16.512, Rocket Propulsion Prof. Manuel Martinez-Sanchez Lecture 33: Performance to LEO

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1 6.5, ocket Propulsion Prof. Manuel Martinez-Sanchez Lecture 33: Performance to LO Calculations for Launches to Low arth Orbit (LO) Ideal arth-to-orbit launch V cosα v ' ' V cosα V cos α V cos α v c V vc v' cosα cosα α α cos cos V V + V / ( η ) η + η ηcos α ' η cos α η cos α η η cosα η + / + η cos α η (increasing f.of α) For α MIN / ( η ) + + η η 6.5, ocket Propulsion Lecture 33 Prof. Manuel Martinez-Sanchez Page of 9

2 Note: Max at η , 6 km (worst altitude) MIN.5363 / F MAX 79 m/ s ε η η + ε ε η + ε + ε + η + ε ε ( + ε ) + ε ε 3 3 ε ε... + ε ε + ε... + ε / + ε ε ε... 4 α η.9.58 (approx..593) / / ( ) η.595 (GO, 4,Km) α.5775 (,98 m/s) α (,965 m/s) α (, m/s) η.395 day, η 6,58 Km (,54 m/s) (,57 m/s) (,766 m/s) η.876 ( Z 9 Km).6387 (8,49 m/s).796 (8,534 m/s).3 (8,856 m/s) η.939 ( Z 6 Km).4399 (85 m/s).595 (8375 m/s).9755 (8676 m/s) η.955 ( Z 3 Km).75 (8,84 m/s).3748 (8 m/s).6953 (8454 m/s) NOT: ε ε / ε ε / So, for LO, mainly (apogee kick) Sticking to α, variation with 6.5, ocket Propulsion Lecture 33 Prof. Manuel Martinez-Sanchez Page of 9

3 /.5 η ffects of arth s otation (a) reduction Ω 463 m / sec β is launch azimuth w.r.t ast ( α, near-horizontal launch) V rocket-imparted V Starting velocity (abs.) is now β v V + Ω + Ω cos So, replaces in previous formulation v 6.5, ocket Propulsion Lecture 33 Prof. Manuel Martinez-Sanchez Page 3 of 9

4 v / + α + cos α + V + Ω + Ω cos β V Ω cos + Ω cos + Ω + β ( β) V Ω cos β + Ω + ( sin β ) Notice rotation reduces even for β 9. The benefit is low for some larger (Westwards launch). For, need v β Ω cosl cos β.56 (for V 8 m/ s ) (for L 8.5, β 9.8 ) xample: For L 8.5, Km, ( 85.8) ( β ) 7 V 47 cosβ sin β ± 3 ± 6 ± 9 ± ± 5 ± 8 V (m/s) 7645 m/s V reduction 47 m/s 355 m/s m/s m/s -95 m/s -35 m/s -47 m/s 6.5, ocket Propulsion Lecture 33 Prof. Manuel Martinez-Sanchez Page 4 of 9

5 (b) Orbit inclination For β (launch due ast), i L. For any other azimuth, higher inclination. v Ω cosl v sin β γ sin γ ( ) Sin γ β γ γ sin γ v V sin cos cos β Cos cos γ sinβ V cosβ + v tan β v + cosβ Given two angles and the side included, find opposite angle ( ) cos i cos 9 γ cos 9 + sin 9 γ sin 9 cos γ 6.5, ocket Propulsion Lecture 33 Prof. Manuel Martinez-Sanchez Page 5 of 9

6 xample: Continuing from previous example, L 8.5 v 47 m/ s, ' 5 Km tan β, cos i.8788 cos γ 47 + V cosβ V from previous table β 3 α 8.54 i ( 66. ) ( 4.4 ) ( 8.5 ) retro-orbits (westwards) slightly different inclination In reality, we probably require the orbit altitude, the orbit inclination and then launch azimuth β must be calculated cos γ cos i cos i γ cos γ + tan cos γ cos i β γ V sin tan cos i V cos β + v sin γ cos i with V v sin β v cos β + V cosβ + v sin β 6.5, ocket Propulsion Lecture 33 Prof. Manuel Martinez-Sanchez Page 6 of 9

7 V + v + v V cosβ + or v v cosβ + v + sin β sin β + β β cos cos γ tan v tan γ V sin β cos β v sin β cos β β β sin sin sin β v + cos β + cos β cos β cos β + sin β sin β cos β β β + + cos β + sin cos cos β tan γ sin β + tan γ sin β sin γ v sin γ + tan β tan β v + sin γ tan β v + cos γ L 8.5 Check: ) γ i 6.8 tan β β , ocket Propulsion Lecture 33 Prof. Manuel Martinez-Sanchez Page 7 of 9

8 tan β v cos i + cos i cos i tan β v cosl cos i + Directly: V v sin β sin β α ( ) v sin γ cos γ V tan β sin γ tan β v cos γ V cos γ v cos γ V and cos γ cos i v CosL Cos i Cos β + tan β CosL Cosi + v CosL v cosl cos i + v CosL Cos i tan β tanα v + cos γ / tan γ + tan β + cos β v + γ cos / 6.5, ocket Propulsion Lecture 33 Prof. Manuel Martinez-Sanchez Page 8 of 9

9 V v cos β tan β v γ cos v + V + tan γ + γ cos / v / v v cos γ v + v cos γ cos γ / cosγ / 6.5, ocket Propulsion Lecture 33 Prof. Manuel Martinez-Sanchez Page 9 of 9

Spherical trigonometry

Spherical trigonometry 1 The spherical Pythagorean theorem Proposition 1.1 On a sphere of radius, any right triangle AC with C being the right angle satisfies cos(c/) = cos(a/) cos(b/). (1) Proof: Let