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Supplement GEOSTATIONARY PERTURBATIONS

INFLUENCE OF ASPHERICITY OF THE :

The gravitational potential of the Earth is no longer µ/r, but varies with . A tangential acceleration is created, depending on the longitudinal location of the , with four points of stable equilibrium:

 two stable equilibrium points (L 75° E, 105° W)

 two unstable equilibrium points ( 15° W, 162° E)

This tangential acceleration causes a drift of the satellite longitude.

Longitudinal drift d'/dt in terms of the longitude about a point of stable equilibrium expresses as:

(d/dt)2 - k cos 2 = constant

Orbits Supplement GEO PERTURBATIONS (CONT'D)

INFLUENCE OF EARTH ASPHERICITY

VARIATION IN THE LONGITUDINAL ACCELERATION OF A GEOSTATIONARY SATELLITE:

Orbits Supplement GEO PERTURBATIONS (CONT'D)

INFLUENCE OF & ATTRACTION

Gravitational attraction by the sun and moon causes the satellite to change with time. The evolution of the inclination vector is mainly a combination of variations:

 period 13.66 days with 0.0035° amplitude

 period 182.65 days with 0.023° amplitude long term drift

The long term drift is given by:

-4 dix/dt = H = (-3.6 sin M) 10 ° /day

-4 diy/dt = K = (23.4 +.2.7 cos M) 10 °/day where M is the moon ascending node longitude:

M = 12.111 -0.052954 T (T: days from 1/1/1950)

2 2 2 2 cos d = H / (H + K ); i/t =  (H + K )

Depending on time within the 18 year period of M

 d varies from 81.1° to 98.9°

 i/t varies from 0.75°/year to 0.95°/year

Orbits Supplement GEO PERTURBATIONS (CONT'D)

INFLUENCE OF SUN

Due to sun radiation pressure, eccentricity arises:

EFFECT OF NON-ZERO ECCENTRICITY

L = difference between longitude of geostationary satellite and (24 hour period orbit with e0)

With non-zero eccentricity the satellite track undergoes a periodic motion about the subsatellite point at perigee.

The MAXIMUM LONGITUDINAL VARIATION Lmax is given by:

Lmax = 114 e = 0.1° (assuming e = 0.001) and is reached 6 HOURS BEFORE/AFTER PERIGEE

Orbits Supplement EFFECT OF NON-ZERO ORBIT INCLINATION

The subsatellite point describes a 'figure of eight' around the desired station.

Given residual inclination i (degrees):

 peak longitude variation (degrees):

-3 2 Lmax =4.36 10 i

 peak variation (degrees):

 lmax = i

-5 typical i = 0.05°, hence Lmax = 10 ° ( negligible compared to other drifts)

Orbits Supplement STATION-KEEPING REQUIREMENTS

The combined non-zero inclination and eccentricity keeps the satellite from being perfectly geostationary (daily movement).

The East-West drift moves the satellite away from its nominal longitude (long term movement)

The mission requires that the satellite which displays daily+long term movements should remain within a specific solid as seen from the center of the earth (window).

Example of a typical window:

 longitude variation L= 0.1°

 latitude variation l = 0.1°

The necessary compensation of forces which drive the satellite outside the window is obtained by accurate firing of reaction equipment () on-board the satellite in order to keep it on if station.

Orbits Supplement SATELLITE DRIFT

EXAMPLE

SIRIO satellite subpoint on June 1 and 2, 1978:

Orbits Supplement ORBIT CORRECTIONS

NORTH/SOUTH STATION KEEPING (CORRECTION OF ORBIT INCLINATION):

Forces acting perpendicular to the orbit plane are delivered by thrusters at the nodes of the orbit.

Cost is of the order of 50 m/s per year.

EAST/WEST STATION KEEPING (CORRECTION OF SATELLITE DRIFT):

Forces acting in the orbital plane, in the direction of satellite or in opposite direction depending on the drift direction, are delivered by thrusters when the satellite is expected to reach the boundaries of the station keeping window.

