They are about 36,000 km away — roughly 85 times farther than the ISS — so their reflected sunlight is thousands of times fainter than a low satellite's.
Several Indian satellites work from geostationary orbit, including EOS-05 (GISAT-1A), launched in September 2026. ESA describes this orbit as 35,786 km above the equator, where one orbit takes 23 hours 56 minutes 4 seconds — the same time the Earth takes to turn once. So these satellites appear to hang still over one spot.
You might expect them to be easy to see: they never move, and many are large. But they never appear in this site's pass lists, and the reason is distance.
The ISS orbits about 420 km up. EOS-05 (GISAT-1A), launched in September 2026, sits about 35,790 km up. That is roughly 85 times farther.
Reflected light falls off with the square of the distance. 85 × 85 ≈ 7,200, so the same satellite at geostationary height would look about 7,000 times fainter than overhead in low orbit. On the magnitude scale that is about 9.6 magnitudes fainter. A geostationary satellite would need to be extraordinarily reflective to reach the naked-eye limit; in practice they need a telescope.
You can still follow them on the globe: pick a geostationary satellite such as EOS-05 or a GSAT and watch its estimated longitude barely change while the ISS crosses whole continents.
Select a geostationary satellite such as a GSAT: its estimated position stays over the equator at almost the same longitude, while the ISS sweeps past. Open the live globe
Use the inverse-square law in magnitude form: Δm = 5 log₁₀(ρ₂ / ρ₁).
From 420 km to 35,786 km: Δm = 5 log₁₀(35,786 / 420) = 5 log₁₀(85.2) ≈ 5 × 1.93 ≈ 9.65 magnitudes. In brightness, 10^(9.65/2.5) ≈ 7,300 times fainter.
Compare with the standard magnitude used by this site, defined at 1,000 km and half-phase. A satellite at 35,786 km is already 5 log₁₀(35.786) ≈ 7.8 magnitudes fainter than its standard magnitude, before any phase effects. Even an object with the ISS's standard magnitude of −1.3 would come out near +6.5 — beyond the roughly +6 limit of a truly dark sky and far beyond the 4.5 limit this site uses for Indian cities.
The actual range from an observer is a little more than the altitude unless the satellite is at the zenith, which only happens on the equator. From Delhi (latitude φ ≈ 28.6°) looking due south at a satellite on the same longitude, the elevation is given by tan E = (cos φ − R/r) / sin φ, with R/r = 6,378 / 42,164 ≈ 0.151: tan E = (0.878 − 0.151) / 0.479 ≈ 1.52, so E ≈ 57°. Farther east or west along the arc, the elevation drops towards the horizon.
The site's candidate filter reads mean motion from columns 53–63 of TLE line 2 and drops objects below 11 revolutions per day, which removes the whole geostationary belt (about 1.0027 rev/day) and the navigation satellites in medium orbits.
Its orbital period matches the Earth's rotation (23 h 56 min) and it orbits over the equator in the same direction the Earth turns, so it keeps pace with the ground below.
Δm = 5 log₁₀(10) = 5 magnitudes, a factor of 100 in brightness.
In the southern sky, along an arc above the equator, because Delhi is about 28.6° north of the equator.