2 / 39 Previous Lecture Approximate solutions for the motion about an oblate planet: The Brouwer model The Cid- Lahulla model
3 / 39 Definition of Orbital Maneuvering Orbital maneuver: the use of the propulsion systems to change the orbit of a spacecraft. It encompasses all orbital changes after insertion required to place a satellite in the desired orbit.
4 / 39 First Orbital Maneuvers Unmanned spacecraft: January 2, 1959, Luna 1: The spacecraft missed the Moon by about 6000 km. But coming even this close required several maneuvers, including circularizing the initial launch orbit and doing midcourse corrections. September 12, 1959, Luna 2: Intentional crash into the lunar surface. Manned spacecraft: March 23, 1965, Gemini 3: A 74s burn gave a ïaďv of 15.5 meters per second. The orbit was changed from 161.2 km x 224.2 km to an orbit of 158 km x 169 km. December 12, 1965: Gemini 6 and 7: First rendezvous. The two Gemini capsules flew around each other, coming within a foot (0.3 meter) of each other but never touching.
5 / 39 Classification Classification Continuous thrust maneuvers. Impulsive maneuvers: mathematical model of a maneuver as an instantaneous change in the spacecraft s velocity. This lecture: Impulsive maneuvers The thrust is applied on a short period of time the change of velocity is approximated to be instantaneous. Very short maneuvering time: the Keplerian model for the orbit is a good approximation.
6 / 39 Impulsive Maneuver: Problem Formulation Input: Output: Consider a spacecraft orbiting on a Keplerian trajectory (T0), characterized by: Specific angular momentum vector h 0. Specific total energy E 0. Consider a desired Keplerian trajectory (T1), characterized by: Specific angular momentum vector h 1. Specific total energy E 1. Determine the conditions to be fulfilled such that a single impulse maneuver at a given moment of time modifies the spacecraft trajectory from (T0) to (T1). Determine the expression of the impulsive velocity vector.
7 / 39 Impulsive Maneuver: Solution Denote by v 0 the velocity at the maneuver moment (before). Denote by v 1 the velocity at the maneuver moment (after). Denote by r 0 the position vector of the point in space where the maneuver is performed. The constants of motion for both orbits are: h 0 = r 0 v 0 E 0 = 1 2 v2 0 µ r 0 h 1 = r 0 v 1 E 1 = 1 2 v2 1 µ r 0 (1) Note that r 0 is the position vector of the orbits intersection
8 / 39 Impulsive Maneuver: Solution The system: r 0 v 1 = h 1 leads to determining v 1 = v 1 (E 1,h 1,r 0 ). v is computed afterwards as: 1 2 v 1 2 µ (2) = E 1 r 0 v = v 1 v 0 (3) Solve the system to obtain v 1!
9 / 39 Impulsive Maneuver: Solution Compatibility condition: 2E 1 + 2µ h2 1 r 0 r0 2 0 (4) Geometrical interpretation: The current point r 0 should belong to the new orbit (E 1,h 1 ) as well Denote: λ = ± The impulsive velocity is: v = h 1 r 0 r 2 0 2E 1 + 2µ r 0 h2 1 r 2 0 (5) + λ r 0 r 0 v 0 (6)
10 / 39 Impulsive Maneuver: Solution The compatibility condition : v = h 1 r 0 r 2 0 2E 1 + 2µ h2 1 r 0 r0 2 0 (7) The impulsive velocity is: ( ± 2E 1 + 2µ ) 1 h2 2 1 r 0 r 0 r0 2 v 0 (8) r 0 The magnitude of the impulse is: v 2 = ( ± ) 2 2E 1 + 2µ h2 1 r 0 r0 2 r 0 v 0 + r 0 ( ) h1 h 2 0, (9) r 0
11 / 39 Impulsive Maneuver: Solution Actually, who is λ? λ = ± 2E 1 + 2µ r 0 h2 1 r 2 0 (10)