Note that part of the satellite drift can be induced while North/South station keeping maneuvers are performed, due to thrusters misalignment relative to the perpendicular to the orbital plane.

Cost is of the order of 2-10 m/s per year depending on the satellite location with respect to the points of equilibrium.

Orbits Supplement ORBIT CORRECTIONS

INCLINATION CONTROL

 iX =  VN cos S / VS ;  iy =  VN sin S / VS where S = satellite right ascension = + +M

Required Velocity Increment to Compensate for Orbit Inclination

Year Velocity increment (m/s)

1980 41.24 1981 42.58 1982 44.19 1983 45.99 1984 47.70 1985 49.16 1986 50.20 1987 50.72 1988 50.62 1989 49.96 1990 48.77 1991 47.21 1992 45.48 1993 43.71 1994 42.15

Orbits Supplement EAST-WEST STATION KEEPING

Orbits Supplement EAST-WEST STATION KEEPING

Longitude drift and east-west corrections (a) when the satellite is located far away from the equilibrium point, and (b) when the satellite is located near the equilibrium point.

Annual velocity increment for east-west station keeping versus station longitude.

Orbits Supplement

EXERCISE: , LAUNCHING & POSITIONING DATA:

Universal G = 6.673 x 10-11 N m2 kg-2 Mass of Earth M = 5.977 x 1024 kg 6 Equatorial Radius of Earth RE = 6.3782 x 10 m Obliquity of the = 23.4°

1. A is placed in a geostationary Earth-orbit at an altitude of 35,788 km. Calculate the following:

(i) the ;

(ii) the maximum period (from geometry alone).

2. What conditions are necessary for an orbit to be geostationary (GEO)? Explain why in practice GEO depart from these conditions.

3. Sketch the trajectory of a practical GEO satellite as seen from a ground-station and explain its shape.

4. By means of a labelled diagram, describe five independent Keplerian Elements which can be used to describe the orbital trajectory of a satellite.

5. A NOAA meteorological satellite is launched into an 870 km altitude, circular, Sun- , with an ascending node at 14.30 local solar time.

(i) Define what is meant by a Sun-synchronous orbit, and explain how the effect occurs.

(ii) Calculate the satellite's orbital velocity

(iii) Calculate the orbital inclination angle

(iv) Find the eclipse period at the Vernal

NOTE: the period of time which an orbit spends in eclipse is given by: T x (1/180°) x cos-1 [(1 - n2)1/2 / sin β]

Where cos-1 (x) is in degrees; n = R / (R + h); R = Earth's radius, h= satellite height; β = the angle between the Sun's rays and the orbit normal; T = orbital period.

Provided cos β < n, otherwise the orbit is not eclipsed at all.

6. A mobile communications satellite is placed in a '' orbit with a semi-major axis of 26,556 km and an eccentricity of 0.71. Calculate the following for the orbit:

(i) perigee height; (ii) apogee radius; (iii) at apogee.

7. Briefly describe the properties of a and explain why it is suitable for communications into high .

CONT/D OVER

Orbits Supplement 8. An orbit has an apogee height of 37,500 km, and inclination of 20° and an eccentricity of 0.7. Calculate:

(i) perigee radius; (ii) orbital period; (iii) nodal regression rate; (iv) rate of change of the argument of perigee.

9. By means of a sketch diagram, explain what is meant by a Hohmann Transfer .

10. Give one advantage and one disadvantage of the Hohmann Transfer over direct ascent from one orbit to another.

11. A satellite is to be deployed from the cargo bay of the STS (), which is in a low-Earth circular of 300 km altitude and 28.5° inclination. The satellite has a boost-motor which gives it an initial -V of 1.942 km s-1 to put the satellite into a . Calculate the following:

(i) the orbital velocity of the satellite in the parking orbit; (ii) the maximum apogee height attainable by the transfer orbit; (iii) the additional delta-V required to circularise the orbit at apogee; (iv) the minimum delta-V required to change the inclination of the orbit to 90°.

