AEM_360_Homework#3

AEM 360 ASTRONAUTICS Homework #3 – Orbits II Fall 2023 Name: none Show all needed work to answer the questions. Upload your answer on Blackboard. Name the file “AEM_360_Homework#2_<last name>.doc” Include any needed figures or equations (typed using equation editor). 1. (10 points) Based on the ground track of the direct (prograde) orbit shown here, what is the inclination AND period of the orbit? Given: The satellite passes through its ascending node at 70°W and again at 50°E longitude. a) Inclination (i) equals the highest latitude of the spacecraft’s orbit: ∴ i = 20 ° b) Period is P = ∆ N 15 ° / hr : First, find nodal displacement: ∆ N = 360 ° −( longitude betweennodes ) ∆ N = 360 ° − ( 50 ° −(− 70 ° ) ) = 240 ° ∴ P = 240 ° 15 ° / hr = 16 hr 1 120 o

2. (10 points) Which hemisphere (North, South) is the apogee of orbit that produces ground track B shown at the right? The apogee of ground track B is in the Northern hemisphere . You can tell because its ground track is scrunched and does not span very far longitudinally, meaning it is travelling relatively slow indicating that it is at the widest point of its elliptical orbit (apogee). 2

Finally, solve for V T1 : V T 1 = √ 2 ( μ Earth R orbit 1 + ε transfer ) V T 1 = √ 2 ( 3 . 986 × 10 5 km 3 s 2 ( 6738 + 400 ) km − 8 . 0855 k m 2 s 2 ) ANSWER : →V T 1 = 9.773 km s b) Find V orbit 1 : V 1 = √ 2 ( μ Earth R orbit 1 + ε orbit 1 ) ,where ε orbit 1 = − μ Earth 2 a orbit 1 ∧ a orbit 1 = R orbit 1 First, find ε orbit 1 : ε orbit 1 = − μ Earth 2 a orbit 1 ε orbit 1 = − 3 . 986 × 10 5 km 3 s 2 2 ( 6738 + 400 ) →ε orbit 1 =− 27.92099 k m 2 s 2 Second, find V 1 : V 1 = √ 2 ( μ Earth R orbit 1 + ε orbit 1 ) V 1 = √ 2 ( 3 . 986 × 10 5 km 3 s 2 ( 6738 + 400 ) km − 27 . 92099 k m 2 s 2 ) ANSWER : →V 1 = 7.472 km s c) Find ΔV 1 : Use relation between all three velocities: ΔV 1 = | V transfer at orbit 1 − V orbit 1 | 4

ΔV 1 = | 9.773 − 7.472 | ANSWER : → ΔV 1 = 2.301 km s 4. (10 points) For the new weather satellite in problem #3 above, continue the calculations to determine the key values for the second part of the Hohman Transfer to a geostationary orbit (orbit 2). Given: R Earth = 6378 km (Earth’s equatorial radius) Orbit 1 altitude = 400 km Orbit 2 radius = 42,160 km µ Earth = 3.986 × 10 5 km 3 / s 2 Find: V orbit 2 , or V 2; V transfer at orbit 2 , or V T2 ; ΔV 2 , total ΔV a) Find V orbit 2 : V 2 = √ 2 ( μ Earth R orbit 2 + ε orbit 2 ) ,where ε orbit 2 = − μ Earth 2 a orbit 2 ∧ a orbit 2 = R orbit 2 First, find ε orbit 2 : ε orbit 2 = − μ Earth 2 a orbit 2 ε orbit 2 = − 3 . 986 × 10 5 km 3 s 2 2 ( 42160 ) km →ε orbit 2 =− 4.2723 k m 2 s 2 5

c) Find ΔV 1 : Use relation between all three velocities: ΔV 2 = | V orbit 2 − V transfer at orbit 2 | ΔV 2 = | 3.219 − 1.655 | ANSWER : → ΔV 2 = 1.564 km s 5. (10 points) Assume the European Space Agency (ESA) launched a remote sensing satellite directly east into a circular low Earth orbit from the Guiana Space Centre, which is located at 5.305°N, 52.834°W (Kourou, French Guiana). The satellite’s mission requires its orbit to lie precisely in the Earth’s equatorial plane. The agency plans to perform a plane change maneuver at the ascending node soon. What is the ΔV simple required to change the orbit’s current inclination to an equatorial inclination? Given: Orbital altitude = 500 km R Earth = 6378 km (Earth’s equatorial radius) µ Earth = 3.986 × 10 5 km 3 / s 2 7 Final Inclination (0 o ) Initial Inclination (5.305 o )

