AGIs EAP Curriculum Orbital Mechanics Lesson 3 Orbital

AGI’s EAP Curriculum Orbital Mechanics Lesson 3 - Orbital Transfers

Introduction § This powerpoint was designed to aid professors in teaching several concepts for an introductory orbital mechanics class. The powerpoint incorporates visuals from Systems Tool Kit (STK) that help students better understand important concepts. § Any relevant parts of the powerpoint (images, videos, scenarios, etc) can be extracted and used independently. § At the end of each lesson there are optional example problems and a tutorial that further expands upon the lesson. Students can complete these as part of a lab or homework assignment. § Here is the link to download all of the supporting scenarios: EAP Scenarios – Username: anonymous Password: (leave blank) 2

Lesson 3: Orbital Transfers § Lesson Overview – – Hohmann transfer General coplanar transfers Simple plane changes Orbit manuever visualization 3

Hohmann Transfer A Hohmann Transfer moves a satellite between two circular, coplanar, and concentric orbits by applying two separate impulsive maneuvers (velocity changes). This type of transfer is the most fuel-efficient. The first impulse is used to bring the satellite out of its original orbit. The satellite then follows a transfer ellipse, known as a Hohmann ellipse, to its apoapsis point located at the radius of the new orbit. A second impulse is then used to return the satellite into a circular orbit at its new radius. Orbit 2 Δv 2 Orbit 1 r 2 Δv 1 Transfer Ellipse Download Scenario: Hohmann transfer Username: anonymous Password: (leave blank) 4

Hohmann Transfer The semi-major axis of the Hohmann ellipse can be determined by the equation at = (r 1 + r 2) / 2, where r 1 is the radius of the original orbit and r 2 is the radius of the new orbit. To find the total change in velocity, Δv. Tot, we must add Δv 1 and Δv 2 together. Orbit 2 Δv 2 Orbit 1 r 2 Δv. Tot = Δv 1 + Δv 2 Δv 1 Transfer Ellipse 5

Hohmann Transfer Δv 1 = v 1 – vcirc 1 Δv 2 = vcirc 2 – v 2 vcirc 1 and vcirc 2 can be found by using the following equation for the velocity of a circular orbit: Δv 2 vcirc 1, circ 2 = √(�� / r 1, 2) v 1 and v 2 can be found by rearranging the energy equation for an orbit: 2 at v 1 vcirc 2 vcirc 1 Δv 1 �� =v 2/2 - �� /r → v 1, 2 = √(2(�� /r 1, 2)), t + �� where the energy of the Hohnman Transfer ellipse is given by �� /2 at t = -�� 6

General Coplanar Transfers Although Hohmann transfers use the minimum amount of energy and fuel to reach a new orbit, they also require the most time. For missions with time constraints, a short transfer time can be achieved at the cost of more fuel. Instead of using an elliptical transfer orbit that just reaches the outer orbit, using a transfer ellipse which extends past the outer orbit will result in faster transfer times. 7

General Coplanar Transfers § vcirc 2 Δv 2 φ2 v 2 8

General Coplanar Transfers § vcirc 2 Δv 2 vcirc 1 v 1 φ2 Δv 1 v 2 Download Scenario: Fast vs Hohmann 9

Simple Plane Transfer For a change in inclination between two circular orbits a burn can be performed at the ascending or descending node of the orbital plane. The Δv for the plane change is given by: Δv = 2 v sin (Δi/2) This implies two important findings: 1) A large inclination change, over 60ᵒ causes Δv > v 2) Inclination changes at slower v, requires less Δv. Therefore inclination manuevers at apoapsis require less Δv. v Δv Δi v 10

End Lesson 3 - Tutorial • Complete tutorial to further explore lesson: Hohmann Transfer Tutorial 11

End Lesson 3 - Exercises True or False: 1. T / F A Hohmann Transfer is the most efficient transfer between two circular orbits. 2. T / F When using a Hohmann transfer to move to a larger circular orbit, the satellite must increase its velocity at the apoapsis of the transfer ellipse to enter the new orbit. 3. T / F If performing a pure inclination change (no other orbital elements change), the old and new orbits will not intersect. Short Answer: 4. We would like to transfer our satellite from a LEO orbit (r LEO = r 1 = 7, 000 km) to a GEO (r GEO = r 2 = 42, 164 km) orbit using a Hohmann Transfer. What is the total change in velocity required? 5. Given the mass of the satellite to be 500 kg and the gravitational force exerted on the satellite by the Earth (F=2. 87× 1031 N), what is the specific potential energy of a satellite, u? 6. We would like to transfer our satellite from a LEO orbit (r LEO = r 1 = 7, 000 km) to a GEO (r GEO = r 2 = 42, 164 km) orbit with a transfer orbit tangent to the LEO orbit and a v 1 = 2. 75 km/s. What is the total Δv required? 12

End Lesson 3 - Answers True or False: 1. T / F A Hohmann Transfer is the most efficient transfer between two circular orbits. 2. T / F When using a Hohmann transfer to move to a larger circular orbit, the satellite must increase its velocity at the apoapsis of the transfer ellipse to enter the new orbit. 3. T / F If performing a pure inclination change (no other orbital elements change), the old and new orbits will not intersect. Short Answer: 4. Given the mass of the satellite to be 500 kg and the gravitational force exerted on the satellite by the Earth (F=2. 87× 1031 N), what is the specific potential energy of a satellite, u? u= 3. 3216× 107 m 2/s 2 5. We would like to transfer our satellite from a LEO orbit (r LEO = r 1 = 7, 000 km) to a GEO (r GEO = r 2 = 42, 164 km) orbit using a Hohmann Transfer. What is the total change in velocity required? See Example on slide 14 6. We would like to transfer our satellite from a LEO orbit (r LEO = r 1 = 7, 000 km) to a GEO (r GEO = r 2 = 42, 164 km) orbit with a transfer orbit tangent to the LEO orbit and a v 1 = 2. 75 km/s. What is the total Δv required? See Example on slide 15 13

End Lesson 3 - Answers 5. Hohmann Transfer Example: We would like to transfer our satellite from a LEO orbit (r. LEO = r 1 = 7, 000 km) to a GEO (r. GEO = r 2 = 42, 164 km) orbit. What is the total change in velocity required? Step 1 - Solve for semi-major axis of the Hohmann transfer. at = (r. LEO + r. GEO) / 2 = (7, 000 + 42, 164) / 2 = 24, 582 km Step 2 - Solve for circular orbit velocities. vcirc 1 = v. LEO = √(�� /r. LEO) = √(3. 986 x 10 5/7, 000) = 7. 546 km/s vcirc 2 = v. GEO = √(�� /r. GEO) = √(3. 986 x 10 5/42, 164) = 3. 075 km/s Step 3 - Determine the energy of the Hohmann transfer. �� /2 at = -3. 986 x 10 5/2*24, 582 = -8. 108 t = -�� Step 4 - Solve for periapsis and apoapsis velocities of the transfer. v 1 = √(2(�� /r 1)) = √(2(-8. 108 + 3. 986 x 10 5/7, 000)) = 9. 883 km/s t + �� v 2 = √(2(�� /r 2)) = √(2(-8. 108 + 3. 986 x 10 5/42, 164)) = 1. 640 km/s t + �� Step 5 - Determine the change in velocities for each maneuver. Then find the total change, Δv. Tot. Δv 1 = v 1 - v. LEO = 2. 337 km/s Δv 2 = v. GEO – v 2 = 1. 435 km/s Δv. Tot = 2. 337 + 1. 435 = 3. 772 km/s 2 at v 1 Δv 2 v vcirc 2 2 vcirc 1 Δv 1 14

End Lesson 3 - Answers 6. Fast Transfer Example: We would like to transfer our satellite from a LEO orbit (r. LEO = r 1 = 7, 000 km) to a GEO (r. GEO = r 2 = 42, 164 km) orbit with a transfer orbit tangent to the LEO orbit and a v 1 = 2. 75 km/s. What is the total Δv required? Step 1 - Solve for circular orbit velocities. vcirc 1 = v. LEO = √(�� /r. LEO) = 7. 546 km/s vcirc 2 = v. GEO = √(�� /r. GEO) = 3. 075 km/s Step 2 - Determine v 1 and v 2 energy of the transfer orbit. Δv 1= 3 km/s = v 1 -vcirc 1 → v 1 = 10. 296 km/s 2 �� /r. LEO = -3. 939 km 2/s 2 t = v 1 /2 - �� 2 �� /r. GEO → v 2 = 3. 321 km/s t = v 2 /2 - �� Step 3 - Determine the flight path angle at location of Δv 2. ht = r 1 v 1 cos(φ1) = 72072 km 2/s ht = r 2 v 2 cos(φ2) → cos(φ2) = 0. 5147 Step 4 - Determine Δv 2 and Δv. Tot. Δv 22 = v 22 + v. GEO 2 – 2 v. GEO cos(φ2) → Δv 2 = 3. 312 km/s Δv. Tot = 2. 75 + 3. 158 = 5. 908 km/s vcirc 2 Δv 2 vcirc 1 v 1 φ2 Δv 1 v 2 Download Scenario: Fast vs Hohmann to compare Δv. Tot =5. 91 km/s vs Δv. Tot. Hohm= 3. 77 km/s To. F = 2. 65 hours vs To. FHohm = 4. 97 hours (To. F=time of flight) 15

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