Equatorial Earth station
For \(\lambda_E=0\), \(b=|\phi_E-\phi_{SS}|\). The original procedure reduces the azimuth to \(+90^\circ\) or \(-90^\circ\), according to whether the satellite lies east or west.
EE 499 · Satellite Technologies
Week 3
Geostationary Orbit & Earth-Station Geometry
From the geostationary condition to antenna look angles and visibility.
By the end: you should be able to distinguish GSO from GEO, derive GEO radius, calculate Earth-station geometry, explain visibility limits, and motivate Molniya orbits.
This chapter is a faithful modern representation of Topic 2 in the original CME 454 notes, with equations typeset and source pages linked.
00 — Why GEO matters
A fixed direction in the sky
A satellite in the geostationary orbit appears fixed in the sky with respect to points on Earth. This is very useful for direct communication: the receiving antenna can be fixed looking in the direction of the satellite, without any need for tracking.
That convenience creates the central engineering task of this week. Given an Earth-station latitude and longitude and a satellite longitude, determine the distance to the satellite, the elevation angle above the local horizon, and the azimuth measured from north.
01 — Geosynchronous versus geostationary
Matching period is necessary, not sufficient
Kepler's third law connects orbital period and semi-major axis: \(a^3\propto T^2\). The handwritten lecture therefore begins with an orbit whose period is 24 hours: the satellite completes one revolution in the same time that Earth completes one revolution about its axis. Such an orbit is called geosynchronous.
However, this condition by itself does not make a satellite appear stationary from Earth. The orbit is geostationary only when three additional conditions are satisfied:
01 — Geosynchronous versus geostationary
Three conditions for a geostationary orbit
Circular\(e\approx0\)
Equatorial\(i=0^\circ\)
ProgradeTravel from west to east
Only one orbital ring satisfies all these conditions, so the original notes describe it as a natural resource whose use must be regulated and divided fairly among nations.
02 — GEO radius and altitude
From period to the equatorial ring
The historical lecture uses a nominal 24-hour argument and pairs it with the familiar GEO radius near \(42{,}164\ \text{km}\). This is preserved as historical context, but those two numbers are not an exact Kepler-law pair.
\[a^3=\frac{\mu}{(2\pi/T)^2},\qquad a=\left(\frac{\mu T^2}{4\pi^2}\right)^{1/3}.\]
02 — GEO radius and altitude
The authoritative GEO radius and altitude
For the current authoritative course calculation, a stationary longitude requires Earth's sidereal rotation period:
\[T_{\text{sidereal}}\approx86{,}164\ \text{s},\qquad \mu_E=398600.4418\ \text{km}^3/\text{s}^2.\]
\[a_{GEO}=\left(\frac{\mu_E T_{\text{sidereal}}^2}{4\pi^2}\right)^{1/3}\approx42{,}164\ \text{km}.\]
02 — GEO radius and altitude
Altitude above the equator
Using the established EE 499 Earth-radius convention \(R_E=6378\ \text{km}\),
\[h_{GEO}=a_{GEO}-R_E\approx42{,}164-6378\approx35{,}786\ \text{km}.\]
Historical convention: \(T=24\times60\times60\ \text{s}\), with \(a_{GSO}\approx42{,}164\ \text{km}\), circumference approximately \(265{,}000\ \text{km}\), and altitude approximately \(35{,}790\ \text{km}\), remembered as roughly \(36{,}000\ \text{km}\).
03 — Earth-station geometry
Three givens, three unknowns
Let \(\lambda_E\) be Earth-station latitude, \(\phi_E\) Earth-station longitude, and \(\phi_{SS}\) satellite longitude. The required outputs are slant range \(d\), elevation \(El\), and azimuth \(Az\). Introduce the geocentric separation angle \(b\):
\[\cos b=\cos(\phi_E-\phi_{SS})\cos\lambda_E.\]
\[d=\sqrt{R_E^2+a_{GSO}^2-2R_Ea_{GSO}\cos b}.\]
03 — Earth-station geometry
Elevation and azimuth
\[\cos(El)=\frac{a_{GSO}\sin b}{d},\qquad \sin A=\frac{\sin|\phi_E-\phi_{SS}|}{\sin b}.\]
The handwritten procedure then selects the azimuth quadrant from the signs of latitude and longitude difference. This sign convention is retained in the source. The modern calculator uses compass azimuth measured clockwise from true north: \(0^\circ\) north, \(90^\circ\) east, \(180^\circ\) south, and \(270^\circ\) west.
