Inclination is the angle between a satellite's orbit and Earth's equator: 0° follows the equator, 90° crosses the poles, and above 90° goes against Earth's spin.
Imagine Earth's equator as a flat ring. A satellite's orbit is another ring, around the same centre. The angle between the two rings is the orbit's inclination.
A satellite's ground track can reach as far north and south as its inclination, and no farther. The ISS orbits at about 51.6°, so it passes over everywhere between about 51.6° north and south.
Indian examples cover the whole range. INSAT-3DS sits at 0.10° in geostationary orbit. The astronomy satellite Astrosat flies at 6.0°, close to the equator. The radar satellite RISAT-2B is at 37.0°. NavIC's tilted satellites are at 29°, so they swing between 29° north and south. Cartosat-3, at 97.5°, is retrograde: its track turns around at 180° − 97.5° = 82.5° north and south.
Satellites built to study the tropics pick low inclinations so they keep passing over the tropics instead of the poles. NASA's rain-measuring TRMM satellite flew at 35°. Megha-Tropiques-1, a tropical climate satellite built by ISRO and the French space agency CNES, flew at 20°.
The cheapest orbit to reach is a low-inclination one launched due east from near the equator, because Earth's spin adds speed for free. Launching from farther north or south, or in another direction, gives a higher inclination. Braeunig gives an example: from Cape Canaveral, at 28.5°N, a due-east launch gets a free 1,471 km/h from Earth's rotation.
Changing inclination after launch is expensive. GSLV-F12 left NVS-01 in a transfer orbit tilted 19.36°; the satellite then spent its own fuel cutting that to the 1.84° it shows today.
Select the ISS and look at the northern and southern turning points of its ground track: they sit near 51.6°, its inclination. Compare ground tracks on the globe
A circular orbit's ground track reaches a highest latitude of i (for i ≤ 90°) or 180° − i (for i > 90°), because the orbit plane's tilt is the largest angle any point on it makes with the equator.
At burnout, the inclination follows from the launch latitude φ and the flight heading (azimuth) β, measured clockwise from north:
cos i = cos φ · sin β
A due-east launch (β = 90°) gives cos i = cos φ, so i = φ. That is the lowest inclination reachable directly from a site; anything lower needs a costly plane change later.
The cost of a plane change is found with the law of cosines. For a turn of Δi with no change in speed:
Δv = 2 v sin(Δi / 2)
Worked numbers:
This is why inclination changes are made where the satellite is slowest, at apogee, and combined with other burns.
It is retrograde: it moves partly westward, against Earth's rotation. Sun-synchronous orbits such as Cartosat-3's (97.5°) are examples.
No. Its ground track only reaches 29° north and south, so places farther from the equator are never directly beneath it.
The plane-change cost is 2v sin(Δi/2), which grows with speed. A satellite is slowest at apogee, so the same turn costs less there.