Satellite Types — Geostationary, Polar, LEO, MEO Classification

Question

What are the different types of satellites based on their orbits, and what are the key properties of each?

Solution — Step by Step

Type	Altitude	Period	Examples
LEO (Low Earth Orbit)	200-2000 km	90-120 min	ISS, Starlink, weather satellites
MEO (Medium Earth Orbit)	2000-35,786 km	2-24 hours	GPS, GLONASS navigation satellites
GEO (Geostationary)	35,786 km (exact)	24 hours	INSAT, weather monitoring, DTH TV
Polar orbit	600-800 km (LEO range)	~100 min	Earth observation, mapping, spy satellites

A geostationary satellite orbits in the equatorial plane with a period exactly equal to Earth’s rotation (24 hours). It appears stationary from the ground.

The altitude is fixed by physics:

T = 2\pi\sqrt{\frac{r^3}{GM_E}} = 24 \text{ hours}

Solving gives $r = 42,164$ km from Earth’s centre, or $h = 35,786$ km above the surface.

This is why all DTH TV dishes point in a fixed direction — the satellite does not move relative to the observer.

Polar satellites orbit over the poles (inclination ~90 degrees). As the Earth rotates beneath them, they scan the entire surface over time.

This makes polar satellites ideal for:

Earth observation and mapping
Remote sensing (agriculture, forestry)
Military reconnaissance
Weather imaging (complete global coverage)

graph TD
    A[Satellite Orbit Classification] --> B{By Altitude}
    B --> C[LEO: 200-2000 km]
    B --> D[MEO: 2000-35786 km]
    B --> E[GEO: 35786 km exact]

    A --> F{By Inclination}
    F --> G[Equatorial: 0 degrees - geostationary]
    F --> H[Polar: ~90 degrees]
    F --> I[Inclined: between 0 and 90 degrees]

    C --> C1[Short period, low latency, needs many satellites for coverage]
    D --> D1[GPS constellation: 24 satellites in 6 planes]
    E --> E1[Appears stationary, covers 1/3 of Earth]

Why This Works

Kepler’s third law ( $T^2 \propto r^3$ ) fixes the relationship between orbital altitude and period. Higher altitude = longer period. At exactly 35,786 km, the period matches Earth’s rotation, creating the geostationary “sweet spot.” Any other altitude gives a different period, so the satellite drifts relative to the ground.

For JEE numerical problems on satellites, remember: geostationary orbit height = 36,000 km (approximately), orbital speed = 3.07 km/s. For any circular orbit, $v = \sqrt{GM/r}$ and $T = 2\pi r/v$ .

Alternative Method

Instead of memorising the geostationary altitude, derive it. You know $T = 86400$ s, $G = 6.67 \times 10^{-11}$ , $M_E = 6 \times 10^{24}$ kg. Solve $r = \left(\frac{GMT^2}{4\pi^2}\right)^{1/3}$ to get $r \approx 42,200$ km. Subtract Earth’s radius (6400 km) to get $h \approx 35,800$ km.

Common Mistake

Students think “geostationary” means the satellite is not moving. It IS moving — at about 3 km/s. But because its angular velocity matches Earth’s rotation, it appears stationary from the ground. Also, a geostationary orbit is ALWAYS in the equatorial plane. A satellite at 36,000 km in a non-equatorial orbit would have the right period but would appear to move north-south during the day — this is called a geosynchronous (not geostationary) orbit.