lecture_34

# lecture_34 - 16.512 Rocket Propulsion Prof Manuel...

This preview shows pages 1–4. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
() 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 =

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 13

lecture_34 - 16.512 Rocket Propulsion Prof Manuel...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online