12. Briefly explain the advantages of possessing a launch site close to the .

13. A communications satellite is to be placed into a geostationary Earth orbit from the Kourou launch site with the following characteristics:

Launch: Dual launch configuration with the ARIANE 3 Geostationary Transfer Orbit (GTO): perigee height 200 km apogee height 35,586 km inclination 7° Launch weight capability: 1140 kg into GTO

Spacecraft Configuration: Body-fixed stabilisation with deployable solar arrays Station keeping accuracy: +/- 0.05° N-S and E-W End-of-Life satellite mass: 510 kg

Propulsion: (Solid Propellant) specific impulse (Isp): 295 seconds Attitude and Orbit control (monopropellant Hydrazine): Isp : 216 seconds Propellant provision for drift orbit control: 13 kg Propellant provision for attitude control: 2.3 kg Propellant provision for E-W station-keeping: 3.2 kg

(i) Briefly describe the typical orbit transfer sequence from launch to final orbit. Calculate the velocity increment (delta-V) required for the injection of the satellite into GEO at the apogee of the GTO.

(ii) Determine the apogee motor propellant mass.

(iii) Estimate the required propellant mass required for station-keeping during the lifetime of the satellite (assume 10 years). State the beginning-of-life satellite mass (dry mass).

Orbits Supplement EXERCISE: ORBITAL MECHANICS, LAUNCHING & POSITIONING -ANSWERS

1. (i) Orbital period of a GEO satellite:

T2 = (4 2 / GM) a3

=> T2 = (4 2 / 3.988 X 1014) ( [35788 + 6378] x 103)3

=> T = 86148 s = 23 hrs 56 mins

(ii) Maximum Eclipse period:

= = R E 6378 / (35788 + 6378)

RE +h => θ = 8.7°

Teclipse = Torbit x (2θ / 360°) => 86164.1 x 17.4 / 360 = 69 mins

(72 mins when the apparent motion of the Sun and penumbra is taken into account)

2. Conditions for GEO:

1. Orbital period must be equal to a sidereal day (i.e. the time taken for the Earth to rotate 360° w.r.t inertial space –note this is 23h 56m 4.1s and not 24 hrs (which is a solar day)).

2. The eccentricity, e, must be zero.

3. The inclination angle, i, must be 0°.

If these latter two conditions are not met, the orbit is geosynchronous but not geostationary.

In practice, "e", and "i" are never exactly zero, because of perturbations to the orbit caused by luni- solar attraction and solar radiation pressure, etc.

Orbits Supplement 3. The daily trajectory will, typically, be a small "figure of eight”, where the satellite oscillates N-S either side of the Equatorial plane by an angle equal to its Inclination angle “δi”, whilst also oscillating E-W along the Equatorial plane either side of the nominal longitude slot by an amount which is a function of its eccentricity, "δe".

4. Define the five Keplerian Elements: (a, e, I, Ω, ω) Size Parameter, any one of the following:

a = semi-major axis -half the long axis of the (elliptical) orbit; T = orbital (anomalistic) period T2 = 42 a3 / GM; n = = 1440 / T, where T is measured in minutes;

Shape parameter:

e = eccentricity = 2c/2a; where 2c is the distance between the foci of the ellipse, or e = 1 – rp/a = ra/a -1, where ra,p = apogee and perigee radius.

Orientation Parameters (all subtended at the geocentre):

i = inclination angle = angle between orbit plane and equatorial plane measured at the ascending node (0° <= i <= 180°).

Ω = Right Ascension of the ascending node (RAAN) = angle measured from the vernal equinox direction (First point in Aries), counter-clockwise to the ascending node along the equatorial plane (0° <= Ω < 360°).

