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lecture_34 - 16.512, Rocket Propulsion Prof. Manuel...

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16.512, Rocket Propulsion Prof. Manuel Martinez-Sanchez Lecture 34: Performance to GEO V Calculations for Launch to Geostationary Orbit (GEO) Idealized Direct GTO Injection (GTO = Geosynchronous Transfer Orbit) Assumptions: - Ignore drag and "gravity" losses - Assume impulsive burns (instantaneous impulse delivery) - Assume all elevations α >0 at launch are acceptable Launch is from a latitude L, directed due East for maximum use of Earth's rotation. The Eastward added velocity due to rotation is then RE E vR c o s L 4 6 3 c o =Ω = s L (m/s) (1) If the launch elevation is α , and the desired velocity after the first burn is V 1 , the rocket must supply a velocity increment 22 11 R 1 R VVv2 V v c o s ∆= + α ( 2 ) The trajectory will then lie in a plane LOI through the Earth's center which contains the local E-W line. In order to be able to perform the plane change to the equatorial plane at GEO, we select the elevation α such as to place the apogee of the transfer orbit (GTO) at the GEO radius 1/3 2 GEO 2 T R 42,200 km 4 ⎛⎞ = ⎜⎟ π ⎝⎠ (T = 24 hr, = 3.986 × 10 µ 14 m 3 /s 2 ) 16.512, Rocket Propulsion Lecture 34 Prof. Manuel Martinez-Sanchez Page 1 of 13 Δ V 1 v 1 α v R
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Since OL is perpendicular to OI, the view in the plane of the orbit is: The polar equation of the trajectory is p r 1e c o s = + θ , > 0 In our case (corresponding to E pR = 2 π θ = ). The elevation is given by 16.512, Rocket Propulsion Lecture 34 Prof. Manuel Martinez-Sanchez Page 2 of 13 North L O R G E O I GTO Q U A T Ω E α EQUATORIAL GEO ORBIT V 1 Fig. 1 R GEO R E r L α V 1 θ o I P Fig. 2
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() 2 /2 dr e sin tan e rd 1e c o s θ=π ⎛⎞ θ ⎜⎟ α= = = θ ⎝⎠ and, in turn, the eccentricity follows from (at θ π = ) E GEO R R = E GEO R e1 R =− and so E GEO R tan 1 0.849 R = ; ( 3 ) 0 40.3 The angular momentum (per unit mass) is E hpR =µ=µ . Equating this to , E1 RV c o s α 1 E V c os R µ ( 4 ) (i.e., the horizontal projection of the launch velocity is the local orbital speed, for any apogee radius, in this case) GEO R Combining (3) and (4), 2 E 1 EG E R V1 1 RR O µ =+ (5) and this can now be substituted in (2): 2 2 E 1R E O R 1 v 2 v ⎡⎤ µµ ⎢⎥ ∆= + − + ⎣⎦ R E R 2 2 E EE G E R Vv 1 R + O ( 6 ) Upon arrival at I, there will have to be a second burn that will simultaneous accelerate the rocket to GEO GEO v R µ =
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lecture_34 - 16.512, Rocket Propulsion Prof. Manuel...

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