GEO: The Geostationary Orbit

Station keeping, and why it ends

The orbit does not hold itself. The earth is not a perfect sphere, and its equatorial bulge is not even: the effect is that the ring has a few longitudes a satellite drifts towards and others it drifts away from, so east-west corrections are needed to hold a slot. The sun and the moon pull the orbital plane out of the equator, so north-south corrections are needed too, and they are the expensive ones, consuming the greater part of the fuel a satellite carries. Solar radiation pressure pushes on the panels.

So a geostationary satellite fires thrusters through its whole life to stay where it is said to be, and its life ends when the fuel does, not when the electronics fail. Before the last of it is gone, the operator raises the satellite into a disposal orbit above the ring, so that a dead satellite does not drift through the slots of living ones.

What it reaches, and what it never will

Figure 105.1 The true ten degree footprints of three geostationary satellites, and the polar bands none of them reaches

The program computes how high the satellite sits in the sky from various places. On the equator, under the satellite, it is overhead. From Mumbai's latitude it is still 67.7 degrees up if the longitudes match. At 60 degrees of latitude it has fallen to 21.9 degrees, and at 71 degrees it is 10.4 degrees, scraping the minimum. A binary search in the program finds the limit exactly: 71.4 degrees north or south, on the satellite's own longitude, and less than that anywhere else.

That limit is absolute. It does not improve with a bigger dish or a stronger transmitter, because the satellite is geometrically too low in the sky, and beyond the limit it is under the horizon altogether. No geostationary satellite, and no number of them, covers the poles. That is the single fact that keeps other orbits in business, and the chapter [LEO and MEO] is largely about the systems that exist because of it.

Along the equator the picture is the opposite. One satellite reaches 71.4 degrees of longitude either side of its own, so neighbours parked 120 degrees apart still overlap by 22.9 degrees. Three satellites cover the inhabited world, exactly as [The History of Satellite Systems] records Clarke working out in 1945.

Pointing a dish

Because the satellite never moves, a receiving dish has two numbers and no motor. The program works them out for a real case: from Mumbai at 19.08 degrees north and 72.88 degrees east, to a satellite parked at 83.0 degrees east, the dish must be raised to an elevation of 64.8 degrees and turned to an azimuth of 151.4 degrees from true north, which is south by south-east. The satellite is 36,305 km away, a little further than the 35,786 km directly below it, because the path runs at a slant.

Set those two angles once, tighten the bolts, and the installation is finished for the life of the dish. Every other orbit needs a tracking mount or an electronically steered array.

What the delay costs

The distance has a price, and it is paid in time. At 36,305 km the one-way trip takes 121.1 ms, so up and down again is 242.2 ms, and a double hop through two satellites is 484.4 ms.

For speech that is noticeable. A speaker hears a reply about a quarter of a second late on one hop, so the natural rhythm of a conversation breaks down and the two ends begin to talk over each other; on two hops it is half a second and the call is hard work. This is why an international call is routed through at most one satellite hop where a choice exists, and why submarine fibre took the traffic, as [Applications of Satellite Systems] sets out.

For data the cost is subtler and often larger. A protocol that may have only one window of data unacknowledged in flight is limited to a window per round trip, no matter how fast the link is. The program computes it over a 272 ms round trip: the classic 64 kB window gives 1.93 Mbit/s, a 256 kB window gives 7.70 Mbit/s, and a megabyte window gives 30.82 Mbit/s. The link's own capacity never appears in that arithmetic. A 50 Mbit/s satellite link can deliver under 2 Mbit/s to a connection with a small window, which is the same reasoning as in [Traditional Transport Control Protocols: TCP and UDP] and the reason satellite links use large windows, window scaling, and sometimes a performance-enhancing proxy that acknowledges locally.

Slots, and why they are scarce

There is only one geostationary ring, and every satellite in it must be far enough from its neighbours that a receiving dish can tell them apart. The program counts what the spacings give: 2 degrees apart, 180 slots, with 1,472 km between neighbours; 1 degree apart, 360 slots; half a degree apart, 720 slots.

Fourteen hundred kilometres sounds like room to spare, and in space it is. The constraint is not space but angle: from the ground, two satellites 2 degrees apart are 2 degrees apart in the sky, and a small dish has a beam wide enough to hear both at once, as [Antennas: Radiators, Dipoles and Radiation Patterns] shows. The dish would receive one satellite's programme with the next one's interference laid over it. So the spacing is set by the smallest dish the service expects, the slots are allocated and coordinated internationally, and a slot over a populous longitude is a genuinely scarce asset.

