Aerospace Engineering - Orbital Mechanics

Full exam

1 of 5 Orbital Mechanics Academic Year 20 20 -20 21 Lecturer: Camilla Colombo Exam 11 June 20 21 Duration: 1.5 h All solution sheets have to be written in clear form in pen . Do not write in pencil. On top of each sheet, please report in capital letters your surname, name and student number. In the solution of the exercise all the analytical and numerical procedure must be reported . Please report the number of the answers from the text of the exercise . Numerical results need to have proper units of measure. Last digit of Person Code (Codice Persona) 0 or 1 2 or 3 4 or 5 6 or 7 8 or 9 Exam Version 1 2 3 4 5 PART 1 (16 PT) E x e r c i s e 1 – O r b i t a l m a n o e u v r e A probe is sent to Mars for an exploration mission. After an interplanetary transfer, the probe is inserted in a hyperbolic trajectory around Mars (Orbit 1), lying on the Mars equatorial plane, with the infinite and perigee velocities , ������∞ and ������������. Once the probe arrives at the perigee of the hyperbolic trajectory, a braking manoeuvre is performed to close the orbit. The orbit after this manoeuvre (Orbit 2) lies on the Mars equatorial plane and has an apocenter radius of 12500 km. Assume that the braking manoeuvre is performed tangentially to the manoeuvring point in the Mars equatorial plane. Now that the probe is on Orbit 2, an apse line rotation manoeuvre is req uired by the scientific mode of the mission, to move the prove on Orbit 3. The apse line rotation manoeuvre consists of a delta -v of ������� , given with a positive angle of 7° with respect to the transversal direction of the local horizon of the radial -transv ersal -normal Version 1 2 3 4 5 Infinite velocity: ������∞ 0.35 km/s 0.38 km/s 0.40 km/s 0.42 km/s 0.46 km/s Perigee velocity: ������������ 4.97 km/s Delta -V ������� 0.08 km/s 0.10 km/s 0.15 km/s 0.2 km/s 0.3 km/s S/C cross section area: �������� 0.5 m^2 0.7 m^2 0.9 m^2 1.0 m^2 1.2 m^2 Spacecraft mass: �������� 42 kg 45 kg 47 kg 50 kg 53 kg S/C drag coefficient: �� 1.2 2 of 5 frame (������,������,ℎ) at the apse line manoeuvring point of Orbit 2. The apse line rotation delta -v is given at a true anomaly of 30 ° on the Orbit 2. Mars Data: Equatorial radius of Mars: ������������=3376.2 km Mars gravitational constant: ������������=4.2828 ⋅10^4 km^3/s^2 (1) Compute the perigee radius and the semi -major axis of the hyperbolic trajectory (Orbit 1). (insert ONLY the NUMBER expressed in km, with 3 decimal digits) (2) Consider the braking maneuver. Compute t he delta -v of the braking maneuver to close Orbit 1 into Orbit 2. (insert ONLY the NUMBER expressed in km/s, with 3 decimal digits) (3) Consider the apse line maneuver. Compute the velocity of the probe at the maneuvering point on the final orbit (Orbit 3) af ter the delta -v. (Provide the velocity on Orbit 3 both in modulus and in vectorial form, in the (r,θ,h) frame. Insert ONLY the NUMBER expressed km/s, with 3 decimal digits. Separate the results by a comma as following ({v3}, {vr, vθ, vh})). (4) Compute the true anomaly of the maneuvering point on Orbit 3. Assume that the apse line rotation happens just after the perigee of Orbit 3. (insert ONLY the NUMBER expressed in degrees, with 3 decimal digits.) Now consider the probe on the Orbit 3 defined in question Q3. Its perigee is affected by the drag perturbation of the Mars atmosphere. Assume that the drag effect is relevant on the probe up to 250 km of altitude. Assume also that the atmospheric density can be considered constant and equal to 1.28 x 10^( -7) kg/ m^3. Consider the following properties of the probe: spacecraft cross section area ��������, drag coefficient ��, spacecraft mass ��������. Assume that the average drag effect on Orbit 3 can be estimated as an energy loss due to the drag force D acting on the probe. The energy loss can be computed as an integral effect: ∫ � ������ ������������ , where r is the orbit radius. Assume that the semi -major axis of Orbit 3 remains constant during the atmospheric passage and assume a