ME 239: Rocket Propulsion Orbital Mechanics (Supplement to Section 4.4) Part 1 of 2 J. M. Meyers, PhD

Transcription

1 ME 239: Rocket Propulsion Orbital Mechanics (Supplement to Section 4.4) Part 1 of 2 J. M. Meyers, PhD Artist s rendering of ESA Mars Express spacecraft Courtesy NASA JPL

2 Orbital Mechanics 1) Introductory Comments 2) Two-body Problem PART I 3) Equations and Constants of Motion 4) Conic Sections 5) Physical Significance of Two-body Trajectories 6) Trajectory Shaping PART II 7) Trajectory Transfers 8) Time of flight 9) Interplanetary Transfers 10) Integrating into rocket propulsion analysis Summary of Orbital Mechanics Relations 2

3 1) Intro Comments This intends to be an introductory treatment of Orbital Mechanics The material herein will suffice for analyses in this course and will ultimately serve as a bridge between propulsion system analysis and space trajectory maneuvering A standalone course exists on the material of Orbital Mechanics offered in the Fall semester 3

4 1) Intro Comments Orbital Mechanicsis essentially an application of Sir Isaac Newton s ( ) law of universal gravitation Before Newton, Johannes Kepler( ) used his own observations along with data from other astronomers (like TychoBrahe ( )) to develop 3 Keplerianlaws of planetary motion: 1. Orbits of planets are ellipses w/ Sun at one focus 2. Line joining a planet to the Sun sweeps out equal areas in equal intervals of time 3. The square of the period of a planet is proportional to the cube of the major axis of its elliptical orbit We ll demonstrate these three points through these lecture notes As this is an introductory treatment for ME 239 we will restrict our analysis to the single plane, two-body problem in two dimensional space. 4

5 1) Intro Comments Every body (mass) in the universe is attracted to and has an attraction for every other body Fortunately, owing to large distances between bodies, we can approximate (quite accurately in many cases) the motion of two masses subject only to their own gravitational forces We will treat these bodies as point masses Newton s Law of Universal Gravitation: To bodies will exert a force on each other that acts along the line joining the bodies and that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. Themagnitudeofthisforcecanberepresentinequationformas: = 5

6 2) Two Body Problem Considering the figure to the right which represents two masses at locations with respect to an inertial references frame (a reference frame with no acceleration). Here, we define the following quantities as: = mass of body 1 = mass of body 2 = mass center of two body system = inetrtial postiion vector of = inetrtial postiion vector of = vector joining two masses #$ = inetrtial postiion vector of mass center 6

7 2) Two Body Problem The gravitational force acting on each body can be written in vector form as: % = Where & ' is the unit vector of and positive in the direction of : & ' & ' = Applying Newton s 2 nd law to : % = ( = ) ) = & ' ) = & ' Similarly: ) = & ' 7

8 2) Two Body Problem We see that from our two-body figure: = ) = ) ) Thus: ) = * +, - Or: ) +. - = 0 This is the Newtonian vector equation of motion describing the motion of a smaller body w.r.t. a larger body. Where the gravitational parameter is defined as:. = * +, 8

9 2) Two Body Problem With the assumption that, becomes very small and the center of the principle attracting body (larger body) can be taken to be the two body system mass center. For example, consider an Earth-Sun 2 body system: ,900 This implies the 2-body mass center (cm) is located 455 km from the Sun s mass center which is 150 million km from Earth. The Sun would then orbit the cm at about 0.1 m/s while the Earth would orbit the cm at about 30,000 m/s to maintain conservation of angular momentum. We will assume for the rest of this orbital mechanics study. This is known as the Central Body Approximation analysis 9

10 2) Two Body Problem The magnitude of this force is termed weight : % =. & ' =. = 7 7 =. This presents us with a relation on how gravity varies with altitude (or distance from the large body mass center) 10

11 2) Two Body Problem Normally, to find & we would perform a mass moment balance: = 0 & = = *, = + Similarly: = + With central body approximation, (i.e. = Earth mass and = spacecraft mass) Thus:. = * +, 3 % = & ' =. & ' where represents the mass of smaller body 11

