = 28996.2X + 15232.7Y + 1289.2Z m/s

      Also,

      v = SQRT[ 28996.22 + 15232.72 + 1289.22 ] = 32,779.2 m/s

   Equations (5.21) and (5.22),

      h = (rY vZ - rZ vY)X + (rZ vX - rX vZ)Y + (rX vY - rY vX)Z
      h = (-1.345206×1011 × 1289.2 - 0 × 15232.7)X + (0 × 28996.2 - 7.079944×1010 × 1289.2)Y 
          + (7.079944×1010 × 15232.7 + 1.345206×1011 × 28996.2)Z
      h = -1.73424×1014X - 9.12746×1013Y + 4.97905×1015Z

      n = -hY X + hX Y
      n = 9.12746×1013X - 1.73424×1014Y

      Also,

      h = SQRT[ (-1.73424×1014)2 + (-9.12746×1013)2 + (4.97905×1015)2 ] = 4.98291×1015
      n = SQRT[ (9.12746×1013)2 + (1.73424×1014)2 ] = 1.95977×1014

   Equation (5.23),

      e = [(v2 - GM / r) × r - (r • v) × v ] / GM

        v2 - GM / r = 32779.22 - 1.327124×1020 / 1.520144×1011 = 2.01451×108

        r • v = 7.079944×1010 × 28996.2 - 1.345206×1011 × 15232.7 + 0 x 1289.2 = 3.80278×1012

      e = [2.01451×108 × (7.079944×1010X - 1.345206×1011Y)
          - 3.80278×1012 × (28996.2X + 15232.7Y + 1289.2Z) ] / 1.327124×1020
      e = 0.106639X - 0.204632Y - 0.000037Z

   Equations (5.24) and (5.25),

      a = 1 / ( 2 / r - v2 / GM )
      a = 1 / ( 2 / 1.520144×1011 - 32779.22 / 1.327124×1020 )
      a = 1.97614×1011 m

      e = SQRT[ 0.1066392 + (-0.204632)2 + (-0.000037)2 ]
      e = 0.230751

   Equations (5.26) though (5.30),

      cos i = hZ / h
      cos i = 4.97905×1015 / 4.98291×1015
      i = 2.255o

      cos  = nX / n
      cos  = 9.12746×1013 / 1.95977×1014
       = 297.76o

      cos  = n • e / (n × e)
      cos  = (9.12746×1013 × 0.106639 - 1.73424×1014 × (-0.204632) + 0 × (-0.000037))
               / (1.95977×1014 × 0.230751)
       = 359.77o

      cos o = e • r / (e × r)
      cos o = (0.106639 × 7.079944×1010 - 0.204632 × (-1.345206×1011) - 0.000037 × 0)
               / (0.230751 × 1.520144×1011)
      o = 0.226o

      cos uo = n • r / (n × r)
      uo = 0 (launch point = ascending node)

   Equations (5.31) and (5.32),

       =  + 
       = 297.76 + 359.77
       = 297.53o

      o =  +  + o
      o = 297.76 + 359.77 + 0.23
      o = 297.76o




PROBLEM 5.6 

For the spacecraft in Problems 5.3 and 5.4, calculate the hyperbolic excess 
velocity at departure, the injection V, and the zenith angle of the departure
asymptote.  Injection occurs from an 200 km parking orbit.  Earth's velocity
vector at departure is 25876.6X + 13759.5Y m/s.

SOLUTION,

   Given:  ro = (6,378.14 + 200) × 1,000 = 6,578,140 m
           r = 0.473265X - 0.899215Y AU
           VP = 25876.6X + 13759.5Y m/s

   From problem 5.4,

      VS = 28996.2X + 15232.7Y + 1289.2Z m/s

   Equation (5.33),

      VS/P = (VSX - VPX)X + (VSY - VPY)Y + (VSZ - VPZ)Z
      VS/P = (28996.2 - 25876.6)X + (15232.7 - 13759.5)Y + (1289.2 - 0)Z
      VS/P = 3119.6X + 1473.2Y + 1289.2Z m/s

   Equation (5.34),

      VS/P = SQRT[ VS/PX2 + VS/PY2 + VS/PZ2 ]
      VS/P = SQRT[ 3119.62 + 1473.22 + 1289.22 ]
      VS/P = 3,683.0 m/s

      V = VS/P = 3,683.0 m/s

   Equations (5.35) and (5.36),

      Vo = SQRT[ V2 + 2 × GM / ro ]
      Vo = SQRT[ 3,683.02 + 2 × 3.986005×1014 / 6,578,140 ]
      Vo = 11,608.4 m/s

      V = Vo - SQRT[ GM / ro ]
      V = 11,608.4 - SQRT[ 3.986005×1014 / 6,578,140 ]
      V = 3,824.1 m/s

   Equation (5.37),

      r = SQRT[ 0.4732652 + (-0.899215)2 + 02 ]
      r = 1.01615 AU

       = arccos[(rX × vX + rY × vY + rZ × vZ) / (r × v)]
       = arccos[ 0.473265 × 3119.6 - 0.899215 × 1473.2 + 0 × 1289.2) / (1.01615 × 3683.0)]
       = 87.677o




PROBLEM 5.7 

For the spacecraft in Problems 5.3 and 5.4, given a miss distance of +18,500 km
at arrival, calculate the hyperbolic excess velocity, impact parameter, and
semi-major axis and eccentricity of the hyperbolic approach trajectory.  Mars'
velocity vector at intercept is -23307.8X + 3112.0Y + 41.8Z m/s.