The orbit, computed

# The geostationary orbit: where it has to be, what it reaches, and what it costs.
import math

GM, R = 3.986004418e14, 6378137.0
C = 299792458.0
SIDEREAL_DAY = 86164.0905      # one turn of the earth against the stars, seconds
SOLAR_DAY = 86400.0            # one turn against the sun

# 1. The altitude is not chosen. It is whatever makes the period one sidereal
#    day, so invert Kepler's third law rather than quoting the answer.
def radius_for_period(T):
    return (GM * (T / (2 * math.pi)) ** 2) ** (1.0 / 3.0)

r_geo = radius_for_period(SIDEREAL_DAY)
print("Where the geostationary orbit has to be:")
print("  the earth turns once against the stars in %.1f s, which is 23 h 56 min 4 s"
      % SIDEREAL_DAY)
print("  the orbit with that period has radius   %8.0f km" % (r_geo / 1000))
print("  so its altitude above the surface is    %8.0f km" % ((r_geo - R) / 1000))
print("  and the satellite travels at %.0f m/s, eastward, over the equator"
      % math.sqrt(GM / r_geo))

# 2. Why the sidereal day and not the 24 hour solar day: the mistake, in km.
r_wrong = radius_for_period(SOLAR_DAY)
print()
print("Why the sidereal day and not the 24 hour day we live by:")
print("  a 24 h period needs a radius of %.0f km, which is %.0f km too high"
      % (r_wrong / 1000, (r_wrong - r_geo) / 1000))
drift_deg = 360.0 * (SOLAR_DAY - SIDEREAL_DAY) / SIDEREAL_DAY
print("  a satellite with a 24 h period slips %.3f degrees of longitude a day," % drift_deg)
print("  which is about %.0f degrees a year: it would not be stationary at all."
      % (drift_deg * 365.25))

# 3. What a geostationary satellite reaches. A user at latitude lat and
#    longitude offset dlon sees it at this elevation.
def look(lat_deg, dlon_deg, r_sat=None):
    """Elevation, azimuth from true north, and slant range, by vectors.

    dlon_deg is the satellite's longitude minus the user's, so a positive
    value puts the satellite east of the user.
    """
    r_sat = r_sat or r_geo
    lat, dlon = math.radians(lat_deg), math.radians(-dlon_deg)
    # the user on a spherical earth, with the satellite's longitude as zero
    u = (R * math.cos(lat) * math.cos(dlon), R * math.cos(lat) * math.sin(dlon),
         R * math.sin(lat))
    s = (r_sat, 0.0, 0.0)
    d = tuple(s[i] - u[i] for i in range(3))
    # the local east, north and up directions at the user
    up = (math.cos(lat) * math.cos(dlon), math.cos(lat) * math.sin(dlon), math.sin(lat))
    east = (-math.sin(dlon), math.cos(dlon), 0.0)
    north = (-math.sin(lat) * math.cos(dlon), -math.sin(lat) * math.sin(dlon), math.cos(lat))
    dot = lambda a, b: sum(a[i] * b[i] for i in range(3))
    e, n, u2 = dot(d, east), dot(d, north), dot(d, up)
    rng = math.sqrt(e * e + n * n + u2 * u2)
    return math.degrees(math.asin(u2 / rng)), math.degrees(math.atan2(e, n)) % 360, rng

print()
print("How high a geostationary satellite sits in the sky, by latitude and longitude offset:")
print("  latitude   same longitude   30 deg away      60 deg away")
for lat in (0, 19, 40, 60, 71, 75):
    cells = []
    for dlon in (0, 30, 60):
        el, _, _ = look(lat, dlon)
        cells.append("%-16s" % ("%.1f deg up" % el if el > 0 else "below the horizon"))
    print("  %5d deg  %s" % (lat, "".join(cells).rstrip()))
print("  the highest latitude that sees it ten degrees up, on its own longitude, is")
lo, hi = 0.0, 89.0
for _ in range(60):
    mid = (lo + hi) / 2
    if look(mid, 0)[0] >= 10.0:
        lo = mid
    else:
        hi = mid
print("  %.1f degrees north or south, so the poles are never covered." % lo)

# 4. Pointing a dish. One real place, one real orbital slot.
print()
print("Pointing a dish from Mumbai, 19.08 N 72.88 E, at a satellite parked at 83.0 E:")
el, az, rng = look(19.076, 83.0 - 72.877)
print("  elevation %.1f degrees, azimuth %.1f degrees from true north, range %.0f km"
      % (el, az, rng / 1000))
print("  the dish is aimed once and bolted: the satellite does not move in that sky.")