stationary (non -rotating) atmosphere. (5) Compute the time spent by the probe in the atmospheric passage in one orbital period (when altitude is < 250 km). (insert ONLY the NUMBER expres sed in minutes, with 3 significant digits.) (6) Compute the energy loss due to the drag force D acting on the probe in one orbital revolution. Provide the result in exponential form in (kg m^2 / s^2). HINT: Be careful with the unit of measurements and ch eck the consistency. RECALL the following relation: ∫ 1 1+ ������cos ������������������ = ( 2 √1−������2)atan (√1−������ 1+������tan (������ 2)) when 0< ������< 1 3 of 5 E x e r c i s e 2 – T i m e M e a s u r e m e n t s a n d S p h e r i c a l G e o m e t r y A spacecraft for astrophysical measurements is orbiting the Earth. A ground station is monitoring the satellite and performed an observation when the true anomaly of the satellite is ������ and the Greenwich Mean Time (GMT) is 24 h, at the summer solstice. At the observation moment, the position and velocity of the spacecraft are in the Earth geocentric frame ������������������� and ������������������� . Earth data: Equatorial radius of the Earth: ������������ = 6378.15 km Earth's gravitational constant: ������������ = 398600 km^3/s^2 Obliquity of the ecliptic: ������ = 23.4° (1) Compute the angular momentum of the spacecraft in vectorial form, in the Earth geocentric frame. (insert ONLY the NUMBER expressed in km^2/s, with 3 decimal digits. ) (2) Compute the right ascension and declination of the satellite at the observation moment. (insert ONLY the NUMBER expressed in degrees, with 3 decimal digits, separated by a comma. ) (3) Compute the ecliptic latitude and longitude of the satellite at the observation moment. (insert ONLY the NUMBER expressed in degrees, with 3 decimal digits, separated by a comma.) (4) Compute the declination and right ascension of the Sun at the observation moment. (insert ONLY the NUMBER expressed in degrees, with 3 significant digits, separated by a comma ). (5) Compute the Local Sidereal Time (LST) of the observation moment. Assume the mean and the actual Sun to coincide. (insert ONLY the NUMBER expressed in hours, with 3 decimal digits. ) Version 1 2 3 4 5 True anomaly ������ 20° 50° 115° 225° 285° Spacecraft position ������������������� ������������������� = {−4885 .976 317 .168 6335 .375 } ������������������� = {−1799 .761 −1951 .968 9387 .003 } ������������������� = {11038 .446 −7031 .936 7218 .728 } ������������������� = {4650 .697 2876 .184 −16865 .639 } ������������������� = {−8936 .362 6553 .116 −8777 .125 } Spacecraft velocity ������������������� ������������������� = {5.160−3.8015.127 } ������������������� = {6.075−3.3002.031 } ������������������� = {3.797−0.733 −3.265 } ������������������� = {−4.0111.6860.340 } ������������������� = {−2.810 −0.1904.912 } 4 of 5 Orbital Mechanics Academic Year 2020 -2021 Lecturer: Camilla Colombo Exam 11 June 2021 Duration: 1.5 h All solution sheets have to be written in clear form in pen. Do not write in pencil. On top of each sheet, please report in capital letters your surname, name and student number. In the solution of the exercise all the analytical and numerical procedure must be reported. Please report the number of the answers from the text of the exercise . Numerical results need to have proper units of measure. Last digit of Person Code (Codice Persona) 0 or 1 2 or 3 4 or 5 6 or 7 8 or 9 Exam Version 1 2 3 4 5 PART 2 – INTERPLA NETARY TRAJECTORY WITH JUPITER FLYBY Note that the DATA section reports the numerical data for the whole exercise. Read the whole exercise till the end before proceeding into its solution. DATA ������������������� = 1.3271 ∙10 11 km3/s2 gravitational constant of the Sun ����������������������� = 1.2669 ⋅10 8 km3/s2 gravitational constant of Jupiter ����������������������� = 7.7857 ⋅10 8 km radius of Jupiter’s orbit around the Sun Vtheta 1 = Trans versal velocity of the spacecraft in Orbit 1 before the flyby [km/s] α = Angle by whi ch the flyby rotates tran sversal velocity in the heliocentric frame [deg] i1 = Inclination of Orbit 1 over the ecliptic [deg] Ω1 = Right ascension of the ascending node of Orbit 1 [deg] u1 = Argument of latitude of the spacecraft in Orbit 1 at the flyby [deg] Table 1 Input data table for Part 2 (flyby) Versions Vtheta 1 [km/s] α [deg] i1 [deg] Ω1 [deg] u1 [deg] Flyby position 1 15.28 5.00 1.24 80.50 10.00 Pericentre 2 14.15 6.00 2.02 47.91 20.00 Pericentre 3 14.15 9.00 3.90 10.00 30.00 Pericentre 4 11.96 8.00 3.08 47.27 15.00 Apocentre 5 10.84 7.00 3.17 10.00 25.00 Apocentre A spacecraft is orbiting on a heliocentric Orbit 1 characterised by inclination i1 (see input data table) and right ascension of the ascending node Ω1 (see input data table) , both given with respect to the Heliocentric Ecliptic Sun -Centre d Inertial Reference Frame. In correspondence of the orbit al position at argument of latitude u1 (see input data table) the spacecraft performs a fly -by with Jupiter (see common data). At the entry fly -by condition on Orbit 1 the spacecraft is at an apse point (see input data table) and the transverse component of the velocity is equal to Vtheta 1 (see input data table). The fly -by is such that the exit transversal velocity Vtheta 2 in the 5 of 5 heliocentric frame is rotated of an angle α around the positive radial direc tion (see input data table) with respect to the entry transversal velocity Vtheta 1. The exit condition in the heliocentric frame is still an apse point on the new orbit modified by the fly -by, Orbit 2. The planet Jupiter is in a circular orbit inclined with respect to the ecliptic p lane. Heliocentric orbit after the fly -by 1) Find the semi -major axis of the heliocentric Orbit 1 before the fly -by [ Output in km with 1 decimal digit ]. 2) Find the eccentricity of the heliocentric Orbit 1 before the fly -by [Output with 4 decimal digits] . 3) Express the components of the heliocentric flyby entry velocity on Orbit 1 , expressed in the Radial – Transversal – Out -of-plane Reference Frame of Orbit 1 [Output in vectorial form in km/s with 2 decimal digits ]. 4) Express the components of the heliocentric flyby exit velocity on Orbit 2 expressed in the Radial – Transversal – Out -of-plane Reference Frame of Orbit 1 [Output in vectorial form in km/s with 2 decimal digits] . Fly -by characterisation 5) Find the components of the variation of relative velocity (delta -vꝏ) provided by the fly -by expressed in the Radial – Transversal – Out -of-plane Reference Frame of Orbit 1 [Output in vectorial form in km/s with 2 decimal digits] . 6) Find the magnitude of the variation of relative orbit velocity ( delta -vꝏ) provided by the fly -by [Output in km/s with 2 decimal digits] . 7) Find the magnitude of the planet velocity at the fly -by [Output in km/s with 2 decimal digits]. 8) Draw the triangle of velocity at the fly -by in the Radial – Transversal – Out -of-plane Reference Frame of Orbit 1. Clearly indicate [Answer on sheet] : a. The reference frame with the direction of the three axes b. Direction and magn itude of the incoming and outgoing heliocentric velocity on Orbit 1 and Orbit 2 c. Direction and magnitude of the incoming and outgoing relative velocity with respect to the planet d. The direction and magnitude of the planet velocity 9) Find the magnitude of velocity at infinity and the turn angle of the fly by hyperbola [Output in km/ s and degrees with 2 decimal digits]. 10) Find the eccentricity of the fly -by hyperbola [Output with 4 decimal digits]. 11) Find the radius of peri centre of the fly -by hyperbola [Output in km with 2 decimal digits]. 12) Find the aiming radius of the fly -by hyperbola [Output in km with 2 decimal digits]. 