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)

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

Hohmann Transfer Δv2 Δv1 Transfer Ellipse Orbit 1 Orbit 2 r1 r2 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. Username: anonymous Password: (leave blank) Download Scenario: Hohmann transfer Username: anonymous Password: (leave blank)

Hohmann Transfer ΔvTot = Δv1 + Δv2 Transfer Ellipse Orbit 1 Orbit 2 r1 r2 The semi-major axis of the Hohmann ellipse can be determined by the equation at = (r1 + r2) / 2, where r1 is the radius of the original orbit and r2 is the radius of the new orbit. To find the total change in velocity, ΔvTot, we must add Δv1 and Δv2 together. ΔvTot = Δv1 + Δv2

Hohmann Transfer Δv1 = v1 – vcirc1 Δv2 = vcirc2 – v2 vcirc1 and vcirc2 can be found by using the following equation for the velocity of a circular orbit: vcirc1,circ2 = √(𝜇/ r1,2) v1 and v2 can be found by rearranging the energy equation for an orbit: 𝜀=v2/2 - 𝜇/r → v1,2 = √(2(𝜀t + 𝜇/r1,2)), where the energy of the Hohnman Transfer ellipse is given by 𝜀t = -𝜇/2at v2 vcirc1 v1 Δv1 vcirc2 Δv2 2 at

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.

General Coplanar Transfers Using more Δv than required to reach the outer orbit will put the satellite in a faster trajectory, however it will also require the satellite to change directions upon arriving at the desired outer orbit. To find the direction change we must first define the local horizontal: The direction perpendicular to the satellite’s position vector. The angle between the satellite’s velocity and the local horizontal is called the flight path angle (φ). For a circular orbit, the satellite’s velocity is perpendicular to its position, thus φ = 0. v2 Δv2 vcirc2 φ2

General Coplanar Transfers Flight path angle can be found by using conservation of angular momentum. Assuming gravity is the only external force acting on the satellites, the specific angular momentum (h) of the satellite is conserved. h= r v cos(φ) Also, using the conservation of energy and some algebraic manipulation yields equations for 𝛥𝑣 for general coplanar transfers. Thus, Hohmann Transfers are just a special case of coplanar transfers where φ = 0. v2 Δv2 vcirc2 φ2 v1 vcirc1 Δv1 Δ𝑣12=𝑣12+𝑣𝑐𝑖𝑟𝑐1−2𝑣1 𝑣𝑐𝑖𝑟𝑐1 𝑐𝑜𝑠(φ1) Δ𝑣22=𝑣22+𝑣𝑐𝑖𝑟𝑐2−2𝑣2 𝑣𝑐𝑖𝑟𝑐2 𝑐𝑜𝑠(φ2) Username: anonymous Password: (leave blank) Download Scenario: Fast vs Hohmann

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. Δv v Δi

End Lesson 3 - Tutorial Complete tutorial to further explore lesson: Hohmann Transfer Tutorial Username: anonymous Password: (leave blank)

End Lesson 3 - Exercises True or False: Short Answer: 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 (rLEO = r1 = 7,000 km) to a GEO (rGEO = r2 = 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 (rLEO = r1 = 7,000 km) to a GEO (rGEO = r2 = 42,164 km) orbit with a transfer orbit tangent to the LEO orbit and a v1 = 2.75 km/s. What is the total Δv required? Username: anonymous Password: (leave blank)

End Lesson 3 - Answers True or False: Short Answer: 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 m2/s2 5. We would like to transfer our satellite from a LEO orbit (rLEO = r1 = 7,000 km) to a GEO (rGEO = r2 = 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 (rLEO = r1 = 7,000 km) to a GEO (rGEO = r2 = 42,164 km) orbit with a transfer orbit tangent to the LEO orbit and a v1 = 2.75 km/s. What is the total Δv required? See Example on slide 15 Username: anonymous Password: (leave blank)

End Lesson 3 - Answers 5. Hohmann Transfer Example: We would like to transfer our satellite from a LEO orbit (rLEO = r1 = 7,000 km) to a GEO (rGEO = r2 = 42,164 km) orbit. What is the total change in velocity required? Step 1 - Solve for semi-major axis of the Hohmann transfer. at = (rLEO + rGEO) / 2 = (7,000 + 42,164) / 2 = 24,582 km Step 2 - Solve for circular orbit velocities. vcirc1 = vLEO = √(𝜇/rLEO) = √(3.986x105/7,000) = 7.546 km/s vcirc2 = vGEO = √(𝜇/rGEO) = √(3.986x105/42,164) = 3.075 km/s Step 3 - Determine the energy of the Hohmann transfer. 𝜀t = -𝜇/2at = -3.986x105/2*24,582 = -8.108 Step 4 - Solve for periapsis and apoapsis velocities of the transfer. v1 = √(2(𝜀t + 𝜇/r1)) = √(2(-8.108 + 3.986x105/7,000)) = 9.883 km/s v2 = √(2(𝜀t + 𝜇/r2)) = √(2(-8.108 + 3.986x105/42,164)) = 1.640 km/s Step 5 - Determine the change in velocities for each maneuver. Then find the total change, ΔvTot. Δv1 = v1 - vLEO = 2.337 km/s Δv2 = vGEO – v2 = 1.435 km/s ΔvTot = 2.337 + 1.435 = 3.772 km/s v2 vcirc1 v1 Δv1 vcirc2 Δv2 2at

End Lesson 3 - Answers 6. Fast Transfer Example: We would like to transfer our satellite from a LEO orbit (rLEO = r1 = 7,000 km) to a GEO (rGEO = r2 = 42,164 km) orbit with a transfer orbit tangent to the LEO orbit and a v1 = 2.75 km/s. What is the total Δv required? Step 1 - Solve for circular orbit velocities. vcirc1 = vLEO = √(𝜇/rLEO) = 7.546 km/s vcirc2 = vGEO = √(𝜇/rGEO) = 3.075 km/s Step 2 - Determine v1 and v2 energy of the transfer orbit. Δv1= 3 km/s = v1-vcirc1 → v1 = 10.296 km/s 𝜀t = v12/2- 𝜇/rLEO = -3.939 km2/s2 𝜀t = v22/2- 𝜇/rGEO → v2 = 3.321 km/s Step 3 - Determine the flight path angle at location of Δv2. ht = r1v1cos(φ1) = 72072 km2/s ht = r2v2cos(φ2) → cos(φ2) = 0.5147 Step 4 - Determine Δv2 and ΔvTot. Δv22 = v22 + vGEO2 – 2 v2 vGEO cos(φ2) → Δv2 = 3.312 km/s ΔvTot = 2.75 + 3.158 = 5.908 km/s v2 Δv2 vcirc2 φ2 v1 vcirc1 Δv1 Download Scenario: Fast vs Hohmann to compare ΔvTot =5.91 km/s vs ΔvTotHohm= 3.77 km/s ToF = 2.65 hours vs ToFHohm = 4.97 hours (ToF=time of flight)

Additional Resources STK Data Federate link for scenario vdf’s: EAP Scenarios Username: anonymous Password: (leave blank) Check out related additional scenarios: Flight Path Angle Eccentric Anomaly Mean Anomaly Visit AGI.com/edu to learn more about the available curriculum and the Educational Alliance Program (EAP) To learn more about STK: Beginner-level video tutorials Complete STK Certification Questions / Comments: Education@agi.com Username: anonymous Password: (leave blank)