ω = argument of perigee = angle measured counter-clockwise from the ascending node to the perigee point, along the orbit plane (0° <= ω < 360°).

[see notes for the diagram(s)]

5. (i) A Sun-synchronous orbit is an orbit which has its rate of nodal regression exactly equal to the rate of rotation of the Earth about the Sun (0.986°/day). This can be achieved only in retrograde orbits, where there is a specific relationship between altitude and inclination angle - i.e. dΩ/dt = -9.95 7/2 2 -2 (RE/a) (1 –e ) cos (i) = 0.986°/day. Typical SSOs are circular with an altitude of ~800km and an inclination of ~98°.

Orbits Supplement The effect occurs because of the bulge at the Earth's Equator, measured by Earth's J2 geopotential harmonic. This bulge exerts a torque on the orbit, which tries to pull the orbit into the Equatorial plane (i.e. tries to reduce the inclination angle to zero). However,

Orbits Supplement because of the angular (gyroscopic) momentum of the orbit, the actual effect of the torque is to cause the orbit plane to precess in space, so that the RAAN changes continuously but the inclination remains unchanged. All orbits (except truly orbits) are affected in this way, and the class of SSOs is simply those orbits which precess at the required rate of 0.986°/day to match the apparent annual motion of the Sun around the sky (i.e. 360° in 365.25 days). The effect of this is that a satellite in SSO crosses the Equator at (very nearly) the same local solar time every orbit.

(ii) For a : v = (GM / r)1/2 => v = 3.988 x 1014 / [(6378 + 870) x 103] v = 7418 ms-1

(iii) For a SSO: 7/2 2 -2 -9.95 (RE/a) (1 - e ) cos (i) = 0.986°/day => cos (i) = 0.986 / [-9.95 x {6378 / (6378+870)} 7/2] i = cos-1 (-0.155) = 98.9°

(iv) At Vernal Equinox, the Sun lies on the Equatorial plane. As the Ascending Node occurs at 14:30 pm local time, the angle to the Sun is given by the following geometry (looking down on Earth from the North Pole):

θ = 360° x 2.5 hr /24 hr = 37.5° => β = 90° - θ = 52.5°

n = R / (R+h) = 6378 / (6378+870) = 0.88 cos (β) = 0.61 => the condition cos β < n is true, so the orbit will be eclipsed.

2 2 3 Torblt = (4  / GM) a => T2 = (42 / 3.988 x 1014) ([870+6378] X 103)3 => T = 6139 s = 102 mins

-1 2 1/2 Teclipse = Torbit x (1/180°) x cos [(1 - n ) / sin (β)]

=> Teclipse = 6139 x 53.2/180 = 1815 s = 30 mins and 15 seconds

Orbits Supplement 6 (i) Perigee height: rp = a (1 - e); hp = rp -RE => rp = 26556 km x (1 - 0.71) = 7701 km => hp = 7701-6378 = 1323 km

(ii) Apogee radius: ra = a (1 + e) => ra =26556 km x (1.71) = 45411 km

(iii) Orbital speed at apogee: v(r) = [GM (2/r – 1/a)]1/2 {vis viva} 1/2 => v(ra) = [GM (2/ra - 1/a)] 14 3 3 1/2 => v(ra) = [3.988 x 10 x { (2/45411 x 10 ) - (1/26556 X 10 ) } ] -1 v(ra) = 1596 ms

(iv) Molniya orbit:

The special properties of the Molniya orbit are that it is an elliptical orbit (e = 0.6 - 0.75), with an orbital period of exactly half a sidereal day, so that the motion of the satellite is semi-synchronised to the rotation of the Earth. However, unlike GEO, two Molniya orbits are completed for every one rotation of Earth.

The second property is that the inclination angle is set at 63.4°, which gives rise to the property that dω/dt = 0. This means that the line of apsides does not rotate within the perifocal plane, and so the apogees occur over the same geographical location on alternate orbits. For any other angle, the orbit does rotate due to the disturbing effect of the of the oblate Earth.

