Altitude of the geostationary orbit

Problem

A telecommunications satellite is placed in geostationary orbit, that is, with period $T = 24\ \text{h}$ above the equator. Determine the altitude $h$ (distance from the Earth’s surface) of the orbit, assuming a circular orbit. Comment on why it is useful for the satellite to appear stationary in the sky.

Data (standard): $M_T = 6{,}0\cdot 10^{24}\ \text{kg}$, $R_T = 6{,}4\cdot 10^6\ \text{m}$, $G = 6{,}67\cdot 10^{-11}\ \text{N}\,\text{m}^2/\text{kg}^2$.

Solution

Setting up. The gravitational force provides the centripetal acceleration of the circular motion of radius $r$ (measured from the Earth’s centre): $\frac{GM_T m}{r^2} = m\,\omega^2 r = m\left(\frac{2\pi}{T}\right)^2 r$

The satellite’s mass $m$ cancels out; solving for $r$: $r^3 = \frac{GM_T\,T^2}{4\pi^2} \;\Rightarrow\; r = \left(\frac{GM_T\,T^2}{4\pi^2}\right)^{1/3}$

Period in seconds. $T = 24\ \text{h} = 24\cdot 3600\ \text{s} = 8{,}64\cdot 10^4\ \text{s}$

Numerical substitution. $r = \left(\frac{6{,}67\cdot 10^{-11}\cdot 6{,}0\cdot 10^{24}\cdot(8{,}64\cdot 10^4)^2}{4\pi^2}\right)^{1/3} = \left(7{,}57\cdot 10^{22}\right)^{1/3} \approx 4{,}23\cdot 10^7\ \text{m}$

Altitude (subtracting the Earth’s radius): $h = r - R_T = 4{,}23\cdot 10^7 - 6{,}4\cdot 10^6$

$\ev{h \approx 3{,}6\cdot 10^7\ \text{m} \approx 36\,000\ \text{km}}$

Why it is useful. With $T=24\ \text{h}$ the satellite rotates in sync with the Earth: seen from the ground it appears stationary in the sky. This lets satellite dishes point in a fixed direction, without tracking the satellite — ideal for telecommunications and television.

Links

Topics: Gravitazione Concepts: Orbite · Forza centripeta · Moto circolare uniforme · Leggi di Keplero Skills: Impostazione simbolica Objects: Satellite