EXAMPLE PROBLEMS |

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 m^{2}. What is the thrust of the engine in a vacuum? SOLUTION, Given: q = 30 kg/s Ve = 3,100 m/s Ae = 0.7 m^{2}Pe = 5 kPa = 5,000 N/m^{2}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 = 3,470K M = 21.40 k = 1.221 q = 500 kg/s Equation (1.27), Pt = Pc × [1 + (k - 1) / 2] 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 m^{-k/(k-1)}Pt = 5.066 × [1 + (1.221 - 1) / 2]^{-1.221/(1.221-1)}Pt = 2.839 MPa = 2.839×10^{6}N/m^{2}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 / 2.839×10^{6}) × SQRT[ (8,314.46 × 3,125) / (21.40 × 1.221) ] At = 0.1756 m^{2}

^{2}k = 1.221 From Atmosphere Properties, Pa = 0.0795 MPa Equation (1.29), Nm^{2}= (2 / (k - 1)) × [(Pc / Pa)^{(k-1)/k}- 1] Nm^{2}= (2 / (1.221 - 1)) × [(5.066 / 0.0795)^{(1.221-1)/1.221}- 1] Nm^{2}= 10.15 Nm = (10.15)^{1/2}= 3.185 Equation (1.30), Ae = (At / Nm) × [(1 + (k - 1) / 2 × Nm^{2})/((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 m^{2}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 m^{2}= 1,756 cm^{2}Dt = 2 × (1,756/)^{1/2}= 47.3 cm = 20^{o}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 cm^{3}From Figure 1.7, Lc = 66 cm (second-order approximation) Equation (1.35), Dc = SQRT[(Dt^{3}+ 24/ × tan × Vc) / (Dc + 6 × tan × Lc)] Dc = SQRT[(47.3^{3}+ 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 m^{2}Equation (1.36), r = a × Pc^{n}r = 5.5 × 5.0^{0.4}= 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: w_{AP}= 0.68 w_{Al}= 0.18 w_{HTPB}= 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: Mo_{1}= 120,000 + 9,000 + 30,000 + 3,000 + 3,000 = 165,000 kg Mf_{1}= 9,000 + 30,000 + 3,000 + 3,000 = 45,000 kg Isp_{1}= 260 s Mo_{2}= 30,000 + 3,000 + 3,000 = 36,000 kg Mf_{2}= 3,000 + 3,000 = 6,000 kg Isp_{2}= 320 s Equation (1.24), C_{1}= Isp_{1}g C_{1}= 260 × 9.80665 = 2,550 m/s C_{2}= Isp_{2}g C_{2}= 320 × 9.80665 = 3,138 m/s Equation (1.39), V_{1}= C_{1}× LN[ Mo_{1}/ Mf_{1}] V_{1}= 2,550 × LN[ 165,000 / 45,000 ] V_{1}= 3,313 m/s V_{2}= C_{2}× LN[ Mo_{2}/ Mf_{2}] V_{2}= 3,138 × LN[ 36,000 / 6,000 ] V_{2}= 5,623 m/s Equation (1.40), V_{Total}= V_{1}+ V_{2}V_{Total}= 3,313 + 5,623 V_{Total}= 8,936 m/s

PROBLEM 3.1 Using the Barrowman method, calculate the location of the center of pressure from the leading edge of a rocket having the dimensions given below. The nose is ogive shaped. SOLUTION, Given: L_{N}= 400 mm d = 200 mm d_{F}= 200 mm d_{R}= 160 mm L_{T}= 120 mm X_{P}= 900 mm C_{R}= 240 mm C_{T}= 120 mm S = 240 mm L_{F}= 247 mm R = 80 mm X_{R}= 120 mm X_{B}= 1,760 mm N = 3 each Equations (3.1) through (3.6), (C_{N})_{N}= 2 X_{N}= 0.466 × L_{N}X_{N}= 0.466 × 400 = 186 mm (C_{N})_{T}= 2 × [(d_{R}/ d)^{2}- (d_{F}/ d)^{2}] (C_{N})_{T}= 2 × [(160 / 200)^{2}- (200 / 200)^{2}] (C_{N})_{T}= -0.72 X_{T}= X_{P}+ L_{T}/ 3 × [1 + (1 - d_{F}/ d_{R}) / (1 - (d_{F}/ d_{R})^{2})] X_{T}= 900 + 120 / 3 × [1 + (1 - 200 / 160) / (1 - (200 / 160)^{2})] X_{T}= 958 mm (C_{N})_{F}= (1 + R / (S + R)) × [4 × N × (S / d)^{2}/ (1 + SQRT[1 + (2 × L_{F}/ (C_{R}+ C_{T}))^{2}])] (C_{N})_{F}= (1 + 80 / (240 + 80)) × [4 × 3 × (240 / 200)^{2}/ (1 + SQRT[1 + (2 × 247 / (240 + 120))^{2}])] (C_{N})_{F}= 8.01 X_{F}= X_{B}+ X_{R}/ 3 × (C_{R}+ 2 × C_{T}) / (C_{R}+ C_{T}) + 1/6 × [C_{R}+ C_{T}- C_{R}× C_{T}/ (C_{R}+ C_{T})] X_{F}= 1760 + 120 / 3 × (240 + 2 × 120) / (240 + 120) + 1/6 × [240 + 120 - 240 × 120 / (240 + 120)] X_{F}= 1,860 mm Equations (3.7) and (3.8), (C_{N})_{R}= (C_{N})_{N}+ (C_{N})_{T}+ (C_{N})_{F}(C_{N})_{R}= 2 - 0.72 + 8.01 = 9.29 X = [(C_{N})_{N}× X_{N}+ (C_{N})_{T}× X_{T}+ (C_{N})_{F}× X_{F}] / (C_{N})_{R}X = [2 × 186 - 0.72 × 958 + 8.01 × 1860] / 9.29 X = 1,570 mm

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 = 3.986005×10^{14}m^{3}/s^{2}Given: r = (6,378.14 + 200) × 1,000 = 6,578,140 m Equation (4.6), v = SQRT[ GM / r ] v = SQRT[ 3.986005×10^{14}/ 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), P^{2}= 4 ×^{2}× r^{3}/ GM P = SQRT[ 4 ×^{2}× r^{3}/ GM ] P = SQRT[ 4 ×^{2}× 6,578,140^{3}/ 3.986005×10^{14}] 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), P^{2}= 4 ×^{2}× r^{3}/ GM r = [ P^{2}× GM / (4 ×^{2}) ]^{1/3}r = [ 86,164.1^{2}× 3.986005×10^{14}/ (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 × 3.986005×10^{14}× 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 × 3.986005×10^{14}× 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 × Vp^{2}) - 1] Ra = 6,578,140 / [2 × 3.986005×10^{14}/ (6,578,140 × 7,850^{2}) - 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 × Vp^{2}/ GM - 1 e = 6,578,140 × 7,850^{2}/ 3.986005×10^{14}- 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 = 89^{o}Equation (4.26), (Rp / r1)_{1,2}= ( -C ± SQRT[ C^{2}- 4 × (1 - C) × -sin^{2}]) / (2 × (1 - C)) where C = 2 × GM / (r1 × v1^{2}) C = 2 × 3.986005×10^{14}/ (6,628,140 × 7,900^{2}) C = 1.927179 (Rp / r1)_{1,2}= ( -1.927179 ± SQRT[ 1.927179^{2}- 4 × -0.927179 × -sin^{2}(89) ]) / (2 × -0.927179) (Rp / r1)_{1,2}= 0.996019 and 1.082521 Perigee Radius, Rp = Rp_{1}= 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 = Rp_{2}= 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 = 89^{o}Equation (4.27), e = SQRT[ (r1 × v1^{2}/ GM - 1)^{2}× sin^{2 }+ cos^{2}] e = SQRT[ (6,628,140 × 7,900^{2}/ 3.986005×10^{14}- 1)^{2}× sin^{2}(89) + cos^{2}(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 = 89^{o}Equation (4.28), tan = (r1 × v1^{2}/ GM) × sin × cos / [(r1 × v1^{2}/ GM) × sin^{2}- 1] tan = (6,628,140 × 7,900^{2}/ 3.986005×10^{14}) × sin(89) × cos(89) / [(6,628,140 × 7,900^{2}/ 3.986005×10^{14}) × sin^{2}(89) - 1] tan = 0.48329 = arctan(0.48329) = 25.794^{o}

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 - v1^{2}/ GM ) a = 1 / ( 2 / 6,628,140 - 7,900^{2}/ 3.986005×10^{14}) ) 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 32^{o}N latitude, 60^{o}W longitude, and the azimuth heading is 86^{o}. Calculate the orbit's inclination, argument of perigee, and longitude of ascending node. SOLUTION, Given: = 86^{o}= 32^{o}_{2}= -60^{o}From problem 4.10, = 25.794^{o}Equation (4.33), cos(i) = cos() × sin() cos(i) = cos(32) × sin(86) i = 32.223^{o}Equations (4.34) and (4.36), tan() = tan() / cos() tan() = tan(32) / cos(86) = 83.630^{o}= - = 83.630 - 25.794 = 57.836^{o}Equations (4.35) and (4.37), tan() = sin() × tan() tan() = sin(32) × tan(86) = 82.483^{o}_{1}=_{2}-_{1}= -60 - 82.483_{1}= -142.483^{o}= Sidereal time at -142.483 longitude, 2000-10-20, 15:00 UT = 7^{h}27' 34" = 111.892^{o}