First, find plane-change angle ( θ ): θ = | i final − i final | θ = 5.305 ° Second, find ε to solve for V initial : ε = − μ 2 a = − μ 2 R ( circular orbit ) ε = − 3.986 × 10 5 km 3 s 2 2 ( 6378 + 500 ) km →ε =− 28.9764 k m 2 s 2 Third, find V initial : V initial = √ 2 ( μ Earth R + ε ) V initial = √ 2 ( 3 . 986 × 10 5 km 3 s 2 ( 6378 + 500 ) km − 28 . 9764 k m 2 s 2 ) V initial = 7 . 613 m s Finally, solve for V simple : V simple = 2 V initial sin ( θ 2 ) V simple = 2 ( 7 . 613 ) sin ( 5 . 305 2 ) ANSWER : →V simple = 0.7046 km s 6. (10 points) In the near future, an automated refueling spacecraft in a circular, equatorial low Earth orbit will need to dock with and refuel a geostationary communications satellite. What is the lead angle, α lead , and the phase angle, ϕ final , for the rendezvous Hohman Transfer? Given: R interceptor = 6828 km R target = 42,160 km 8

Semimajor axis for Ceres, a Ceres = 414,010,000 km = 4.1401 x 10 8 km m Ceres = 9.393 x 10 20 kg m Sun = 1.9891 x 10 30 kg R SOI = a planet ( m planet m Sun ) 2 5 R SOI ,Ceres =( 4 . 1401 x 10 8 km ) ( 9 . 393 x 10 20 kg 1 . 9891 x 10 30 kg ) 2 5 ANSWER : →R SOI ,Ceres = 77157.61 km 8. (10 points) In the patched-conic approximation for interplanetary transfers, what is the hyperbolic excess velocity (only) of a spacecraft leaving Mars to return to Earth. Given: Altitude at Mars is 1000 km Return parking orbit (Earth) will be altitude of 1000 km departure : v ∞ Mars = | V transfer at Mars − V Mars | arrival : v ∞ Earth = | V Earth − V transfer at Earth | First, find v ∞ Earth : v Earth = √ 2 ( μ Sun R ¿ Earth + ε Earth ) ,w here ε Earth = − μ sun 2 a Earth 10

v Earth = √ 2 ( 1 . 327 ∗ 10 11 1 . 496 ∗ 10 8 − 1 . 327 ∗ 10 11 2 . 992 ∗ 10 8 ) →v Earth = 29 . 783 km s V transfer at Earth = √ 2 ( μ Sun R ¿ Earth + ε transfer ) ,w hereε transer = − μ sun R ¿ Earth + R ¿ Mars V transfer at Earth = √ 2 ( 1 . 327 ∗ 10 11 1 . 496 ∗ 10 8 − 1 . 327 ∗ 10 11 1 . 496 ∗ 10 8 + 2 . 278 ∗ 10 8 ) →V transfer at Earth = 32 . 724 km s Next, find v ∞ Earth : →v ∞ Earth = | 29.783 − 32.724 | = 2.941 km s Finally, solve for hyperbolic arrival velocity at Earth: v h yperbolic @Eart h = √ 2 ( μ Earth R parkat Earth + ε ∞ Earth ) ,where ε ∞Earth = V ∞Earth 2 2 v h yperbolic @Eart h = √ 2 ( 3 . 986 × 10 5 1000 + ( 2 . 941 ) 2 2 ) ANSWER : →v hyperbolic@ Earth = 28.388 km / s 9. (10 points) Your team just received their first telemetry from your satellite via our local ground station at the University of Alabama. The Built by Bama Sat (B3Sat) is a 6U cubesat (10 kg, 10 x 20 x 30 cm). Assume 11