12 2) Two Body Problem Summary of Assumptions 1. Larger body is a homogeneous sphere and can be treated as a point mass with attractive force concentrated at center. 2. Smallerbodyisalsoapointmass. 3. Mass of larger body is much greater than that of the smaller body, thus attractive force of smaller body can be neglected and the 2-body mass center can be assumed to be at the center of the larger body. 4. Gravitational attraction of any other bodies may be ignored. 5. At the moment no other forces acting on the body (Lift, Drag, Thrust, solar radiation pressure(srp), etc ). Because of this the motion, or trajectory, of the satellite/body will be of constant energy. 12

13 3) Equations and Constants of Motion Let s define an inertial reference system using both Cartesian and polar coordinates. The term, 9, is defined as the position angle an is an angular measurement between the focal line of the orbit/trajectory and radial vector to smaller body. Recall: ) +. - = 0 Let s do some math! : ) +. -* :, = 0 Since: = & ' 1 0 : = < <= & ' = :& ' + <& ' <= : = : & ' & ' + & ' <& ' <= : = : Similarly: : ) = > >: =??: 13

14 3) Equations and Constants of Motion Substituting inourscalarvalues wecanarriveatascalarrelationas: : ) +. -* :, = 0??: +.: - = 0? <? <= =. < <=?<? =. < Integrating via separation of variables we find that:? 2. Where@ is the constant of integration. Specific Kinetic Energy Specific Total Energy NOTE,Aisthemassspecific totalenergyofthesystem.normallyb wouldbeusedbutthattermistooclosetooureccentricityterm.? 2. Specific Potential Energy 14

15 3) Equations and Constants of represents the total energy of a given orbit/trajectory which is constant provided no other mechanism is added energy or taking energy away(thrust, aerodynamic forces, s.r.p.) Thus, if kinetic energy of the trajectory increases, then the potential energy of the trajectory must decrease to satisfy constant energy > C C D Specific Kinetic Energy Specific Total Energy? 2. Specific Potential Energy 15

16 4) Conic Sections All space flight trajectories follow conic section paths: Circle Ellipse Parabola Hyperbola A general representation of for conic sections in polar coordinates (with the origin as the principle force) is well known: = F 1+EcosG E Conic Section Where: F semi-latus rectum E eccentricity. It determines the shape and type of conic section and thus the trajectory of the smaller body <1 Ellipse 0 Circle 1 Parabola >1 Hyperbola 16

17 4) Conic Sections Thus: = F 1+EcosG = I 1 E 1+EcosG Not derived here but trust me Using F = K. and E = We can come to a very important =. 2I This shows that the orbital energy (which is constant) is inversely proportional to the major axis and directly proportional to the gravitational parameter Trajectory >0 (positive) <0 (negative) Orbits(circles and ellipses) 0 Parabolic <0 (negative) >0 (positive) Hyperbolic 17

18 4) Conic Sections Comparison of all conic section geometries Note that: B E L 9 M N (to be discussed later) 18

19 4) Conic Sections Elliptical Orbit Notation and Details Note that: B E L 9 M N (to be discussed later) 19

20 4) Conic Sections Comparison of Elliptical Orbit and Circular Orbit 20

21 4) Conic Sections Hyperbolic Trajectory Notation and Details The hyperbola is the primary trajectory for escape from gravitational attraction of a larger body(planet, moon, Sun) This is also the trajectory of interest when a spacecraft enters or reenters the sphere of gravitaional attraction 21

22 4) Conic Sections Position Angle The position angle can be found at any point on the trajectory using the following relation G = cos O 1 E F 1 This shows that when the smaller body is at the semilatusrectum (G = 90 ) then the distance is equal to the semilatus rectum F. 22

23 5) Physical Significance of Two-body Trajectories Eccentricity The relationship between the eccentricity and energy can be seen from the expression: E = Since K is always greater than zero, the sign of E determines the value of E and thus the shape of the trajectory (circle, ellipse, hyperbola, parabola) 23

24 5) Physical Significance of Two-body Trajectories Elevation Angle The elevation angle, N, is the angle between the local horizon and the velocity vector. Recall that the velocity vector is ALWAYS tangent to the local trajectory path! 24