12 / 39 Impulsive Maneuver: Solution Recall from the Keplerian equations of motion (polar coordinates): λ is the radial velocity on the new orbit: ṙ 2 = 2E + 2µ r h2 r 2 (11) λ = r 0 v 1 r 0 (12) The magnitude of the impulsive velocity : ( v 2 = λ r ) 0 v 2 ( ) 0 h1 h 2 0 + r 0 r 0 }{{}}{{} change of radial velocity change of angular momentum (13)
13 / 39 Impulsive Maneuver: Solution The LVLH frame (Local-Vertical-Local-Horizontal): i x = ˆr 0 ; i y = ĥ 0 ˆr 0 ; i z = ĥ 0 (14) The inertial impulsive velocity v = v in expressed in the LVLH frame: v in x = λ ˆr 0 v 0 ; v in y = h 1 ĥ 0 h 0 r 0 ; v in z = h 1 v 0 h 0 (15) The impulsive velocity v LVLH with respect to LVLH is: v LVLH = v in ω r 0, ω = h 0 r0 2 (16) v LVLH x = v in x ; v LVLH y = h 1 ĥ 0 r 0 ; v LVLH z = v in z (17)
14 / 39 Impulsive Maneuver: Solution Note that: v in x = ± 2E 1 + 2µ h2 1 r 0 r0 2 ± 2E 0 + 2µ r 0 h2 0 r 2 0 The sign + : the radial velocity is positive (radial distance increasing) The sign : the radial velocity is negative (radial distance decreasing) Note that: ( v 2 = λ r ) 0 v 2 0 r 0 }{{} ( v in x ) 2 ( ) h1 h 2 0 + r 0 }{{} ( v in y ) 2 +( v in z ) 2 (18) (19)
15 / 39 Optimizing the Impulsive Maneuver Minimize v 2 x : v 2 x = 2(E 1 + E 0 ) + 2µ h2 1 + h2 0 r 0 r0 2 ±2 2E 1 + 2µ h2 1 r 0 r0 2 2E 0 + 2µ h2 0 r 0 r0 2 (20) sign should be ". one of the aaa terms is 0 both aaa terms are 0 (maneuver at the common pericenter/apocenter) It would be ideal to have v 2 x = 0.
16 / 39 Optimizing the Impulsive Maneuver Minimize ( v in ) 2 ( ) y + v in 2 z : ( ) ( ) v in 2 ( ) y + v in 2 h1 h 2 0 z = (21) r 0 Apparently, r 0 should be at its maximum value (maneuver at apocenter) However...
The Oberth Effect The Oberth Effect When travelling at high speed, a rocket engine generates more useful energy than one at low speed. Hermann Oberth, 1894 1989 German-Romanian physicist 17 / 39
Coplanar maneuvers Coplanar maneuver: The above means: h 0 h 1 = 0 (22) v z = 0 (23) It is reasonable to assume that h 0 and h 1 have the same orientation, and denote: The impulsive velocity has now the expression: v = αh 0 r 0 r 2 0 α = h 1 h 0 (24) ( ± 2E 1 + 2µ α2 h 2 ) 1 2 1 r 0 r 0 r0 2 v 0 (25) r 0 18 / 39
19 / 39 Coplanar maneuvers Decomposed in the non-orthogonal frame (r 0,v 0 ) : [ v = λ r ] 0 v 0 r0 + (α 1)v 0 (26) r 0 r 0 If the term associated with r 0 is null: tangential maneuver Coplanar maneuvers do not change the plane of the orbit The only orbital elements which can be changed by coplanar single-impulsive maneuver: a, e, ω, M The closed-form expression of v allows to obtain specific algorithms for each required orbit change
20 / 39 Example: Parking Orbit A spacecraft is launched from the surface of the Earth. At the apogee, it is necessary to apply a maneuver to circularize the orbit. Input data: a 0, e 0 (semimajor axis, eccentricity) of the launching orbit. Determine v!
21 / 39 Example: Parking Orbit The expression of the magnitude of v (tangential burn at apogee): ( ) µ 1 1 e0 v = a0 (1 + e 0 ) (27)
22 / 39 Problem 1: Circular to Elliptic Circular to Elliptic Maneuver A spacecraft is orbiting on a circular orbit of radius R. Determine the tangential impulsive velocity which should be applied such that the spacecraft will move on an orbit with a semimajor axis a and eccentricity e. Discussion.
23 / 39 Problem 2: Elliptic to Circular Elliptic to Circular Maneuver A spacecraft is orbiting on an elliptic orbit of semimajor axis a and eccentricity e. Determine the tangential impulsive velocity which should be applied at the apogee/perigee such that that the spacecraft will move on a circular orbit of radius R. Discussion.