13) Prove symbolically that the delta -v provided by the fly -by corresponds to that of a plane change manoeuvre [Answer on sheet] . Heliocentric orbit after the fly -by 14) Find when the s/c on Orbit 2 will meet again the planet after the fly -by. Express the result by showing k and j integer numbers that corresponds , respectively, to the number of revolution s of the spacecraft on Orbit 2 and the number of revolution s of the planet on their own orbit s. 15) Express the eccentricity vector of Orbit 2 in the Heliocentric Ecliptic Inertial Reference Frame [Outpu t in vectorial form with 4 decimal digits]. Part 1 =============== EX 1 ================================ =============================================== Case 1 Initial Data: V_inf_m = 0.350 km/s V_perigee = 4.970 km/s DV_aps = 0.080 km/s A_sc = 0.500 m2 m_sc = 42.000 kg Q1) ------------------ perigee radius: 3485.011 km semi major axis: -349616.327 km Q2) ------------------ DV breaking: 0.586 km/s Q3) ------------------ eccentricity 2: 0.564 flight path angle 2: 10.728 deg semi-major axis 2: 7992.506 km semi major axis 3 : 9162.789 km eccentricity 3 : 0.618 perigee altitude 3 : 121.210 velocity at manoeuvre point 3: 4.326 km/s velocity at manoeuvre point 3: [0.800, 4.252, 0.000] km/s Q4) ------------------ true anomaly at manoeuvre point 3: 28.065 deg Q3) ------------------ true anomaly at entrance: -24.900 deg true anomaly at exit: 24.900 deg Q5) ------------------ Time in the atmospheric passagae: 11.661 min Period of orbit 3: 7.397 h Q6) ------------------ Energy loss: 2306653.297 kg m2/s2 (J) =============================================== Case 2 Initial Data: V_inf_m = 0.380 km/s V_perigee = 4.970 km/s DV_aps = 0.100 km/s A_sc = 0.700 m2 m_sc = 45.000 kg Q1) ------------------ perigee radius: 3488.119 km semi major axis: -296592.798 km Q2) ------------------ DV breaking: 0.588 km/s Q3) ------------------ eccentricity 2: 0.564 flight path angle 2: 10.724 deg semi-major axis 2: 7994.060 km semi major axis 3 : 9516.415 km eccentricity 3 : 0.632 perigee altitude 3 : 127.135 velocity at manoeuvre point 3: 4.344 km/s velocity at manoeuvre point 3: [0.802, 4.269, 0.000] km/s Q4) ------------------ true anomaly at manoeuvre point 3: 27.626 deg Q3) ------------------ true anomaly at entrance: -24.148 deg true anomaly at exit: 24.148 deg Q5) ------------------ Time in the atmospheric passagae: 11.277 min Period of orbit 3: 7.829 h Q6) ------------------ Energy loss: 3158746.456 kg m2/s2 (J) =============================================== Case 3 Initial Data: V_inf_m = 0.400 km/s V_perigee = 4.970 km/s DV_aps = 0.150 km/s A_sc = 0.900 m2 m_sc = 47.000 kg Q1) ------------------ perigee radius: 3490.337 km semi major axis: -267675.000 km Q2) ------------------ DV breaking: 0.590 km/s Q3) ------------------ eccentricity 2: 0.563 flight path angle 2: 10.721 deg semi-major axis 2: 7995.168 km semi major axis 3 : 10537.991 km eccentricity 3 : 0.667 perigee altitude 3 : 135.915 velocity at manoeuvre point 3: 4.392 km/s velocity at manoeuvre point 3: [0.808, 4.317, 0.000] km/s Q4) ------------------ true anomaly at manoeuvre point 3: 26.601 deg Q3) ------------------ true anomaly at entrance: -22.876 deg true anomaly at exit: 22.876 deg Q5) ------------------ Time in the atmospheric passagae: 10.593 min Period of orbit 3: 9.123 h Q6) ------------------ Energy loss: 3931341.303 kg m2/s2 (J) =============================================== Case 4 Initial Data: V_inf_m = 0.420 km/s V_perigee = 4.970 km/s DV_aps = 0.200 km/s A_sc = 1.000 m2 m_sc = 50.000 kg Q1) ------------------ perigee radius: 3492.671 km semi major axis: -242789.116 km Q2) ------------------ DV breaking: 0.592 km/s Q3) ------------------ eccentricity 2: 0.563 flight path angle 2: 10.718 deg semi-major axis 2: 7996.335 km semi major axis 3 : 11820.531 km eccentricity 3 : 0.702 perigee altitude 3 : 144.235 velocity at manoeuvre point 3: 4.440 km/s velocity at manoeuvre point 3: [0.813, 4.365, 0.000] km/s Q4) ------------------ true anomaly at manoeuvre point 3: 25.667 deg Q3) ------------------ true anomaly at entrance: -21.675 deg true anomaly at exit: 21.675 deg Q5) ------------------ Time in the atmospheric passagae: 9.951 min Period of orbit 3: 10.838 h Q6) ------------------ Energy loss: 4228627.670 kg m2/s2 (J) =============================================== Case 5 Initial Data: V_inf_m = 0.460 km/s V_perigee = 4.970 km/s DV_aps = 0.300 km/s A_sc = 1.200 m2 m_sc = 53.000 kg