25 5) Physical Significance of Two-body Trajectories The Elliptical Q 0 2. Q 0 E Q 1 I R 0 The KE is always less than the PE as E is constant and less than zero This means that the body cannot escape from gravitational attraction of the larger body 25

26 5) Physical Significance of Two-body Trajectories Elliptical Analysis Recall from our polar equations of the apsides: S = F 1+E Apside distances V = F 1 E Substituting in F = I 1 E S = $TU = I 1 E V = $VW = I 1+E E = I = V S V + S Velocity at apsides We can take advantage of the fact that BOTH the angular momentum and energy is constant through the elliptical =? 2. =. 2F K = S? S = V? V? S = K S = V? V S =? $VW? V = K V = S? S V =? $TU OR? = 26

27 5) Physical Significance of Two-body Q 0and = OX ' E = 0 I = The Circular Orbit? 2. =. 2 Which can be solved for the velocity for a circular satellite([)? #Y =. = 7 As gravity varies with it is safer to stick with the./relation. The circular satellite relation can illustrate several salient points that are also valid for elliptical orbits. First, as? #Y decreases increases going to zero as the radius approaches infinity The period of the circular satellite (\ #Y ), which is the time it takes to make one orbital revolution is simply: \ #Y = 2]? #Y = 2]./ \ #Y = 2] -. 27

28 5) Physical Significance of Two-body Trajectories The Parabolic = 0 E = 1 I =? 2. = 0 At any point on the trajectory the KE is equal to the negative PE The smaller body in this case has just enough KE to follow the trajectory to infinity Thus, the velocity at = will be zero. Of course this will never happen as there will ultimately be another large body to add energy to the trajectory. The velocity at a specified is called the escape velocity (B[) and found to be:?^y# = 2. = 27 = 2? #Y MINIMAL velocity to escape planet s gravitational influence. It is interesting to see that the escape velocity at a given altitude is always about 40% ( 2) greater than the circular orbit at that same location 28

29 5) Physical Significance of Two-body Trajectories The Hyperbolic R 0 E R 1 I Q 0? 2. R 0 Here the KE is always greater than the PE If any velocity is applied that is great than the escape velocity (defined on the previous slide) then the trajectory is hyperbolic So why are all interplanetary trajectories hyperbolic? It is impossible to deliver a velocity that is exactly Vesc always design to be a little higher to ensure actual escape Window of opportunity for planetary intercept may require larger velocity leaving planetary body There will be residual velocity at planetary intercept which is required for orbital capture 29

30 5) Physical Significance of Two-body Trajectories Summary E I Energy Equation Elliptical <0 <1 >0 Circular. 2 0 Parabolic 0 1 Hyperbolic >0 >1 <0?? =. 2? 2. = 0? 2. It is important to note that for any trajectory type, once the energy is known (no matter how determined) the velocity at any point on that trajectory can be found from:? = 30

31 Summary Orbital Mechanics Relations = F 1+EcosG _ = cos O 1 E F 1 K = S? S = V? V? S = K S = V? V S =? $VW F = I 1 E = K. E = = V S V + S V + S = 2I? V = K V = S? S V =? $TU V = $VW = I*1+E, S = $TU = I*1 =? 2. =. 2F? = #Y =. \ #Y = 2] = 2] -? #Y. K =?cosn N = cos O K? = tano Esin_ 1+cos_ E I Energy Equation Elliptical <0 <1 >0 Circular. 2 0 Parabolic 0 1 Hyperbolic >0 >1 <0?? =. 2? 2. = 0? 2. = 2. = 27 = 2? #Y? = ln*1 b, = ln* c / d, = ln*1/mr,? ' =?^Y# +?` ' Orberth maneuver? = 7 c g YS ln*1 b, = 7 c g YS ln* c / d, = 7 c g YS ln*1/mr, 31

32 References 1. G. P. Sutton and O. Biblarz, Rocket Propulsion, Inc. John Wiley and Sons, Inc., F. J. Hale, Introduction to Space Flight, Prentice Hall, V. A. Chobotov, Orbital Mechanics, AIAA Education Series, M. D. Griffin and J. R. French, Space Vehicle Design, AIAA Education Series,