Question

STS 133, Space Shuttle Discovery’s last flight, was launched on February 24, 2011 at 16:53 EST from the Kennedy Space Center into an initial 208 x 232 km altitude orbit at an inclination of 51.6 degrees.

What was the launch azimuth?

Answer 1

The problem involves the concept of of launch azimuth. The launch azimuth is inertial frame is easy to calculate. But since earth is rotating the actual azimuth should be little less than inertial launch azimuth.

Apart from that knowledge of orbital velocity in an elliptical orbit is also required. The detailed explanation is given in below images. Hope it would be clear to you..

Tabes en launch Azimuth It is the ongle between to the north direction & projection of initial plane of orbit onto the location for lounch. Taunch azimuth = sin cos Ctorget inclination) 4 l cos ( launch sile lotitude) Mere Angle of inclination of orbile: 51.6° And Latitude of Kennedy space centre = 28 5729 B = sin' I cos (51.80 7 I cos (28.57290)] Be sint (0.707 28901) - ß = 45:01 But this only mu for inestial fromes. Since earth is rotating wrl Shuttle the effect of rotation must be considered. had been The correct azimuth would be that which Calcu leated in rotating from B Let it be formula for B T ß fan TV jota I Voty! Voorle = velocity in x-disen - art rotating frame Wolly velocity in y diren wor rotating frame

t het Venonat be veleity vects from incital from o be earth rotational velocity vector at equator be velocity vector in rotahong promet Voor hot - Yota i forth solation ( Veenh) turor y ĵ rector = 465m/sec Se formula for Isota - Verbil. Sin (3) - Veurth Carlo - o launch Site Cattitude (Vortly = Vorbite sais (B) Where Vexbilla arbitul velocity of stittelite in elliptical cebit calculated at opogee (forthest point) Zn given question altitude of orbite is 20 8X 282 km Here 29=232 > a = 116 C 26=208 = b =104 fix apo gee distance som centre is A = ata where eccentricity (e) = a a ca va ² 62 A 1167 / 1162 - 1042 = 167.38 km - = 167 38 x 10m

e / 116² 1042 -0.44 Velocity of saltellite at apogee should be - Eva) = Lampe Hej - Jamfice) Q = 2.674 X10 " " " m kg + 2 mass of earth (M) = 5.9 72 X 1024 kg Vase 6475x15" x 5.972 4100 forinto t x 6-614) 715.972) x 10 167.38 X 103 Vo = 1 6:674 X 5.972 0.56 x 10 m/sec -- + 167.34__ Va = 3651.7 m/sec Dobit velocity ax_ apoges (Van) = 365 7.7m/sec Now Calculating (Vers & Lundly bykla (Vorbil) Sin (8) - Vearth Cord = (3751,7) sin(45.01 T-465 cos (2857297 = 2174.213 m/sec = 3651-74 cos (45-019 Vook) y = (vorbit ) Cos(8) 25 AUGLA 2581,691m/se

EL No C acheol_azimuth ange_wn _relating fome AB = Fon / 2174.213 ) ( 2581.691) = lon' to 842156) = 40010 B = 40.10 So wazimuth angle in inertial frome was () = 45.00 But since earth is rotating the actual azimuth ] Angle (B) is rotating prome should be 40-01 T = 40.00 So the launch azimuth should be around 4o. 01