# 5. What the distance costs in time.
print()
print("The delay a geostationary link pays:")
one_way = rng / C
print("  one way, Mumbai to the satellite:        %6.1f ms" % (one_way * 1000))
print("  up and down again, one hop:              %6.1f ms" % (2 * one_way * 1000))
print("  a double hop, through two satellites:    %6.1f ms" % (4 * one_way * 1000))
print("  a speaker hears a reply a quarter of a second late on one hop, and half a")
print("  second late on two, which is why a double hop is avoided in a phone call.")

# 6. What the delay costs a protocol that waits for acknowledgements.
print()
print("What that delay costs TCP, which may only have a window in flight:")
rtt = 2 * one_way + 0.030            # the satellite hop plus 30 ms of terrestrial tails
for window_kb, label in ((64, "the classic 64 kB window"),
                         (256, "a 256 kB window"),
                         (1024, "a 1 MB window")):
    rate = window_kb * 1024 * 8 / rtt
    print("  %-24s gives %6.2f Mbit/s over a %.0f ms round trip"
          % (label, rate / 1e6, rtt * 1000))
print("  the link's own speed never enters this: the window and the round trip decide it.")

# 7. How many satellites fit, and how far apart they really are.
print()
print("Orbital slots along the geostationary ring:")
circumference = 2 * math.pi * r_geo
for sep in (2.0, 1.0, 0.5):
    print("  %.1f degrees apart: %3d slots in the ring, %5.0f km between neighbours"
          % (sep, int(360 / sep), circumference * sep / 360 / 1000))
print("  they look close from the ground, so a small dish with a wide beam hears both:")
print("  that, not space, is what makes a slot scarce.")

# 8. Three satellites, and the gap they leave.
print()
print("Three satellites, 120 degrees apart, at a 10 degree minimum elevation:")
e = math.radians(10.0)
gamma = math.degrees(math.acos(R / r_geo * math.cos(e)) - e)
cap = 2 * math.pi * R ** 2 * (1 - math.cos(math.radians(gamma)))
print("  each covers %.1f degrees of earth-central angle, or %.1f million km2"
      % (gamma, cap / 1e12))
print("  so one reaches %.1f degrees of longitude either side of its own along the equator"
      % gamma)
print("  and neighbours parked 120 degrees apart still overlap by %.1f degrees there,"
      % (2 * gamma - 120))
print("  but nothing above %.1f degrees of latitude is covered, however many are launched."
      % lo)

Advantages and disadvantages

The examination asks for these by name, so here they are as a list, each one traceable to a number above.

Advantages. The satellite appears fixed, so ground antennas need no tracking and no motor, and installation is cheap. There is no handover caused by the satellite moving, so a link lasts as long as the equipment does. Three satellites cover the inhabited world. The footprint is huge, about a third of the earth, which is what makes broadcasting economic. There is almost no Doppler shift for a fixed user, because the satellite does not move relative to the dish, which removes the problem [Multipath, Fading and the Doppler Effect] describes. The satellite is always available, with no gaps to schedule around.

Disadvantages. The delay is about a quarter of a second round trip, which speech notices and acknowledged protocols pay for. No coverage above about 71 degrees of latitude, and none at all at the poles. The distance means high path loss, so both ends need power and gain, and a handheld terminal cannot close the link at broadcast frequencies. Launching to 35,786 km is expensive, and the satellite must be large and long-lived to be worth it. Slots and frequencies are scarce and must be coordinated. Station keeping consumes fuel, and the fuel decides the satellite's life. And a single satellite is a single point of failure for everything under it.

Distinctions

What it does not mean

Geostationary does not mean motionless. The satellite travels at about 3,075 metres a second; it only appears fixed because the earth turns underneath it at the same rate.

It does not stay put by itself. Without station keeping it drifts in longitude and its plane tilts, so it fires thrusters throughout its life and its life ends with the fuel.

The period is not 24 hours. It is the sidereal day, and using 24 hours would put the satellite 77 km too high and let it slip a degree of longitude a day.

Three satellites do not cover the earth. They cover the inhabited world between about 71 degrees north and south. The poles are outside every geostationary footprint that exists or ever will.

A bigger dish does not extend the coverage. Beyond the latitude limit the satellite is too low in the sky or below the horizon, and gain cannot raise it.

The delay is not a fault of the equipment. It is the distance divided by the speed of light, and no improvement in electronics will reduce it.