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 t_{O}= 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 / a^{3}] n = SQRT[ 3.986005×10^{14}/ 7,500,000^{3}] n = 0.00097202 rad/s Equation (4.38), M - Mo = n × (t - t_{O}) t = t_{O}+ (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 t_{O}= 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 - t_{O}) M = Mo + n × (t - t_{O}) M = 1.37113 + 0.00097202 × (1,200 - 0) M = 2.53755METHOD #1, Low Accuracy:Equation (4.42), ~ M + 2 × e × sin M + 1.25 × e^{2}× sin 2M ~ 2.53755 + 2 × 0.1 × sin(2.53755) + 1.25 × 0.1^{2}× sin(2 × 2.53755) ~ 2.63946 = 151.2 degreesMETHOD #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 - e^{2}) / (1 + e × cos ) r = 7,500,000 × (1 - 0.1^{2}) / (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[ 3.986005×10^{14}× (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 × sin^{2}i) / n_{Moon}= 0.00169 × (4 - 5 × sin^{2}51.6) / 15.5_{Moon}= 0.000101 deg/day_{Sun}= 0.00077 × (4 - 5 × sin^{2}i) / n_{Sun}= 0.00077 × (4 - 5 × sin^{2}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 J_{2}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}= -2.06474×10^{14}× a^{-7/2}× (cos i) × (1 - e^{2})^{-2}_{J2}= -2.06474×10^{14}× (7,500)^{-7/2}× (cos 28.5) × (1 - (0.1)^{2})^{-2}_{J2}= -5.067 deg/day_{J2}= 1.03237×10^{14}× a^{-7/2}× (4 - 5 × sin^{2}i) × (1 - e^{2})^{-2}_{J2}= 1.03237×10^{14}× (7,500)^{-7/2}× (4 - 5 × sin^{2}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 m^{2}m = 1,000 kg C_{D}= 2.67 From Atmosphere Properties, = 2.62×10^{-12}kg/m^{3}H = 58.2 km Equation (4.6), V = SQRT[ GM / a ] V = SQRT[ 3.986005×10^{14}/ 6,778,140 ] V = 7,669 m/s Equations (4.53) through (4.55), a_{rev}= (-2 × × C_{D}× A × × a^{2}) / m a_{rev}= (-2 × × 2.67 × 8 × 2.62×10^{-12}× 6,778,140^{2}) / 1,000 a_{rev}= -16.2 m P_{rev}= (-6 ×^{2}× C_{D}× A × × a^{2}) / (m × V) P_{rev}= (-6 ×^{2}× 2.67 × 8 × 2.62×10^{-12}× 6,778,140^{2}) / (1,000 × 7,669) P_{rev}= -0.0199 s V_{rev}= ( × C_{D}× A × × a × V) / m V_{rev}= ( × 2.67 × 8 × 2.62×10^{-12}× 6,778,140 × 7,669) / 1,000 V_{rev}= 0.00914 m/s Equation (4.56), L ~ -H / a_{rev}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: r_{A}= (6,378.14 + 200) × 1,000 = 6,578,140 m From problem 4.3, r_{B}= 42,164,170 m Equations (4.58) through (4.65), a_{tx}= (r_{A}+ r_{B}) / 2 a_{tx}= (6,578,140 + 42,164,170) / 2 a_{tx}= 24,371,155 m Vi_{A}= SQRT[ GM / r_{A}] Vi_{A}= SQRT[ 3.986005×10^{14}/ 6,578,140 ] Vi_{A}= 7,784 m/s Vf_{B}= SQRT[ GM / r_{B}] Vf_{B}= SQRT[ 3.986005×10^{14}/ 42,164,170 ] Vf_{B}= 3,075 m/s Vtx_{A}= SQRT[ GM × (2 / r_{A}- 1 / a_{tx})] Vtx_{A}= SQRT[ 3.986005×10^{14}× (2 / 6,578,140 - 1 / 24,371,155)] Vtx_{A}= 10,239 m/s Vtx_{B}= SQRT[ GM × (2 / r_{B}- 1 / a_{tx})] Vtx_{B}= SQRT[ 