24 / 39 Two-impulse maneuvers A single-impulse maneuver usually does not lead to the desired orbit Two or more impulsive maneuvers are necessary Example: GTO to GEO: 2 impulsive maneuvers
25 / 39 The Hohmann Transfer A two-impulse tangential maneuver between elliptic or circular orbits. Hohmann Conjecture (1924) The energy-optimal transfer orbit between two circular orbits of radii R 1 and R 2 is an elliptic orbit with: r min = min(r 1, R 2 ) r max = max(r 1, R 2 ) Proven for circular orbits by Lawden (1952) Proven for coaxial ellipses by Thompson (1986) Walter Hohmann, 1880 1945 German engineer
The Hohmann Transfer 26 / 39
27 / 39 The Hohmann Transfer The semimajor axis of the transfer orbit: a = R 0 + R 2 (28) 2 The tangential impulsive velocities are: ( ) µ 2R 2 v 1 = 1 u (29) R 0 + R 2 R 0 µ v 2 = R 2 ( 1 2R 0 R 0 + R 2 ) u (30) Hohmann transfer is reversible: no restrictions were imposed on R 0, R 2 Total transfer time: Kepler s third law: T coasting = π µ ( R0 + R 2 2 ) 3/2 (31)
28 / 39 Example: GTO to GEO Initial circular orbit: 322 km altitude (32) First impulse: v 1 = 2.4195 km/s (33) Transfer orbit semimajor axis: a = 24,432 km (34) Second impulse: v 2 = 1.4646 km/s (35) Final circular orbit: GEO 35, 786 km altitude (36) Time on the transfer orbit: T = 19,003 sec = 5.27 hours (37)
29 / 39 Bi-Elliptic Transfer Maneuver A three-impulse tangential maneuver between two circular orbits. Ary Abramovich Sternfeld, 1905 1980 Polish-Soviet engineer
Bi-Elliptic Transfer Maneuver 30 / 39
31 / 39 Bi-Elliptic Transfer Maneuver The impulsive velocities are: v 1 = 2µ R 0 µ a 1 µ R 0 (38) 2µ 2µ v 2 = µ R a 2 2µ v 2 = µ R 3 a 2 µ (39) R a 1 2µ (40) R 3 The coasting orbits are characterized by: a 1 = R 0 + R ; T 1 = π ( ) R0 + R 3/2 (41) 2 µ 2 a 2 = R + R 3 ; T 2 = π ( R3 + R 2 µ 2 ) 3/2 (42)
32 / 39 Bi-Elliptic Transfer Maneuver The optimal total v for the bi-elliptic transfer is achieved when: R (43) It is called the bi-parabolic transfer The case is purely theoretical the transfer time is infinity
33 / 39 Bi-Elliptic versus Hohmann The expressions for the total v are: Hohmann two-impulsive maneuver: v H = vh = 1 2(1 α) 1 (44) v 0 α α (1 + α) Bi-elliptic three-impulsive maneuver: v BE = vbe v 0 = 2(α + β) 1 + α αβ α 2 (1 β) (45) β (1 + β) Where: α = R f inal R initial ; (46) β = R R initial (bi-elliptic only) (47)
Bi-Elliptic versus Hohmann 34 / 39
Bi-Elliptic versus Hohmann 35 / 39
36 / 39 Bi-Elliptic versus Hohmann It depends on the ratio of the radii of the inner and outer orbits; threshold: R f inal R initial = 11.94 (48) For many practical applications (LEO to GEO), the two-impulse transfer is more economical. It is also the case for interplanetary transfers from Earth to all planets except the outermost three. Time of flight: the bi-parabolic transfer requires an infinite transfer time.
37 / 39 This Lecture: Orbital Maneuvers 1 General framework: one impulsive maneuver Conditions of compatibility Closed form expression of the impulsive maneuver Discussion Coplanar maneuvers: Two-impulse maneuver: The Hohmann transfer Three-impulse maneuver: Bi-Elliptic Transfer Comparison
38 / 39 Next Lecture: Orbital Maneuvers 2 Non-coplanar maneuvers The Lambert problem Maneuver efficiency Rendezvous and stationkeeping