SOLUTION,

   Given:  d = 18,500 km / 149.597870×106 = 0.000123664 AU
           r = 0.066842X + 1.561256Y + 0.030948Z AU
           VP = -23307.8X + 3112.0Y + 41.8Z m/s
           
   From Basics Constants,

      GM of Mars = 4.282831×1013 m3/s2

   From problem 5.4,

      VS = -21147.0X + 3994.5Y - 663.3Z m/s

   Equation (5.33),

      VS/P = (VSX - VPX)X + (VSY - VPY)Y + (VSZ - VPZ)Z
      VS/P = (-21147.0 + 23307.8)X + (3994.5 - 3112.0)Y + (-663.3 - 41.8)Z
      VS/P = 2160.8X + 882.5Y - 705.1Z m/s

   Equation (5.34),

      VS/P = SQRT[ VS/PX2 + VS/PY2 + VS/PZ2 ]
      VS/P = SQRT[ 2160.82 + 882.52 + (-705.1)2 ]
      VS/P = 2,438.2 m/s

      V = VS/P = 2,438.2 m/s

   Equations (5.38.A) and (5.38.B),

      dx = -d × ry / SQRT[ rx2 + ry2 ]
      dx = -0.000123664 × 1.561256 / SQRT[ 0.0668422 + 1.5612562 ]
      dx = -0.000123551 AU 

      dy = d × rx / SQRT[ rx2 + ry2 ]
      dy = 0.000123664 × 0.066842 / SQRT[ 0.0668422 + 1.5612562 ]
      dy = 0.0000052896 AU 

   Equation (5.39),

       = arccos[(dx × vx + dy × vy) / (d × v)]
       = arccos[(-0.000123551 × 2160.8 + 0.0000052896 × 882.5) / (0.000123664 × 2,438.2)]
       = 150.451o

   Equations (5.40) through (5.42),

      b = d × sin 
      b = 18,500 × sin(150.451)
      b = 9,123.6 km

      a = -GM / V2
      a = -4.282831×1013 / 2,438.22
      a = -7.2043×106 m = -7,204.3 km

      e = SQRT[ 1 + b2 / a2 ]
      e = SQRT[ 1 + 9,123.62 / -7,204.32 ]
      e = 1.6136





PROBLEM 5.8 

As a spacecraft approaches Jupiter, it has a velocity of 9,470 m/s, a flight
path angle of 39.2 degrees, and a targeted miss distance of -2,500,000 km. At
intercept, Jupiter's velocity is 12,740 m/s with a flight path angle of 2.40
degrees.  Calculate the spacecraft's velocity and flight path angle following
its swing-by of Jupiter.

   Given:  VP = 12,740 m/s
           P = 2.40o
           VSi = 9,470 m/s
           Si = 39.2o
           d = -2,500,000 km
           
   From Basics Constants,

      GM of Jupiter = 1.26686×1017 m3/s2

   Equations (5.44) and (5.45),

      VP = (VP × cos P)X + (VP × sin P)Y
      VP = (12740 × cos(2.40))X + (12740 × sin(2.40))Y
      VP = 12729X + 533Y m/s

      VSi = (VSi × cos Si)X + (VSi × sin Si)Y
      VSi = (9470 × cos(39.2))X + (9470 × sin(39.2))Y
      VSi = 7339X + 5985Y m/s

   Equations (5.46) and (5.47),

      VS/Pi = ((VSi)X - VPX)X + ((VSi)Y - VPY)Y
      VS/Pi = (7339 - 12729)X + (5985 - 533)Y
      VS/Pi = -5390X + 5452Y m/s

      VS/P = SQRT[ (VS/Pi)X2 + (VS/Pi)Y2 ]
      VS/P = SQRT[ (-5390)2 + 54522 ]
      VS/P = 7,667 m/s

      V = VS/P = 7,667 m/s

   Equation (5.48),

      i = arctan[ (VS/Pi)Y / (VS/Pi)X ]
      i = arctan[ 5452 / -5390 ]
      i = 134.67o

   Equations (5.40) through (5.42),

      b = d × sin 
      b = -2,500,000 × sin(134.67)
      b = -1,777,900 km 

      a = -GM / V2
      a = -1.26686×1017 / 76672
      a = -2.1552×109 m = -2,155,200 km

      e = SQRT[ 1 + b2 / a2 ]
      e = SQRT[ 1 + (-1,777,900)2 / (-2,155,200)2 ]
      e = 1.2963

   Equation (4.80),

      sin(/2) = 1 / e
       = 2 × arcsin( 1 / 1.2963 )
       = 100.96o

   Equation (5.49),

      f = i - 
      f = 134.67 - 100.96
      f = 33.71o

   Equation (5.50),

      VS/Pf = (VS/P × cos f)X + (VS/P × sin f)Y
      VS/Pf = (7667 × cos(33.71))X + (7667 × sin(33.71))Y
      VS/Pf = 6378X + 4255Y m/s

   Equations (5.51) and (5.52),

      VSf = ((VS/Pf)X + VPX)X + ((VS/Pf)Y + VPY)Y
      VSf = (6378 + 12729)X + (4255 + 533)Y
      VSf = 19107X + 4788Y m/s

      VSf = SQRT[ (VSf)X2 + (VSf)Y2 ]
      VSf = SQRT[ 191072 + 47882 ]
      VSf = 19,698 m/s

   Equation (5.53),

      Sf = arctan[ (VSf)Y / (VSf)X ] 
      Sf = arctan[ 4788 / 19107 ] 
      Sf = 14.07o 

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Penulis : Drs.Simon Aenold Julian Jacob