3.986005×10^{14}× (2 / 42,164,170 - 1 / 24,371,155)] Vtx_{B}= 1,597 m/s V_{A}= Vtx_{A}- Vi_{A}V_{A}= 10,239 - 7,784 V_{A}= 2,455 m/s V_{B}= Vf_{B}- Vtx_{B}V_{B}= 3,075 - 1,597 V_{B}= 1,478 m/s V_{T}= V_{A}+ V_{B}V_{T}= 2,455 + 1,478 V_{T}= 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: r_{A}= (6,378.14 + 200) × 1,000 = 6,578,140 m r_{B}= 42,164,170 m a_{tx}= 30,000 × 1,000 = 30,000,000 m Equations (4.66) through (4.68), e = 1 - r_{A}/ a_{tx}e = 1 - 6,578,140 / 30,000,000 e = 0.780729 = arccos[(a_{tx}× (1 - e^{2}) / r_{B}- 1) / e ] = arccos[(30,000,000 × (1 - 0.780729^{2}) / 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), Vi_{A}= SQRT[ GM / r_{A}] Vi_{A}= SQRT[ 3.986005×10^{14}/ 6,578,140 ] Vi_{A}= 7,784 m/s Vf_{B}= SQRT[ GM / r_{B}] Vf_{B}= SQRT[ 3.986005×10^{14}/ 42,164,170 ] Vf_{B}= 3,075 m/s Vtx_{A}= SQRT[ GM × (2 / r_{A}- 1 / a_{tx})] Vtx_{A}= SQRT[ 3.986005×10^{14}× (2 / 6,578,140 - 1 / 30,000,000)] Vtx_{A}= 10,388 m/s Vtx_{B}= SQRT[ GM × (2 / r_{B}- 1 / a_{tx})] Vtx_{B}= SQRT[ 3.986005×10^{14}× (2 / 42,164,170 - 1 / 30,000,000)] Vtx_{B}= 2,371 m/s V_{A}= Vtx_{A}- Vi_{A}V_{A}= 10,388 - 7,784 V_{A}= 2,604 m/s Equation (4.69), V_{B}= SQRT[ Vtx_{B}^{2}+ Vf_{B}^{2}- 2 × Vtx_{B}× Vf_{B}× cos ] V_{B}= SQRT[ 2,371^{2}+ 3,075^{2}- 2 × 2,371 × 3,075 × cos(46.876)] V_{B}= 2,260 m/s Equation (4.65), V_{T}= V_{A}+ V_{B}V_{T}= 2,604 + 2,260 V_{T}= 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[ a_{tx}^{3}/ GM ] TOF = (2.11688 - 0.780729 × sin(2.11688)) × SQRT[ 30,000,000^{3}/ 3.986005×10^{14}] 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[ 3.986005×10^{14}/ 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: r_{A}= (6,378.14 + 200) × 1,000 = 6,578,140 m r_{B}= 42,164,170 m = 28 degrees From problem 4.19, Vf_{B}= 3,075 m/s Vtx_{B}= 1,597 m/s V_{A}= 2,455 m/s Equation (4.74), V_{B}= SQRT[ Vtx_{B}^{2}+ Vf_{B}^{2}- 2 × Vtx_{B}× Vf_{B}× cos ] V_{B}= SQRT[ 1,597^{2}+ 3,075^{2}- 2 × 1,597 × 3,075 × cos(28)] V_{B}= 1,826 m/s Equation (4.65), V_{T}= V_{A}+ V_{B}V_{T}= 2,455 + 1,826 V_{T}= 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: i_{i}= 30 degrees i = 75 degrees i_{f}= 32 degrees f = 80 degrees Equation (4.75), a1 = sin(i_{i})cos(i) = sin(30)cos(75) = 0.129410 a2 = sin(i_{i})sin(i) = sin(30)sin(75) = 0.482963 a3 = cos(i_{i}) = cos(30) = 0.866025 b1 = sin(i_{f})cos(f) = sin(32)cos(80) = 0.0920195 b2 = sin(i_{f})sin(f) = sin(32)sin(80) = 0.521869 b3 = cos(i_{f}) = 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 lat_{1}= arctan(c3 / (c1^{2}+ c2^{2})^{1/2}) lat_{1}= arctan(0.0230928 / (-0.0423757^{2}+ -0.0300543^{2})^{1/2}) lat_{1}= 23.965 degrees long_{1}= arctan(c2 / c1) + 90 long_{1}= arctan(-0.0300543 / -0.0423757) + 90 long_{1}= 125.346 degrees lat_{2}= -23.965 degrees long_{2}= 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), V_{esc}= SQRT[ 2 × GM / r ] V_{esc}= SQRT[ 2 × 3.986005×10^{14}/ 6,578,140 ] V_{esc}= 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 = 4.282831×10^{13}m^{3}/s^{2}Given: r = 100,000 × 1,000 = 100,000,000 m v = 5,140.0 m/s = -85.300^{o}Equations (4.30) and (4.32), e = SQRT[ (r × v^{2}/ GM - 1)^{2}× cos^{2}+ sin^{2}] e = SQRT[ (100,000,000 × 5,140^{2}/ 4.282831×10^{13}- 1)^{2}× cos^{2}(-85.3) + sin^{2}(-85.3) ] e = 5.0715 a = 1 / ( 2 / r - v^{2}/ GM ) a = 1 / ( 2 / 100,000,000 - 5,140^{2}/ 4.282831×10^{13}) a = -1,675,400 m Equations (4.80) through (4.85), sin(/2) = 1 / e = 2 × arcsin( 1 / 5.0715 ) = 22.744^{o}cos = -1 / e = arccos( -1 / 5.0715 ) = 101.37^{o}= arccos[ (a × (1 - e^{2}) - r) / (e × r) ] = arccos[ (-1,675,400 × (1 - 5.0715^{2}) - 100,000,000) / (5.0715 × 100,000,000) ] = -96.633^{o}b = -a / tan(/2) b = 1,675.4 / tan(22.744/2) b = 8,330.0 km r_{o}= a × (1 - e) r_{o}= -1,675.4 × (1 - 5.0715) r_{o}= 6,821.4 km p = a × (1 - e^{2}) p = -1,675.4 × (1 - 5.0715^{2}) 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 = 75^{o}Equations (4.43) through (4.45), r = a × (1 - e^{2}) / (1 + e × cos ) r = -1,675,400 × (1 - 5.0715^{2}) / (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))] = 64.729^{o}v = SQRT[ GM × (2 / r - 1 / a)] v = SQRT[ 4.282831×10^{13}× (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}= 15^{o}= 120^{o}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 - t_{O}= SQRT[(-a)^{3}/ GM ] × [(e × sinh F - F) - (e × sinh Fo - Fo)] t - t_{O}= SQRT[(36,000,000)^{3}/ 3.986005×10^{14}] × [(1.1823 × sinh(1.10023) - 1.10023) - (1.1823 × sinh(0.07614) - 0.07614)] t - t_{O}= 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: V_{bo}= 11,500 m/s From problem 4.25, V_{esc}= 11,009 m/s Equation (4.88), V_{}^{2}= V_{bo}^{2}- V_{esc}^{2}V_{}= SQRT[ 11,500^{2}- 11,009^{2}] V_{}= 3,325 m/s

PROBLEM 4.30 Calculate the radius of Earth's sphere of influence. SOLUTION, From Basics Constants, D_{sp}= 149,597,870 km M_{P}= 5.9737×10^{24}kg M_{S}= 1.9891×10^{30}kg Equation (4.89), R_{Earth}= D_{sp}× (M_{P}/ M_{S})^{0.4}R_{Earth}= 149,597,870 × (5.9737×10^{24}/ 1.9891×10^{30})^{0.4}R_{Earth}= 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: r_{A}= 1.000 AU r_{B}= 1.524 AU a_{tx}= 1.300 AU × 149.597870×10^{9}m/AU = 194.48×10^{9}m From Basics Constants, GM of Sun = 1.327124×10^{20}m^{3}/s^{2}Equations (4.66) and (4.67), e = 1 - r_{A}/ a_{tx}e = 1 - 1.0 / 1.3 e = 0.230769 = arccos[(a_{tx}× (1 - e^{2}) / r_{B}- 1) / e ] = arccos[(1.3 × (1 - 0.230769^{2}) / 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[ a_{tx}^{3}/ GM ] TOF = (2.41383 - 0.230769 × sin(2.41383)) × SQRT[ (194.48×10^{9})^{3}/ 1.327124×10^{20}] 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}= 146.488^{o}t_{2}-t_{1}= 194.77 days_{t}= 0.5240^{o}/day Equation (5.1), = (_{2}-_{1}) -_{t}× (t_{2}-t_{1}) = 146.488 - 0.5240 × 194.77 = 44.43^{o}

