Presented by: Jamie Quinnell Jean Moiso Gus Mashensic

Definition of Hohmann Transfer: Transfer between two circular orbits, sharing a focus and in the same orbital plane Things to note: 1. Path of spacecraft is elliptical 2. Position does NOT change during impulsive maneuver 3. First impulse is at periapsis and the second is at apoapsis 4. Period of an elliptical orbit is defined as Therefore, the time it takes for a hohmann transfer can be found to be Δt xfer = ½ T =

Hohmann Transfer Visual

Vis-Viva: Energy of an Orbit ξ = ½ v 2 - = Solve for v to get: Also,

Application: Satellite needs to go from LEO (Low Earth Orbit) to GEO (Geosynchronous Earth Orbit) Problem Statement: A satellite is to undergo a Hohmann Transfer from a LEO of r i = 7000km to a GEO of r f = 42164km. 1. What is the time required for this transfer? 2. Rank the values of energy from most negative to least negative for orbit 1, the Hohmann elliptical orbit and orbit What is the Δv required to complete this transfer?

So, = s ≈ 5 hours and 19 minutes What is the time required for this transfer? μ = GM (Earth Gravitational Parameter) μ = x 10 5 km 3 /s 2 1.

Rank the values of energy from most negative to least negative for orbit 1, the hohmann elliptical orbit and orbit 2. ξ 1 = - μ/2a 1 a 1 = r i = 7000km ξ H = - μ/2a H a H = 24582km ξ 2 = - μ/2a 2 a 2 = r f = 42164km From observation, the energies only differ due to the value of a. Therefore you can rank the energies without doing any calculations. ξ 1, ξ H, ξ 2 2.

What is the Δv required to complete this transfer? ∆V 1 = V p - V 1 ∆V 2 = V 2 - V a 3.

Calculate all velocities needed. V 1 = sqrt(μ/r 1 ) V 1 = 7.546km/s V 2 = sqrt(μ/r 2 ) V 2 = 3.075km/s V p = sqrt[2(ξ H + μ/r p )] V p = sqrt[2((-μ/2a H ) + μ/r 1 )] V p = 9.883km/s V a = sqrt[2(ξ H + μ/r a )] V a = sqrt[2((-μ/2a H ) + μ/r 2 )] V a = 1.641km/s

∆V 1 = V p -V 1 ∆V 1 = 2.337km/s ∆V 2 = V 2 - V a ∆V 2 = 1.434km/s ∆V total = | ∆V 1 | + | ∆V 2 | ∆V total = 3.771km/s

We can relate this to Dynamics by using Linear Impulse. M i V i + Σ∫F dt = M f V f Where a force (F) is applied over time (dt) at the beginning and end of the orbit change. Unfortunately Solving a problem using this method can get very messy due to: 1. The mass of the satellite is changing due to the loss of propellant as the thrusters are fired. 2. It is more involved to calculate the forces the thrusters apply than to just consider energy. 3. You must also include the time interval the thrusters are applied. Is it impossible? No Is it difficult? YES!

References Dr. Melton lecture notes Video: