1. ORBITAL MECHANICS: HOW OBJECTS MOVE IN SPACE FROM KEPLER FIRST LAW: A SATELLITE REVOLVES IN AN ELLIPTICAL ORBIT AROUND A CENTER OF ATTRACTION POSITIONED AT ONE FOCI OF THE ELLIPSE. SECOND LAW: THE RATE OF TRAVEL ALONG THE ORBIT IS DIRECTLY PROPORTIONAL TO THE AREA OF SWEEP IN THE ELLIPSE. THIRD LAW: PERIOD OF THE ORBIT SQUARED IS PROPORTIONAL TO THE MEAN DISTANCE TO CENTER CUBED. SATELLITE CENTER OF ATTRACTION (FOCI) APOGEE PERIGEE AREA OF SWEEP EARTH EXAMPLE ORBIT PLANE

2. ORBITAL MECHANICS: WHY OBJECTS MOVE THE WAY THEY DO V = ORBITAL VELOCITY Fg = GRAVITATION FORCE Fc = CENTRIPETAL FORCE DUE TO REVOLUTION ALTITUDE ORBITAL PATH NEWTONIAN THEORY Fg = G M m / r 2 Fc = m V 2 / r A SATELLITE MAINTAINS ITS ORBIT WHEN Fc = Fg G = UNIVERSAL GRAVITY CONSTANT M = MASS OF EARTH m = MASS OF SATELLITE r = DISTANCE EARTH CENTER TO SATELLITE IN NEAR CIRCULAR ORBITS THE ORBITAL VELOCITY IS ABOUT CONSTANT. IN HIGHLY ELLIPTICAL ORBITS THE SATELLITE SPEEDS UP TO MAX VELOCITY AT PERIGEE AND SLOWS DOWN TO MIN VELOCITY AT APOGEE.

3. Newton’s Laws A body remains at rest or in constant motion unless acted upon by external forces The time rate of change of an object’s momentum is equal to the applied force For every action there is an equal and opposite reaction The force of gravity between two bodies is proportional to the product of their masses and inversely proportional to the square of the distances between them.

4. Acceleration, Time, Distance F = ma V f = V 0 + at s = V 0 t + at 2 2

5. Vector Addition V1V1 V2V2 V3V3 V 1 + V 3 = V 2 Vector addition is done by adding the two head to tail vectors to equal the tail to tail and head to head vector Law of Sines a / sin A = b / sin B = c / sin C Law of Cosines a 2 = b 2 + c 2 - 2bc Cos A b 2 = a 2 + c 2 - 2ac Cos B c 2 = a 2 + b 2 - 2ab Cos C c b a C B A

6. Trigonometric Functions sine = opposite/ hypotenuse = y/r cosine = adjacent / hypotenuse = x/r tangent = opposite / adjacent = y/x -1< sin or cos <+1 - infinity < tangent < + infinity sin 0 = 0 cos 0 = 1 sin 90 = 1 cos 90 = 0 tan 0 = 0 tan 90 = + infinity y x  r r 2 = x 2 + y 2 sin 2  + cos 2  = 1 y x

7. Rocket Engines Liquid Propellant –Mono propellant Catalysts –Bi-propellant Solid Propellant –Grain Patterns Hybrid Nuclear Electric Performance Energy Safety Simplicity Expanding Gases Thrust Termination Restart

8. Specific Heat Specific Heat : the amount of heat that enters or leaves a unit mass while the substance changes one degree in temperature. –c = Btu per lbsm - degree Rankine –c p : specific heat at constant pressure –c v : specific heat at constant volume –k : ratio of specific heats k = cpcp cvcv > 1> 1 c = dQ w dT

9. Specific Impulse I sp = 9.797 k k - 1 TcmgTcmg ( )( ) [ ] 1 - ( ) P e k-1 P c k P e : Nozzle Exit Pressure (psi) < 14.7 psi P c : Combustion Chamber Pressure (psi)6,000 psi T c : Combustion Chamber Temperature (degrees Rankine)5,000 o R m g : average molecular weight of combustion products (lb/mole) I sp = W. F 2H 2 + O 2 2H 2 O 18 lbs/mole = m g

10. Launch Velocity Losses Gravity losses Pitch over to get correct velocity vector alignment for orbital insertion Drag from atmosphere Not instantaneous application of velocity Losses are between 15 and 17 % of  V

11.  V = I sp x g x ln MR Rocket Equation Mass Ratio Specific Impulse Thrust MR = m initial m final I sp = W. F W. g VeVe F = + A e ( P e - P a ) Rocket Formulas

12. Three Stage Booster Weights (lbs.) 1st Stage 35,000 365,000 100 280400,000 599,000 2.56 2nd Stage 10,000 125,000 120 290135,000 199,000 2.69 3rd Stage 4,000 50,000 80 250 54,000 4.57 Payload 10,000 StructurePropellants Burn Time (sec) I sp (sec) Stage Weight MR  V 1 = (280)(32.2)ln (2.56) = 8,475 ft/sec  V 2 = (290)(32.2)ln (2.69) = 9,238 ft/sec  V 3 = (250)(32.2)ln (4.57) = 12,232 ft/sec V l = 29,945 ft/sec V posigrade = 29,535 ft/sec V retrograde = 30,183 ft/sec Can place payload in posigrade orbit, but not in retrograde orbit, All three stages

13. ORBIT FORMULAS ELLIPTICAL & CIRCULAR ORBITS 2a = r A + r P c a c = a - rPrP = = a - r P a  1 - 1 - 2r P r A + r P r A - r P r A + r P eccentricity a } b } } c Apogee Perigee r’ r.. r’ + r = 2a if r = r’ then at that point 2r = 2a r = a b = a 2 - c 2    a rPrP = rPrP rArA a b c

14. PHYSICAL E = Specific Energy H = Specific Momentum  = Universal gravitational attraction mr Trajectory m 2 m1m1 V r r =  H2H2 1 + 2EH 2 2 cos k = H2H2  = 2EH 2   2 1 + GEOMETRIC abcabc E < 0,= 0 for a circle E < 0, 0 << 1 for an ellipse = 1 for a parabola > 1 for a hyperbola Polar coordinates for any conic section pages 32, 33, 34 “Handout” CONSTANTS FOR ORBIT        m2m2

15. CIRCULAR ORBIT Period= 2  r V  r V= P2P2 = 4 2 r 2  r P2P2 = 4 2 r 3  ELLIPTICAL ORBIT P= 2  a V a = mean distance from focus = = semi major axis P2P2 = 4  a3a3 2  P2P2 = (2.805 x 10 15 )a 3 units= sec 2 ft 3 KEPLER’S THIRD LAW Orbital Period  

16. EARTH SATELLITES Eccentricity = Major Axis = 2a Minor Axis = 2b E = specific energy H = specific angular momentum E = V 2 M 2 r =  2a 2 r H = Vr cos V=  a  r p = a - c r a = a + c r p + r a = 2a c 2 = a 2 - b 2 for elliptical orbits from (1) V =  r  c a =    for circular orbits a = r therefore

17. Coordinate Systems Cartesian Coordinates –Abscissa = x –Ordinate = y –(x,y) Polar Coordinates –Radius Vector = r –Vectorial Angle =  –(r,   r +y +x -x -y

18. Description of Orbit Right Ascension –Measured eastward from the vernal equinox In Spring when the sun’s center crosses the equatorial plane once thought to be aligned with the first point of the constellation Aries Inclination Argument of Perigee Two of the following –Eccentricity –Perigee –Apogee

19. Orbit Calculations aa b b x 2 y 2 a 2 b 2 + = 1 rprp rara c r r’ r + r’ = 2a Ellipse is the curve traced by a point moving in a plane such that the sum of its distances from the foci is constant.  c a a 2 = b 2 + c 2

20. INCLINATION FUNCTION OF LAUNCH AZIMUTH AND LAUNCH SITE LATITUDE cos i (inclination) = cos (latitude) sin (azimuth) a zimuth lat. N S 270 o North = 0 degrees Azimuth 90 o 180 o AZIMUTH cos i + cos (lat) sin (az) sin 90 o = 1 sin 0 o = 0 sin 180 o = 0 sin 270 o = -1 launch azimuth from 180 o to360 o = retrograde orbit launch azimuth from 0 o to 180 o = posigrade orbit East South West

21. Celestial Equator Argument of Perigee (  ) Perigee Inclination Right Ascension  Orbit Trace Celestial Sphere

22. ORBITAL MECHANICS: GROUND TRACES GROUND TRACES THE POINTS ON THE EARTH’S SURFACE OVER WHICH A SATELLITE PASSES AS IT TRAVELS ALONG ITS ORBIT PRINCIPLE : GROUND TRACE IS THE RESULT OF THE ORBITAL PLANE BEING FIXED AND THE EARTH ROTATING UNDERNEATH IT AMPLITUDE OF GROUND TRACE (LATITUDE RANGE) IS EQUAL TO THE ORBITAL INCLINATION MOVEMENT OF GROUND TRACE IS DICTATED BY THE SATELLITE ALTITUDE AND THE CORRESPONDING TIME FOR IT TO COMPLETE ONE ORBIT ORBIT 4 ORBIT 3 ORBIT 2 ORBIT 1 EQUATORIAL ORBIT INCLINED ORBIT ANGLE OF INCLINATION (0 DEG. FOR EQUATORIAL) MULTIPLE ORBITS SAT EARTH MOTION BENEATH SATELLITE

23. ORBITAL MECHANICS: SPECIFIC ORBITS AND APPLICATIONS POLAR (100- 700 NM AT 80 - 100 DEG. INCLINATION) –SATELLITE PASSES THROUGH THE EARTH'S SHADOW AND PERMITS VIEWING OF THE ENTIRE EARTH’S SURFACE EACH DAY WITH A SINGLE SATELLITE SUN SYNCHRONOUS (80 - 800 NM AT 95 - 105 DEG INCLINATION) –PROCESSION OF ORBITAL PLANE SYNCHRONIZED WITH THE EARTH’S ROTATION SO SATELLITE IS ALWAYS IN VIEW OF THE SUN –PERMITS OBSERVATION OF POINTS ON THE EARTH AT THE SAME TIME EACH DAY SEMISYNCHRONOUS (10,898 NM AT 55 DEG INCLINATION) –12 HR PERIODS PERMITTING IDENTICAL GROUNDTRACES EACH DAY HIGHLY INCLINED ELLIPTICAL (FIXED PERIGEE POSITION) –SATELLITE SPENDS A GREAT DEAL OF TIME NEAR THE APOGEE COVERING ONE HEMISPHERE –CLASSICALLY CALLED “MOLNIYA ORBIT” BECAUSE OF ITS HEAVY USE BY THE RUSSIANS FOR NORTHERN HEMISPHERE COVERAGE GEOSYNCHRONOUS (GEO) (CIRCULAR, 19,300 NM AT 0 DEG INCLINATION) –24 HR PERIOD PERMITS SATELLITE POSITIONING OVER ONE POINT ON EARTH. –ORBITAL PERIOD SYNCHRONIZED WITH THE EARTH’S ROTATION (NO OTHER ORBIT HAS THIS FEATURE)

24. (1 radian = 57.3 degrees) LINEAR MOTION S = r  ft V tangential = r  ft/sec a tangential = r  ft/sec 2 Linear and Angular Motion ff =  o +  t  =  o t +  t 2 2 Distance Velocity Acceleration  avg = ff oo tftf toto radians sec 2 ff  avg = tftf toto radians sec ff S/r radians  = ANGULAR MOTION

25.   srsr s r radians

26. CONSERVATION OF: ENERGY MOMENTUM V22V22 rr = constant = F Specific Energy# #Specific means per 1b mass Angular Momentum = mr 2 w mr 2 w = constant = mH H = V r cos  Specific Angular Momentum VrVr V ^ - 

27. PHYSICAL E = Specific Energy H = Specific Momentum  = Universal gravitational attraction mr Trajectory m 2 m1m1 V r r =  H2H2 1 + 2EH 2 2 cos k = H2H2  = 2EH 2   2 1 + GEOMETRIC abcabc E < 0,= 0 for a circle E < 0, 0 << 1 for an ellipse = 1 for a parabola > 1 for a hyperbola CONSTANTS FOR ORBIT        m2m2

28. r 2 + (V t  t) 2 = (r + A) 2 A = -- V r  t a r  t 2 2 but for circular orbit V r = 0 arar  t 2 r 2 + (V t  t) 2 = r + 2 r 2 + V t 2  t 2 = r 2 + r a r  t 2 + ar2ar2  t 4 4 subtract r 2 from each side of the equation V t 2  t 2 = r a r  t 2 + ar2ar2 4 divided by  t 2 Vt2Vt2 = ra r + a r  t 2 4 as  t 0 Vt2Vt2 = r a r arar = Vt2Vt2 r VttVtt r r CENTRIPETAL ACCELERATION () 2 A  t 4 then or

29. ESCAPE VELOCITY E = 0 V 2 2 r  0 =  r2 V2V2 E 1 =  V escape = 2 r = (2) 14.075x10 15 ft 3 sec 2 20.9 x 10 6 ft = 36,700 ft sec