04 — Antenna look angles
Calculate a fixed pointing direction
Modern calculator convention: \(R_E=6378\ \text{km}\), \(a_{GEO}=42164\ \text{km}\).
- Longitude difference
- Slant range
- Elevation
- Azimuth clockwise from north
Calculating…
Procedure from the original lecture notes · Topic 2, pp. 4–6 →
05 — Special look-angle cases
Keep the geometry visible
Directly below the satellite
For \(\lambda_E=0\) and \(\phi_E=\phi_{SS}\), \(b=0\), \(d=a_{GSO}-R_E\), and \(El=90^\circ\). Azimuth is meaningless because the satellite is overhead.
Same longitude, nonzero latitude
For \(\phi_E=\phi_{SS}\) and \(\lambda_E\ne0\), \(b=|\lambda_E|\). Azimuth is \(0^\circ\) from a southern station or \(180^\circ\) from a northern station.
06 — Polar-mount antennas
One mechanical rotation axis
Many commercial antennas can be rotated on one axis only. In this case, the original procedure first adjusts the elevation to a fixed value \(El_0\):
\[\cos(El_0)=\frac{a_{GSO}}{d}\sin\lambda_E.\]
The antenna elevation is kept constant at this angle. Azimuthal rotation is then calculated using the same geometry as before, with
\[\sin A=\frac{\sin|\phi_E-\phi_{SS}|}{\sin b},\qquad b=\cos^{-1}\!\left[\cos(\phi_E-\phi_{SS})\cos\lambda_E\right].\]
07 — Limits of GEO visibility
The horizon sets a hard boundary
At any location, the antenna elevation must be greater than zero—above the horizon. At the equator, the zero-elevation limit gives a maximum theoretical longitude separation of approximately \(81^\circ\):
\[\cos(\phi_E-\phi_{SS})\ge\frac{R_E}{a_{GSO}},\qquad |\phi_E-\phi_{SS}|_{\max}\approx81^\circ.\]
07 — Limits of GEO visibility
Practical elevation and the latitude limit
In practice, the minimum elevation is higher than \(0^\circ\) to avoid obstruction by surrounding objects and to reduce ground reflections. The allowed longitude difference therefore becomes smaller.
At the north and south poles, a zero-elevation look direction runs parallel to the equator and never meets the GSO ring; the required elevation would be below the horizon. Thus the maximum allowable longitude difference is widest at the equator and decreases as the magnitude of Earth-station latitude increases, reaching zero before the pole.
07 — Limits of GEO visibility
Maximum longitude separation versus latitude
Original handwritten geometry · Topic 2, p. 11 → · View original handwritten figure · Topic 2, p. 13 →
08 — Beyond GEO: Molniya
A high-latitude orbital solution
The problem of limited GSO visibility affected Russia strongly because much of its territory lies in the extreme north. To address this problem, Molniya orbits were used: inclined, highly elliptical orbits with the following historical characteristics.
The historical notes state that apogee reaches geostationary-orbit distance and perigee is as low as roughly \(300\ \text{km}\) above sea level. They also list an inclination of approximately \(65^\circ\). These source statements are retained above and in the linked pages.
| Eccentricity | \(e=0.74\) |
|---|---|
| Semi-major axis | \(a=26{,}600\ \text{km}\) |
| Inclination | approximately \(65^\circ\) |
| Period | approximately 12 hours, or two revolutions per day |
| Argument of perigee | \(\omega=270^\circ\) |
08 — Beyond GEO: Molniya
Modern consistency check
Modern consistency check: the characteristic Molniya critical inclination is approximately \(63.4^\circ\). Applying the listed historical \(a=26{,}600\ \text{km}\) and \(e=0.74\) gives
\[r_p=a(1-e)=6916\ \text{km},\qquad r_a=a(1+e)=46{,}284\ \text{km},\] \[h_p=r_p-R_E=538\ \text{km},\qquad h_a=r_a-R_E=39{,}906\ \text{km}.\]Thus the listed elements imply neither exactly \(300\ \text{km}\) perigee altitude nor exactly the \(42{,}164\ \text{km}\) GEO radius at apogee. The historical values are approximate and are not mutually exact. The physical conclusion remains: by Kepler's second law the satellite moves slowly near its high northern apogee, giving a long dwell, and multiple satellites can sustain coverage.
09 — Check your understanding
Connect condition, geometry, and coverage
Concept check
A geosynchronous satellite has the correct orbital period but an inclination of \(10^\circ\). Is it geostationary?
EE 499 · Week 3
Week 3 complete
- period
- stationarity
- radius
- look angles
- visibility
- high-latitude coverage