Q1) ------------------ perigee radius: 3497.691 km semi major axis: -202400.756 km Q2) ------------------ DV breaking: 0.596 km/s Q3) ------------------ eccentricity 2: 0.563 flight path angle 2: 10.712 deg semi-major axis 2: 7998.845 km semi major axis 3 : 15703.405 km eccentricity 3 : 0.775 perigee altitude 3 : 159.774 velocity at manoeuvre point 3: 4.537 km/s velocity at manoeuvre point 3: [0.824, 4.461, 0.000] km/s Q4) ------------------ true anomaly at manoeuvre point 3: 24.027 deg Q3) ------------------ true anomaly at entrance: -19.437 deg true anomaly at exit: 19.437 deg Q5) ------------------ Time in the atmospheric passagae: 8.771 min Period of orbit 3: 16.596 h Q6) ------------------ Energy loss: 4747945.939 kg m2/s2 (J) =============== EX 2 ================================ =============================================== Case 1 Initial Data: rr = [-4885.976, 317.168, 6335.375 ] km vv = [5.160, -3.801, 5.127 ] km/s theta = 20.000 deg Q1) ------------------ hh = [25708.243, 57741.659, 16936.028 ] km2/s Q2) ------------------ alpha = 176.286 deg delta = 52.302 deg Q3) ------------------ b = 47.893 deg l = 155.516 deg Q4) ------------------ delta_sun = 23.400 deg alpha_sun = 90.000 deg Q7) ------------------ LST = 270.000 deg =============================================== Case 2 Initial Data: rr = [-1799.761, -1951.968, 9387.003 ] km vv = [6.075, -3.300, 2.031 ] km/s theta = 50.000 deg Q1) ------------------ hh = [27017.815, 60683.006, 17798.745 ] km2/s Q2) ------------------ alpha = 227.323 deg delta = 74.207 deg Q3) ------------------ b = 53.479 deg l = 251.940 deg Q4) ------------------ delta_sun = 23.400 deg alpha_sun = 90.000 deg Q7) ------------------ LST = 270.000 deg =============================================== Case 3 Initial Data: rr = [11038.446, -7031.936, 7218.728 ] km vv = [3.797, -0.733, -3.265 ] km/s theta = 115.000 deg Q1) ------------------ hh = [28248.263, 63446.637, 18609.337 ] km2/s Q2) ------------------ alpha = 327.501 deg delta = 28.879 deg Q3) ------------------ b = 14.856 deg l = 319.823 deg Q4) ------------------ delta_sun = 23.400 deg alpha_sun = 90.000 deg Q7) ------------------ LST = 270.000 deg =============================================== Case 4 Initial Data: rr = [4650.697, 2876.184, -16865.639 ] km vv = [-4.011, 1.686, 0.340 ] km/s theta = 225.000 deg Q1) ------------------ hh = [29412.990, 66062.658, 19376.634 ] km2/s Q2) ------------------ alpha = 31.734 deg delta = -72.036 deg Q3) ------------------ b = -53.958 deg l = 63.524 deg Q4) ------------------ delta_sun = 23.400 deg alpha_sun = 90.000 deg Q7) ------------------ LST = 270.000 deg =============================================== Case 5 Initial Data: rr = [-8936.362, 6553.116, -8777.125 ] km vv = [-2.810, -0.190, 4.912 ] km/s theta = 285.000 deg Q1) ------------------ hh = [30522.051, 68553.649, 20107.259 ] km2/s Q2) ------------------ alpha = 143.747 deg delta = -38.381 deg Q3) ------------------ b = -22.688 deg l = 133.249 deg Q4) ------------------ delta_sun = 23.400 deg alpha_sun = 90.000 deg Q7) ------------------ LST = 270.000 deg Part 2 Version a orbit 2 e orbit 2 V^ - [km/s] V^+ [km/s] DV [km/s] |DV| [km/s] |v_P| [km/s] alpha_P [deg] v_inf [km/s] 1 1235331571.6 0.3697 [0;15.28;0] [0;15.22;1.33] [0; -0.06;1.33] 1.33 13.06 2.50 2.31 2 943316109.2 0.1746 [0;14.15;0] [0;14.07;1.48] [0; -0.08;1.48] 1.48 13.06 3.00 1.31 3 943316109.2 0.1746 [0;14.15;0] [0;13.98;2.21] [0; -0.17;2.21] 2.22 13.06 4.50 1.53 4 670708310.4 0.1608 [0;11.96;0] [0;11.84;1.66] [0; -0.12;1.66] 1.67 13.06 4.00 1.40 5 594042704.1 0.3106 [0;10.84;0] [0;10.76;1.32] [0; -0.08;1.32] 1.32 13.06 3.50 2.33 Version delta [deg] e flyby [ -] r_p flyby [km] Delta flyby [km] [k_SC,k_P] e_vec [ -] 1 33.57 3.4628 58574320.3 78848693.6 [1,2] [-0.0032;0.3697;0.0014] 2 69.13 1.7625 56703866.4 107929126.3 [3,4] [0.0657;0.1618;0.0021] 3 93.16 1.3768 20431463.0 51315908.8 [3,4] [0.1338;0.1121;0.0059] 4 73.12 1.6787 43837321.8 87089217.2 [5,4] [-0.0749; -0.1423; -0.0022] 5 32.97 3.5237 58799259.1 78722568.7 [3,2] [-0.2545; -0.1780; -0.0073]