Quick revision

• Altitude 35,786 km, radius 42,164 km, over the equator, eastward, circular, zero inclination. The period is the sidereal day, 23 h 56 min 4 s.
• A 24 h period would need 42,241 km and would slip 0.986 degrees of longitude a day.
• Geostationary needs period, circularity, zero inclination and direction; geosynchronous needs only the period.
• Latitude limit 71.4 degrees at a 10 degree minimum elevation, on the satellite's own longitude: the poles are never covered.
• One satellite reaches 71.4 degrees of longitude either side, so three 120 degrees apart overlap by 22.9 degrees at the equator and cover the inhabited world.
• From Mumbai to a satellite at 83 degrees east: elevation 64.8 degrees, azimuth 151.4 degrees, range 36,305 km.
• Delay: 121.1 ms one way, 242.2 ms one hop, 484.4 ms double hop.
• Throughput over a 272 ms round trip: 1.93 Mbit/s with a 64 kB window, 7.70 with 256 kB, 30.82 with 1 MB.
• Slots: 2 degrees apart gives 180 slots and 1,472 km of separation; the limit is the angle a small dish can resolve, not the space.
• Advantages: fixed antenna, no handover, huge footprint, three satellites, no Doppler, always available. Disadvantages: quarter-second delay, no polar coverage, high path loss, expensive launch, scarce slots, station keeping, single point of failure.

Test yourself

1. Why is the geostationary altitude 35,786 km? Because the satellite must go round in exactly the time the earth takes to turn once against the stars, which is the sidereal day of 23 hours 56 minutes 4 seconds. Kepler's third law gives the period from the orbit's radius; inverting it for that period gives a radius of 42,164 kilometres from the earth's centre, and subtracting the earth's radius leaves 35,786 kilometres above the surface. The altitude is therefore arithmetic, not a design choice.

2. What is the difference between a geostationary and a geosynchronous orbit? A geosynchronous orbit only has to have a period of one sidereal day, so the satellite returns to the same place in the sky once a day. A geostationary orbit must also be circular, lie in the equatorial plane and run eastward. Without circularity the satellite runs ahead and falls behind, oscillating east and west; with any inclination it swings north and south and traces a figure of eight. Only when all four conditions hold does the satellite appear motionless, which is what lets a dish be fixed.

3. Why is the sidereal day used rather than the 24 hour day? Because the sidereal day is one true rotation of the earth, while the solar day is about four minutes longer, since the earth must turn a little past one rotation to bring the sun overhead again as it moves along its own orbit. A satellite matched to 24 hours would be 77 kilometres too high, would slip 0.986 degrees of longitude each day, and would drift right round the earth in a year, so it would not be stationary over anything.

4. Why can a geostationary satellite never serve the poles? Because it sits over the equator, so the further a user is from the equator the lower it appears. The chapter's search finds the limit exactly: at 71.4 degrees of latitude, on the satellite's own longitude, the satellite is only ten degrees above the horizon, and beyond that it falls below the usable angle and then below the horizon. The limit is geometric, so no dish size, transmitter power or number of satellites can change it.

5. List the advantages and disadvantages of GEO. Advantages: the satellite appears fixed, so ground antennas need no tracking and installation is cheap; there is no handover caused by satellite movement; three satellites cover the inhabited world; the footprint of about a third of the earth makes broadcasting economic; there is almost no Doppler shift for a fixed user; and the satellite is always available. Disadvantages: a round trip delay of about 242 milliseconds, which speech notices and acknowledged protocols pay for; no coverage above about 71 degrees of latitude and none at the poles; very high path loss, so both ends need power and gain; an expensive launch and a large satellite; scarce, internationally coordinated slots and frequencies; station keeping that consumes fuel and so decides the satellite's life; and a single satellite as a single point of failure.

6. Why does a 50 Mbit/s geostationary link often deliver far less than 50 Mbit/s? Because a protocol that may only have one window of unacknowledged data in flight can send a window per round trip and no more, whatever the link's speed. Over the chapter's 272 millisecond round trip, a 64 kilobyte window yields 1.93 Mbit/s, a 256 kilobyte window 7.70 Mbit/s and a one megabyte window 30.82 Mbit/s. The remedy is a larger window, window scaling, or a proxy that acknowledges data locally, not a faster link.

7. Why are geostationary slots scarce when the satellites are more than a thousand kilometres apart? Because what matters on the ground is angle, not distance. Two satellites two degrees apart along the ring are 1,472 kilometres apart in space but only two degrees apart as seen from a dish, and a small dish has a beam wide enough to receive both at once, so one satellite's signal interferes with the other's. The usable spacing is therefore set by the smallest dish the service expects, the ring holds only as many slots as that allows, and a slot over a populous longitude is genuinely valuable.