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- 0.899215YAU. Mars' postion vector at intercept is 0.066842X+ 1.561256Y+ 0.030948ZAU. Calculate the parameter and semi-major axis of the transfer orbit. SOLUTION, Given: t = 207 daysr= 0.473265_{1}X- 0.899215YAUr= 0.066842_{2}X+ 1.561256Y+ 0.030948ZAU GM = 1.327124×10^{20}m^{3}/s^{2}= 1.327124×10^{20}/ (149.597870×10^{9})^{3}= 3.964016×10^{-14}AU^{3}/s^{2}From vector magnitude,r= SQRT[ 0.473265_{1}^{2}+ (-0.899215)^{2}]r= 1.016153 AU_{1}r= SQRT[ 0.066842_{2}^{2}+ 1.561256^{2}+ 0.030948^{2}]r= 1.562993 AU From vector dot product, = arccos[ (0.473265 × 0.066842 - 0.899215 × 1.561256) / (1.016153 × 1.562993) ] = 149.770967_{2}^{o}Equations (5.9), (5.10) and (5.11), k = r_{1}× r_{2}× (1 - cos ) k = 1.016153 × 1.562993 × (1 - cos(149.770967)) k = 2.960511 AU = r_{1}+ r_{2}= 1.016153 + 1.562993 = 2.579146 AU m = r_{1}× r_{2}× (1 + cos ) m = 1.016153 × 1.562993 × (1 + cos(149.770967)) m = 0.215969 AU Equations (5.18) and (5.19), p_{i}= k / ( + SQRT(2 × m)) p_{i}= 2.960511 / (2.579146 + SQRT(2 × 0.215969)) p_{i}= 0.914764 AU p_{ii}= k / ( - SQRT(2 × m)) p_{ii}= 2.960511 / (2.579146 - SQRT(2 × 0.215969)) p_{ii}= 1.540388 AU Since < , 0.914764 < p < Equation (5.12), Select trial value, p = 1.2 AU a = m × k × p / [(2 × m -^{2}) × p^{2}+ 2 × k × × p - k^{2}] a = 0.215969 × 2.960511 × 1.2 / [(2 × 0.215969 - 2.579146^{2}) × 1.2^{2}+ 2 × 2.960511 × 2.579146 × 1.2 - 2.960511^{2}] a = 1.270478 AU Equations (5.5), (5.6) and (5.7), f = 1 - r_{2}/ p × (1 - cos ) f = 1 - 1.562993 / 1.2 × (1 - cos(149.770967)) f = -1.427875 g = r_{1}× r_{2}× 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/r_{1}- 1/r_{2}] = 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 - r_{1}/ a × (1 - f) ] E = arccos[ 1 - 1.016153 / 1.270478 × (1 + 1.427875) ] E = 2.798925 radians Equation (5.16), t = g + SQRT[ a^{3}/ GM ] × (E - sin E) t = 3,666,240 + SQRT[ 1.270478^{3}/ 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), p_{n+1}= p_{n}+ (t - t_{n}) × (p_{n}- p_{n-1}) / (t_{n}- t_{n-1}) p_{n+1}= 1.3 + (207 - 178.9588) × (1.3 - 1.2) / (178.9588 - 247.4647) p_{n+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:r= 0.473265_{1}X- 0.899215YAUr= 0.066842_{2}X+ 1.561256Y+ 0.030948ZAUr= 1.016153 AU_{1}r= 1.562993 AU p = 1.250633 AU a = 1.320971 AU = 149.770967_{2}^{o}Equations (5.5), (5.6) and (5.7), f = 1 - r_{2}/ p × (1 - cos ) f = 1 - 1.562993 / 1.250633 × (1 - cos(149.770967)) f = -1.329580 g = r_{1}× r_{2}× 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/r_{1}- 1/r_{2}] = 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 - r_{1}/ p × (1 - cos ) = 1 - 1.016153 / 1.250633 × (1 - cos(149.770967)) = -0.514536 Equation (5.3),v= (_{1}r- f ×_{2}r) / g_{1}v= [(0.066842 + 1.329580 × 0.473265) / 3,591,258]_{1}X+ [(1.561256 + 1.329580 × -0.899215) / 3,591,258]Y+ [(0.030948 + 1.329580 × 0) / 3,591,258]Zv= 0.000000193828_{1}X+ 0.000000101824Y+ 0.00000000861759ZAU/s × 149.597870×10^{9}v= 28996.2_{1}X+ 15232.7Y+ 1289.2Zm/s Equation (5.4),v= ×_{2}r+ ×_{1}v_{1}v= [-8.795872×10_{2}^{-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]Zv= -0.000000141359_{2}X+ 0.0000000267017Y- 0.00000000443406ZAU/s × 149.597870×10^{9}v= -21147.0_{2}X+ 3994.5Y- 663.3Zm/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 eitherr&_{1}vor_{1}r&_{2}v– we will use_{2}r&_{1}v. Given:_{1}r= (0.473265_{1}X- 0.899215YAU) × 149.597870×10^{9}m/AU = 7.079944×10^{10}X- 1.345206×10^{11}Ymr= 1.016153 × 149.597870×10_{1}^{9}= 1.520144×10^{11}m GM = 1.327124×10^{20}m^{3}/s^{2}From problem 5.4,v= 28996.2_{1}X+ 15232.7Y+ 1289.2Zm/s Also,v= SQRT[ 28996.2^{2}+ 15232.7^{2}+ 1289.2^{2}] = 32,779.2 m/s Equations (5.21) and (5.22),h= (r_{Y}v_{Z}- r_{Z}v_{Y})X+ (r_{Z}v_{X}- r_{X}v_{Z})Y+ (r_{X}v_{Y}- r_{Y}v_{X})Zh= (-1.345206×10^{11}× 1289.2 - 0 × 15232.7)X+ (0 × 28996.2 - 7.079944×10^{10}× 1289.2)Y+ (7.079944×10^{10}× 15232.7 + 1.345206×10^{11}× 28996.2)Zh= -1.73424×10^{14}X- 9.12746×10^{13}Y+ 4.97905×10^{15}Zn= -h_{Y}X+ h_{X}Yn= 9.12746×10^{13}X- 1.73424×10^{14}YAlso,h= SQRT[ (-1.73424×10^{14})^{2}+ (-9.12746×10^{13})^{2}+ (4.97905×10^{15})^{2}] = 4.98291×10^{15}n= SQRT[ (9.12746×10^{13})^{2}+ (1.73424×10^{14})^{2}] = 1.95977×10^{14}Equation (5.23),e= [(v^{2}- GM /r) ×r- (r • v) ×v] / GMv^{2}- GM /r= 32779.2^{2}- 1.327124×10^{20}/ 1.520144×10^{11}= 2.01451×10^{8}r • v= 7.079944×10^{10}× 28996.2 - 1.345206×10^{11}× 15232.7 + 0 x 1289.2 = 3.80278×10^{12}e= [2.01451×10^{8}× (7.079944×10^{10}X- 1.345206×10^{11}Y) - 3.80278×10^{12}× (28996.2X+ 15232.7Y+ 1289.2Z) ] / 1.327124×10^{20}e= 0.106639X- 0.204632Y- 0.000037ZEquations (5.24) and (5.25), a = 1 / ( 2 /r-v^{2}/ GM ) a = 1 / ( 2 / 1.520144×10^{11}- 32779.2^{2}/ 1.327124×10^{20}) a = 1.97614×10^{11}me= SQRT[ 0.106639^{2}+ (-0.204632)^{2}+ (-0.000037)^{2}]e= 0.230751 Equations (5.26) though (5.30), cos i =h/_{Z}hcos i = 4.97905×10^{15}/ 4.98291×10^{15}i = 2.255^{o}cos =n/_{X}ncos = 9.12746×10^{13}/ 1.95977×10^{14}= 297.76^{o}cos =n • e/ (n×e) cos = (9.12746×10^{13}× 0.106639 - 1.73424×10^{14}× (-0.204632) + 0 × (-0.000037)) / (1.95977×10^{14}× 0.230751) = 359.77^{o}cos_{o}=e • r/ (e×r) cos_{o}= (0.106639 × 7.079944×10^{10}- 0.204632 × (-1.345206×10^{11}) - 0.000037 × 0) / (0.230751 × 1.520144×10^{11})_{o}= 0.226^{o}cos u_{o}=n • r/ (n×r) u_{o}= 0 (launch point = ascending node) Equations (5.31) and (5.32), = + = 297.76 + 359.77 = 297.53^{o}_{o}= + +_{o}_{o}= 297.76 + 359.77 + 0.23_{o}= 297.76^{o}

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+ 13759.5Ym/s. SOLUTION, Given: r_{o}= (6,378.14 + 200) × 1,000 = 6,578,140 mr= 0.473265X- 0.899215YAUV= 25876.6_{P}X+ 13759.5Ym/s From problem 5.4,V= 28996.2_{S}X+ 15232.7Y+ 1289.2Zm/s Equation (5.33),V= (VS_{S/P}_{X}- VP_{X})X+ (VS_{Y}- VP_{Y})Y+ (VS_{Z}- VP_{Z})ZV= (28996.2 - 25876.6)_{S/P}X+ (15232.7 - 13759.5)Y+ (1289.2 - 0)ZV= 3119.6_{S/P}X+ 1473.2Y+ 1289.2Zm/s Equation (5.34),V= SQRT[ VS/P_{S/P}_{X}^{2}+ VS/P_{Y}^{2}+ VS/P_{Z}^{2}]V= SQRT[ 3119.6_{S/P}^{2}+ 1473.2^{2}+ 1289.2^{2}]V= 3,683.0 m/s V_{S/P}_{}=V= 3,683.0 m/s Equations (5.35) and (5.36), V_{S/P}_{o}= SQRT[ V_{}^{2}+ 2 × GM / r_{o}] V_{o}= SQRT[ 3,683.0^{2}+ 2 × 3.986005×10^{14}/ 6,578,140 ] V_{o}= 11,608.4 m/s V = V_{o}- SQRT[ GM / r_{o}] V = 11,608.4 - SQRT[ 3.986005×10^{14}/ 6,578,140 ] V = 3,824.1 m/s Equation (5.37),r= SQRT[ 0.473265^{2}+ (-0.899215)^{2}+ 0^{2}]r= 1.01615 AU = arccos[(r_{X}× v_{X}+ r_{Y}× v_{Y}+ r_{Z}× v_{Z}) / (r × v)] = arccos[ 0.473265 × 3119.6 - 0.899215 × 1473.2 + 0 × 1289.2) / (1.01615 × 3683.0)] = 87.677^{o}

