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ESCAPE VELOCITY Example AAE 439 ESCAPEESCAPE VELOCITYVELOCITY !Kinetic energy equals work required to overcome gravity. Ekin = Epot = Wgravity 1 2 Ekin = 2 m vz !Newton’s Law of Gravity: Epot = m g E !z G !ME m µE m !Earth’s Surface: F = m g = = r +z r +z 2 grav E r2 r2 E E r E E W = F !dr = m g E !dr r2 gravity " " E r2 E rE rE !Any distance r > rE: F = m g grav E r2 1 z 2g r z !Velocity Requirement: m v2 = m g r v = E E 2 z E E r + z z r + z E E 2µ !Escape Velocity: z ! " v = 2 g r = E esc E E r E !Escape Velocity for selected bodies: !Earth: 11.2 km/s !Sun: 616 km/s !Moon: 2.4 km/s Ch3 –30 AAE 439 ExampleExample 19,300 Nm 400 Nm Earth Ch3 –31 AAE 439 3.53.5 ORBITORBIT MANEUVERINGMANEUVERING Ch3 –32 AAE 439 HOHMANNHOHMANN TRANSFERTRANSFER !Hohmann Transfer: Minimum energy elliptical transfer path for co-planar transfer between two circular orbits. !!v Requirement: !vtotal = !vA + !vB = ) 1 1 1 1 , 1 + # 2 1 & 2 # 1 & 2 # 2 1 & 2 # 1 & 2 . = µ 2 " " + " " + % r a ( % r ( % r a ( % r ( . + $ A tx ' $ A ' $ B tx ' $ B ' . * Transfer Ellipse - Ch3 –33 AAE 439 HohmannHohmann TTransferransfer EquationsEquations !Given the initial and final circular orbit radii, we can calculate the velocity changes required to transfer between the circular orbits. Orbit A: 6,567 km Orbit B: 42,160 km Period of an orbit [min] a3 P = 2! µ Obit Frequency [rad/s] µ !0 = 3 a Ch3 –34 AAE 439 OTHEROTHER COPLANARCOPLANAR ORBITORBIT CHANGESCHANGES !One-Tangent Burn between two Circular, Coplanar Orbits !Transfer requires less time to complete than Hohmann Transfer. !Total Velocity Change: !vtotal = !vA + !vB !vA = vtxA " viA 2 2 !vB = vfB + vtxB " 2vfB vtxB cos# Variable Hohmann Transfer One-Tangent Burn rA 6,570 km 6,570 km rB 42,200 km 42,200 km atx 24,385 km 28,633 km !vtotal 3.935 km/s 4.699 km/s Time-of-Flight 5.256 hr 3.457 hr !Spiral Transfer between two Circular, Coplanar Orbits !Constant low-thrust burn results in a spiral transfer. !v = !v " !v total f i Correction Ch3 –35 AAE 439 SURFACESURFACE VELOCITYVELOCITY !Launch into Easterly Direction uorbit KSC usurface ! Equator North Pole Orbit 2! r "cos# !Surface Velocity: u = E surf 1 day !Overall !v for LEO !v = uorb " usurf Ch3 –36 AAE 439 HohmannHohmann TTransferransfer !Example: !Initial Orbit: 5,000 km !Final Orbit: 40,000 km !Let’s examine and understand the meaning of some of the individual terms in the Hohmann-Transfer equation. Ch3 –37 AAE 439 ORBITORBIT PLANEPLANE CHANGECHANGE !Simple Plane Change: Orbit-Plane (Inclination) Change !Direction change of velocity vector, !Maneuver with component of !v perpendicular to orbital plane. !Law of Cosines: v v v f " i = f = !v = 2v sin 2 i !Plane and Altitude Change: !Hohmann & Simple Plane Change !Plane Change & Tangential Burn f 2 2 !v = vi + vf " 2vivf cos# i !Circularization at Apogee µ r v = Earth 2µ p µ 1" e !v = vCircOrbit " vapogee CO vapogee = ! = ! r CO ra + rp ra a 1+ e Ch3 –38 AAE 439 !!vv BudgetBudget forfor OrbitOrbit PlanePlane ChangesChanges !Example: Change low altitude (h=185 km) inclined orbit (i = 28°) to equatorial orbit at same altitude !!v=3.8 km/s !At ro=6563 km, v=7.8 km/s, velocity increment requirement is nearly half the orbital velocity! !For an inclination change of 60o, the plane change !v is approximately equal to the ideal !v required to get to orbit. !Plane changes can be very expensive maneuvers: !Do plane change when velocity is lowest (at the apogee). !Sometimes it is more economical to boost vehicle to a higher orbit, perform plane change at low v (apogee), and then return to lower orbit. Ch3 –39 AAE 439 PegasusPegasus LaunchLaunch VehicleVehicle !Effects of Inclination and Orbit Height on Payload Technical Data: http://www.astronautix.com/lvs/pegasus.htm Ch3 –40.
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