Orbits in Geographic Context. Instantaneous Time Solutions Orbit Fixing in Geographic Frame Classical Orbital Elements

Transcription

1 Orbits in Geographic Context Instantaneous Time Solutions Orbit Fixing in Geographic Frame Classical Orbital Elements

2 Instantaneous Time Solutions Solution of central force motion, described through two geometric parameters a & e, represents a static picture over one time period. However, in practice, we need to know the spacecraft position and velocity vectors at any time instant, which involves determining the six scalar unknowns from six initial conditions, generated through the ascent mission. In reality, it is found to be more convenient, if we can also describe the instantaneous motion of spacecraft with respect to the parameters of the conic section.

3 Instantaneous Orbit Parameters All earth bound missions require that orbits are referenced in the geographical context of earth and hence, we need to create a framework for fixing the orbits in relation to earth. In this regard, it is worth noting that earth has a time system that is related to its rotation on its axis as well as its revolution around the sun. Therefore, we need to fix the conic section (or orbit) in relation to a point on earth, which is typically referenced through time coordinate.

4 Coordinate Fixing Consider the following ECEF frame in which the position and velocity of a spacecraft are defined.

5 Coordinate Fixing X axis points outwards from centre of Earth along the line joining the centre and 1 st point of Aries (Vernal Equinox), in the equatorial plane. Y axis points outwards from the centre of Earth along the line perpendicular to X axis, but lying in the equatorial plane. Z axis points towards outward normal to the equatorial plane, towards geographical north. Various orbital quantities can then be referred to the above coordinate system.

6 Classical Orbital Elements Consider the orbital figure as below.

7 Classical Orbital Parameters Group-1: (Motion within Orbital Plane) a Semi-major axis (Orbit Size) e Orbit Eccentricity (Orbit Shape & Type of Conic) T 0 Time of Perigee Passage (Location in Orbit) Group-2: (Orientation of Orbital Plane in ECEF) Ω - Right Ascension of Line of Nodes (Longitude?) i Inclination of the Orbit (Latitude?) ω - Argument of the Perigee (Global Time Reference)

8 Orbit Parameters Definitions Orbital parameters are related to initial conditions. µ 1 µ r a = Semi-major Axis; e = r ɺ H Eccentricity 2ε µ r k H ˆn= (or n= k H ) Ascending Node Vector k H 1 ( ˆ ) iˆ. n Ω =cos i. nˆ = cos 1 Right Ascension n 1 k H i = cos Orbit Inclination H 1 nˆ e 1 n e ω = cos = cos Argument of e Perigee n e

9 Orbital Parameter Features Inclination i is in the range 0 i 180 o. i = 0 o Equatorial Orbit; i = 90 o Polar Orbit. 0 o i 90 o Prograde Orbit (rotation in the same direction as Earth s rotation). i > 90 o Retrograde Orbit. Orbit inclination depends on the latitude of the launch site. (Why?) i = 0 o or 180 o represents a singularity for the definition of the right ascension, Ω. (Why?) For a perfectly circular orbit, argument of perigee, ω is indeterminate. (Why?)

10 Orbital Angular Location In most orbital missions, spacecraft injection occurs at the closest point to earth or perigee, which provides an important time reference for the orbital mission. In cases where the injection point is not the perigee, it is useful to obtain the angular location with respect to the perigee point. Angular location θ of the spacecraft in the orbit, with reference to vector e, can be defined as below. 1 e r θ ( t) = cos e r

11 Orbital Elements Example Find the orbital elements of a satellite injected at 1500 GMT on July 23, 1991 as follows. r = i j k 10 m; ( ˆ ˆ ˆ) 6 v = iˆ ˆj kˆ ( m / s) Use Following Normalization: 6 P DU = ; TU = = s; 2π SU = DU TU = m s µ = DU TU = 3 2 / 7905 / ; 1 /1 1

12 Orbital Elements Example The orbital elements are as follows. r = 0.8iˆ ˆj kˆ ; v = 0.4iˆ 0.8 ˆj + 0.6kˆ r = 1.12 DU ; v = 1.08 SU ; 2 2 ε = 0.31 DU / TU ; a = 1.61 DU; e = 0.163iˆ ˆj kˆ ; e = 0.32 o i = 42.6 ; n = 0.28iˆ ˆj ; Ω = 110 ; ω = 10.3 ; θ = 26.8 o o o

13 Summary Classical orbital parameters help us define the orbit in a Classical orbital parameters help us define the orbit in a geographic context.