Satellites and circular orbits

The simplest kind of orbit is the circular one: the satellite keeps a constant distance $r$ from the centre. Here gravity does just one thing, curving the trajectory, and the physics reduces to a single equation. The gravitational force provides exactly the centripetal acceleration required by circular motion:

$\frac{GMm}{r^2} = m\frac{v^2}{r} \;\Rightarrow\; v = \sqrt{\frac{GM}{r}}$

Orbital velocity

$\ev{v_\text{orb} = \sqrt{\frac{GM}{r}}}$ Compared with the escape velocity at the same distance: $v_\text{orb} = v_\text{fuga}/\sqrt{2}$. Escaping requires $\sqrt{2}$ times the speed needed to orbit.

A counter-intuitive fact emerges from the formula: the higher the orbit (larger $r$), the slower the satellite moves. Low satellites zip along, distant ones proceed calmly. This is why the Space Station, in low orbit, travels at $7{,}7\;\text{km/s}$, while a geostationary satellite, much further away, moves at only $3\;\text{km/s}$.

Geostationary satellite

A geostationary satellite has period $T = 24\;\text{h} = 86\,400\;\text{s}$: it must stay fixed relative to a point in the sky. From Kepler’s third law: $r^3 = \frac{GM_T T^2}{4\pi^2} \;\Rightarrow\; r \approx 4{,}2\cdot 10^7\;\text{m}$ that is, about $35\,800\;\text{km}$ above the surface (Earth’s radius $6\,400\;\text{km}$). The corresponding orbital velocity is: $v_\text{orb} = \sqrt{\frac{GM_T}{r}} \approx 3{,}1\;\text{km/s}$

The geostationary orbit is unique: there is only one radius for which the period is exactly 24 hours. This is why satellite dishes can point fixed at one spot in the sky: the satellite there appears motionless.

Links

Topics: Gravitazione Concepts: Orbite · Velocità di fuga · Leggi di Keplero Objects: Satellite

Related exercises: Vero o falso su energia e orbite · Altitudine dell’orbita geostazionaria · Dimostrazione della velocità di fuga