Geostationary Orbits

Many communication satellites around Earth follow a geostationary orbit
This is a specific type of orbit in which the satellite:
- Remains directly above the equator, therefore, it always orbits at the same point above the Earth’s surface
- Moves from west to east (same direction as the Earth spins)
- Has an orbital time period equal to Earth’s rotational period of 24 hours

Geostationary satellites are used for telecommunication transmissions (e.g. radio) and television broadcast
A base station on Earth sends the TV signal up to the satellite where it is amplified and broadcast back to the ground to the desired locations
The satellite receiver dishes on the surface must point towards the same point in the sky
- Since the geostationary orbits of the satellites are fixed, the receiver dishes can be fixed too

Geostationary orbit satellite, downloadable AS & A Level Physics revision notes

Satellite in geostationary orbit remain directly above the equator, and a circular orbit equal to that of the Earth

Calculate the distance above the Earth's surface that a geostationary satellite will orbit.

Mass of the Earth = 6.0 × 10²⁴ kg

Radius of the Earth = 6400 km = 6400 × 10³ m

Answer:

Step 1: State Kepler's Third Law

$T^{2} = \frac{4 π^{2} r^{3}}{G M}$

Step 2: Rearrange to make radius the subject

$r = \sqrt[3]{\frac{G M T^{2}}{4 π^{2}}}$

Step 3: Substitute in the known values to calculate

$r = \sqrt[3]{\frac{(6.67 \times 10^{- 11}) \times (6.0 \times 10^{24}) \times {(86400)}^{2}}{4 π^{2}}}$

$r = 4.2 \times 10^{7} m (2 s . f .)$

Step 4: Determine the distance above Earth's surface

$Distance above Earth' s surface = r - radius of Earth$

$Distance above Earth' s surface = (4.2 \times 10^{7}) - (6400 \times 10^{3})$

$Distance above Earth' s surface = 3.56 \times 10^{7} m$

Geostationary Orbits (CIE A Level Physics)