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+ 41.8Zm/s. SOLUTION, Given: d = 18,500 km / 149.597870×10^{6}= 0.000123664 AUr= 0.066842X+ 1.561256Y+ 0.030948ZAUV= -23307.8_{P}X+ 3112.0Y+ 41.8Zm/s From Basics Constants, GM of Mars = 4.282831×10^{13}m^{3}/s^{2}From problem 5.4,V= -21147.0_{S}X+ 3994.5Y- 663.3Zm/s Equation (5.33),V= (VS_{S/P}_{X}- VP_{X})X+ (VS_{Y}- VP_{Y})Y+ (VS_{Z}- VP_{Z})ZV= (-21147.0 + 23307.8)_{S/P}X+ (3994.5 - 3112.0)Y+ (-663.3 - 41.8)ZV= 2160.8_{S/P}X+ 882.5Y- 705.1Zm/s Equation (5.34),V= SQRT[ VS/P_{S/P}_{X}^{2}+ VS/P_{Y}^{2}+ VS/P_{Z}^{2}]V= SQRT[ 2160.8_{S/P}^{2}+ 882.5^{2}+ (-705.1)^{2}]V= 2,438.2 m/s V_{S/P}_{}=V= 2,438.2 m/s Equations (5.38.A) and (5.38.B), d_{S/P}_{x}= -d × r_{y}/ SQRT[ r_{x}^{2}+ r_{y}^{2}] d_{x}= -0.000123664 × 1.561256 / SQRT[ 0.066842^{2}+ 1.561256^{2}] d_{x}= -0.000123551 AU d_{y}= d × r_{x}/ SQRT[ r_{x}^{2}+ r_{y}^{2}] d_{y}= 0.000123664 × 0.066842 / SQRT[ 0.066842^{2}+ 1.561256^{2}] d_{y}= 0.0000052896 AU Equation (5.39), = arccos[(d_{x}× v_{x}+ d_{y × }v_{y}) / (d × v)] = arccos[(-0.000123551 × 2160.8 + 0.0000052896 × 882.5) / (0.000123664 × 2,438.2)] = 150.451^{o}Equations (5.40) through (5.42), b = d × sin b = 18,500 × sin(150.451) b = 9,123.6 km a = -GM / V_{}^{2}a = -4.282831×10^{13}/ 2,438.2^{2}a = -7.2043×10^{6}m = -7,204.3 km e = SQRT[ 1 + b^{2}/ a^{2}] e = SQRT[ 1 + 9,123.6^{2}/ -7,204.3^{2}] 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: V_{P}= 12,740 m/s_{P}= 2.40^{o}VS_{i}= 9,470 m/s S_{i}= 39.2^{o}d = -2,500,000 km From Basics Constants, GM of Jupiter = 1.26686×10^{17}m^{3}/s^{2}Equations (5.44) and (5.45),V= (V_{P}_{P}× cos_{P})X+ (V_{P}× sin_{P})YV= (12740 × cos(2.40))_{P}X+ (12740 × sin(2.40))YV= 12729_{P}X+ 533Ym/sVS= (VS_{i}_{i}× cos S_{i})X+ (VS_{i}× sin S_{i})YVS= (9470 × cos(39.2))_{i}X+ (9470 × sin(39.2))YVS= 7339_{i}X+ 5985Ym/s Equations (5.46) and (5.47),VS/P= ((VS_{i}_{i})_{X}- VP_{X})X+ ((VS_{i})_{Y}- VP_{Y})YVS/P= (7339 - 12729)_{i}X+ (5985 - 533)YVS/P= -5390_{i}X+ 5452Ym/sV= SQRT[ (VS/P_{S/P}_{i})_{X}^{2}+ (VS/P_{i})_{Y}^{2}]V= SQRT[ (-5390)_{S/P}^{2}+ 5452^{2}]V= 7,667 m/s V_{S/P}_{}=V= 7,667 m/s Equation (5.48),_{S/P}_{i}= arctan[ (VS/P_{i})_{Y}/ (VS/P_{i})_{X}]_{i}= arctan[ 5452 / -5390 ]_{i}= 134.67^{o}Equations (5.40) through (5.42), b = d × sin b = -2,500,000 × sin(134.67) b = -1,777,900 km a = -GM / V_{}^{2}a = -1.26686×10^{17}/ 7667^{2}a = -2.1552×10^{9}m = -2,155,200 km e = SQRT[ 1 + b^{2}/ a^{2}] e = SQRT[ 1 + (-1,777,900)^{2}/ (-2,155,200)^{2}] e = 1.2963 Equation (5.49), = 2 × arcsin( 1 / e ) = 2 × arcsin( 1 / 1.2963 ) = 100.96^{o}Equation (5.50),_{f}=_{i}-_{f}= 134.67 - 100.96_{f}= 33.71^{o}Equation (5.51),VS/P= (V_{f}_{S/P}× cos_{f})X+ (V_{S/P}× sin_{f})YVS/P= (7667 × cos(33.71))_{f}X+ (7667 × sin(33.71))YVS/P= 6378_{f}X+ 4255Ym/s Equations (5.52) and (5.53),VS= ((VS/P_{f}_{f})_{X}+ VP_{X})X+ ((VS/P_{f})_{Y}+ VP_{Y})YVS= (6378 + 12729)_{f}X+ (4255 + 533)YVS= 19107_{f}X+ 4788Ym/sVS= SQRT[ (VS_{f}_{f})_{X}^{2}+ (VS_{f})_{Y}^{2}]VS= SQRT[ 19107_{f}^{2}+ 4788^{2}]VS= 19,698 m/s Equation (5.54), S_{f}_{f}= arctan[ (VS_{f})_{Y}/ (VS_{f})_{X}] S_{f}= arctan[ 4788 / 19107 ] S_{f}= 14.07^{o}