Explain why geostationary satellite must be at 36000 km altitude

Question

Why must a geostationary satellite orbit at an altitude of approximately 36,000 km above the Earth’s surface? Derive the expression for the orbital radius of a geostationary satellite.

(JEE Main 2022, similar pattern)

Solution — Step by Step

A geostationary satellite must appear stationary relative to the Earth’s surface. This means its orbital period must equal 24 hours (the Earth’s rotation period), and it must orbit in the equatorial plane in the same direction as the Earth’s rotation.

T = 24 \text{ hours} = 86400 \text{ s}

For a satellite of mass $m$ orbiting at radius $r$ from the Earth’s centre:

\frac{GMm}{r^2} = \frac{mv^2}{r} = mr\omega^2

Since $\omega = \frac{2\pi}{T}$ , we get:

\frac{GM}{r^2} = r \cdot \frac{4\pi^2}{T^2}

Rearranging:

r^3 = \frac{GMT^2}{4\pi^2}

r = \left(\frac{GMT^2}{4\pi^2}\right)^{1/3}

Substituting $G = 6.674 \times 10^{-11}$ N m $^2$ kg $^{-2}$ , $M = 5.972 \times 10^{24}$ kg, and $T = 86400$ s:

r \approx 42,200 \text{ km}

The Earth’s radius $R_E \approx 6,400$ km. Altitude $h = r - R_E$ :

h = 42,200 - 6,400 = \mathbf{35,800 \approx 36,000 \text{ km}}

This is the only altitude where the orbital period matches Earth’s rotation period. Any closer and the satellite orbits faster; any farther and it orbits slower.

Why This Works

The key insight is that for circular orbits, the orbital period depends only on the orbital radius (for a given central body). There is a unique one-to-one relationship between $r$ and $T$ — Kepler’s third law tells us $T^2 \propto r^3$ .

Since we need $T = 24$ hours exactly, only one value of $r$ satisfies this condition. The physics forces this altitude on us — we have no freedom to choose differently. This is why all geostationary satellites (INSAT, GSAT series) sit in the same orbital ring at ~36,000 km.

Alternative Method — Using $g$ and $R_E$ Directly

We can avoid plugging in $G$ and $M$ separately by using $GM = gR_E^2$ :

r^3 = \frac{gR_E^2 T^2}{4\pi^2}

r = \left(\frac{9.8 \times (6.4 \times 10^6)^2 \times (86400)^2}{4\pi^2}\right)^{1/3} \approx 42,200 \text{ km}

This trick of replacing $GM$ with $gR^2$ saves calculation time in numericals and avoids handling the awkward value of $G$ . JEE toppers use this substitution routinely.

Common Mistake

Many students confuse orbital radius with altitude. The formula gives $r$ (distance from Earth’s centre), but the question often asks for $h$ (height above surface). Always subtract $R_E = 6400$ km at the end. Missing this step gives the wrong answer of 42,200 km instead of ~36,000 km.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using ggg and RER_ERE​ Directly

Common Mistake

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Alternative Method — Using $g$ and $R_E$ Directly