7. (i) Perigee radius: ra = ha + RE => ra = 37500 km + 6378 km = 43878 km ra = a (1+e) => a = ra / (1+e) = 43878 /1.7 = 25810 km rp = a (1-e) = 25810 x 0.3 = 7743 km

(ii) Orbital period: 2 2 3 Torbit = (4  / GM) a => T2 = (4 2 / 3.988 x 1014) (25810 X 103)3 => T = 41256 s = 11 hrs 27 mins 36 seconds

(iii) Nodal regression rate: (i.e. rate of change of RAAN): 7/2 2 -2 dΩ/dt = -9.95 (RE/a) (1 - e ) cos (i) °/day => dΩ/dt = -9.95 (6378 / 25810)7/2 x (1 - 0.49)-2 cos (20°) = -0.27 °/day i.e. 0.27°/day regression

(iv) Rate of change of argument of perigee: 7/2 2 -2 2 dω/dt = (-4.97) (RE/a) (1 - e ) [1 - 5 cos (i)] => dω/dt = -4.97 (6378 / 25810)7/2 x (0.51)-2 [1 - 5 cos2(20°)] = 0.49°/day

{you may note that dω/dt is related to dΩ /dt by the factor [1 - 5cos2(i)] / [2 cos(i)] }

Orbits Supplement 8. A is shown below:

This HTO is the minimum energy solution for transferring from one orbit to another, The HTO is tangential to both the initial and final orbits, and requires two impulsive "burns" to complete the transfer. The first

change in velocity (Δ V1) moves the vehicle from the initial orbit to the HTO, and the second change in

velocity (Δ V2) moves the vehicle from the HTO to the final desired orbit. In both cases the ΔV is simply the difference in speed between the initial or final orbital velocity, and that for the appropriate position (apogee or perigee) in the HTO.

If the final orbit is bigger than the initial orbit, then the changes in velocity are in the forward direction (i.e. the vehicle needs to increase its speed at both engine firings). If the final orbit is smaller than the initial orbit, the firings cause the vehicle's speed to decrease. Both firings must be impulsive - that is the duration of the burns should be negligibly small compared with the orbital periods.

9. Advantage of HTO: Minimum ΔV required => less fuel needed than for direct ascent. Disadvantage of HTO: Manoeuvre takes longer to complete than direct ascent.

10. (i) Parking orbit velocity: 1/2 Vcirc = (GM / R) 14 1/2 -1 => Vcirc = (3.988 x 10 / (6378+300) x 103) = 7728 ms

(ii) Maximum apogee height: -l -1 ΔV = 1942 ms => vp = 7728 + 1942 = 9670 ms rp = 6678 km v(r) = [GM (2/r -1/a)1/2 {vis viva} 2 => v p = 2GM/rp – GM/a 2 => a = GM / (2GM/rp – v p) = 3.988 x 1014 / [(2 x 3.988x1014 / 6678 x 103) -96702] = 15381 km

ra + rp = 2a; ha = ra - RE => ha = (2a - rp) - RE = 2 x 15381 - 6678 - 6378 = 17706 km

Orbits Supplement (iii) ΔV for circularisation: 1/2 va = [GM (2/ra- l/a)] 14 3 3 1/2 => va = [ 3.988 x 10 {(2 / 24084 x 10 ) - (1/15381 x 10 )} ] = 2681 ms-1 1/2 Vcirc = (GM / ra) = [3.988 x 1014 / 24084 x 103]1/2 = 4069 ms-1 => ΔV = 4069 -2681 = 1388 ms-1

(iv) Minimum ΔV for plane change to 90°:

To minimise ΔV, the plane change should be done when the vehicle is at its smallest velocity (i.e. at apogee).