R = 36130.4 ( 1 −( 0.81235 ) 2 ) 1 +( 0.81235 ) cos ( 115.3 ° ) = 18821.725 km V = √ 2 ( μ R + ε ) ,where ε = − μ 2 a V = √ 2 ( 3.986 × 10 5 18821.725 − 3.986 × 10 5 2 ( 36130.4 ) ) →V = 5.597 km s h = ( R ) ( V ) sin ( θ ) ,where θis 90 ° sinceV is perpendicular ¿ R h = 18821.725 ∗ 5.597 ∗ 1 h = 105345.195 k m 2 s Find state vector r: cos θ ^ u h 2 μ 1 1 + e cos θ ( ¿¿ p + sin θ ^ u q ) ,use v for angle . r = ¿ 105345.195 ¿ ¿ ¿ 2 ¿ ¿ ¿ r = ¿ →r = ⟨ − 18263.06,38572.36,0 ⟩ km Find state vector V: − sin θ ^ u ¿ θ e + cos ¿ μ h ¿ v = ¿ 13

− sin ( 2.013 ) ^ u ¿ ( 2.013 ) 0.81235 + cos ¿ 3.986 × 10 5 105345.195 ¿ ¿ →v = ⟨ − 3.4198,1.4545,0 ⟩ m s Conversion matrix from inertial to perifocal: [ Q ] X ´ x = [ cos243.8 − sin 243.8 0 sin 243.8 cos243.8 0 0 0 1 ][ 1 0 0 0 cos112.74 − sin 112.74 0 sin112.74 cos112.74 ][ cos355.41 − sin 355.41 0 sin 355.41 cos355.41 0 0 0 1 ] [ Q ] X ´ x = [ − .4415 .8973 0 − .8973 − .4415 0 0 0 1 ][ 1 0 0 0 − .3866 − .9223 0 .9223 − .3866 ][ .9968 .08 0 − .08 .9968 0 0 0 1 ] [ Q ] X ´ x = [ − 0.4123 − 0.3811 − 0.8276 − 0.9081 0.09835 0.4072 − 0.0738 0.9194 − 0.3866 ] Conversion matrix from perifocal to inertial: [ Q ] ´ x X = [ − 0.4123 − 0.3811 − 0.8276 − 0.9081 0.09835 0.4072 − 0.0738 0.9194 − 0.3866 ] − 1 = [ − 0.4123 − 0.9080 − 0.0738 − 0.3810 0.09835 0.9192 − 0.8275 0.4071 − 0.3866 ] Transform to geocentric frame: { ´ r } X = [ Q ] ´ x X { ´ r } ´ x = [ − 0.4123 − 0.9080 − 0.0738 − 0.3810 0.09835 0.9192 − 0.8275 0.4071 − 0.3866 ][ − 18263.06 38572.36 0 ] = [ − 27493.84 10751.82 30815.49 ] ( km ) { ´ v } X = [ Q ] ´ x X { ´ v } ´ x = [ − 0.4123 − 0.9080 − 0.0738 − 0.3810 0.09835 0.9192 − 0.8275 0.4071 − 0.3866 ][ − 3.4198 1.1454 0 ] = [ 0.3700 1.4156 3.2962 ] ( km / s ) 10. (10 points) From #9, determine the orbit position in 5 hours from the time stamp in the TLE. Use Kepler’s equation. 14

Fifth guess, E future = 2.3915 : 1.9393 +( 0.81235 ) sin ( 2.3915 ) = 2.4931 Sixth guess, E future = 2.4931 : 1.9393 +( 0.81235 ) sin ( 2.4931 ) = 2.4300 Seventh guess, E future = 2.4300 : 1.9393 +( 0.81235 ) sin ( 2.4300 ) = 2.4700  Eventually, E future = 2.4550 rad . Solve for v future : v future = cos − 1 ( cos ( E future ) − e 1 − ecos ( E future ) ) 2.455 cos ( ¿ − 0.81235 ) ¿ ( ¿ 1 −( 0.81235 ) cos ( 2.455 ) ¿ ) ¿ v future = cos − 1 ¿ ANSWER : →v future = 2.9126 rad 16