ΔV (plane change only) = 2 v sin (θ/2); where θ is the angle through which the orbit plane must be rotated (i.e. the Δi) Minimum ΔV = 2 va sin (90° - 28.5° I 2) = 2 x 2681 x sin (30.75°) = 2742 ms-1

In practice, the plane change and circularisation burn would be combined:

2 2 1/2 ΔV = [ vcirc + va - 2vcirc va cos (90° - 28.5°) ] = [40692 + 26812 -2 x 4069 x 2681 x cos (61.5°) ]1/2 = 3652 ms-1

11. A launch due East (i .e. a launch of 0°) takes maximum advantage of the velocity imparted by the rotating Earth, v = ωr cos (latitude) = 6378 km x 2 / 86164 s (note: sidereal day) = 465 ms-1 i.e. the already effectively has this velocity (eastwards) before lift off, and so fuel is saved.

The site also gives direct access to all orbital inclinations, as the minimum inclination angle for the orbit is set by the latitude of the launch site.

12. (i) The Ariane 3 lifts off from the launch pad and quickly accelerates to the desired orbital injection velocity. This velocity is reached at 200 km, whereupon the satellite is released from the last stage of the rocket at what will become the perigee of the GTO. The satellite coasts up to the apogee of the GTO, losing velocity all the time. At apogee the satellite's kick-motor is fired to simultaneously change the inclination angle of the orbit to 0o, and to boost the velocity so that the satellite enters the circular GEO. (In practice, this final orbit is not GEO, but is a nearly stationary drift orbit. In this orbit, the satellite is stabilised, and the solar panels are deployed. The satellite is then slowly manoeuvred to the desired orbital slot in the GEO arc).

The ΔV for injection from GTO to GEO (with plane change) is: a = (ra + rp) / 2 = (ha + hp + 2RE) / 2 = (200 + 35536 + 2 x 6378) / 2 = 24246 km 1/2 va = [GM (2/ra - 1/a)] 14 3 3 1/2 -1 => va = [ 3.988 x 10 {(2 / 35536+6378 x 10 ) - (1/24246 x 10 )} ] = 1607 ms 1/2 Vcirc = (GM / ra) = [3.988 x 1014 / 35536 + 6378 x 103]1/2 = 3085 ms-1 2 2 1/2 ΔV = [vcirc + va – 2vcirc va cos (90° - 28.5°)] = [30852 + 16072 - 2 x 3085 x 1607 x cos (7°)]1/2 = 1503 ms-1

Orbits Supplement (ii) Apogee motor propellant mass:

Note: Propellant quantities quoted are "per year" values -i.e. Propellant provision for drift orbit control = 13 kg per year Propellant provision for attitude control = 2.3 kg per year Propellant provision for E-W station keeping = 3.2 kg per year

ΔV = g0 Isp In [Minitial / Mfinal]

Taking the maximum mass lift capability of the rocket = 1140 kg, then

Mfinal = Minitial / exp (ΔV / g0 Isp ) => Mfinal = 1140 / exp (1503 / 9.807 x 295) = 678 kg => Mpropellant = Minitial - Mfinal = 1140 - 678 = 462 kg

(iii) Estimate the station keeping propellant requirements (10 years):

E-W drift control ~ 2-10 ms-1 year-1 => (worst case) 100 ms-1 over 10 years. N-S control ~ 50 ms-1 year-1 => 500 ms-1 over 10 years.

Total ΔV ~ 600 ms-1

ΔV = g0 Isp In [Minitial / Mfinal] Minitial = Mfinal exp (ΔV / g0 Isp) Mfinal= EOL mass = 510 kg => Minitial = 510 x exp (600 / 9.807 x 216) = 677 kg => Mpropellant= Minitial -Mfinal= 677 - 510 = 167 kg

Wet mass (BOL) = 1140 kg - 462 kg (apogee kick motor propellant) - 167 kg (station keeping propellant over 10 years) -23 kg (ACS propellant over 10 years) => Dry mass (BOL) = 1140 - 652 = 488 kg