3.7.2.4
Geostationary orbits have a fixed position above the equator
Gravitational Fields & Orbits — AQA A-Level Physics
Key Definition
Synchronous orbit — An orbit where the orbiting body has a time period equal to the rotational period of the body being orbited, and orbits in the same direction of rotation.
Key Definition
Geostationary orbit — A specific synchronous orbit that is directly above the equator, in the plane of the equator, moving west to east, with a period of exactly 24 hours.
- A geostationary satellite always stays above the same point on the Earth's surface.
- Used for telecommunications and TV broadcasting. Satellite dishes point at a fixed position in the sky.
- The orbital radius is approximately 36 000 km above the Earth's surface (42 000 km from Earth's centre).
Low-Earth orbits
- Low-Earth orbit (LEO) satellites are much closer to the surface (typically 200-2000 km).
- Polar orbits pass over the north and south poles. The Earth rotates beneath them, so they eventually cover the whole surface.
- LEO is used for high-resolutionThe smallest change in a quantity that an instrument can detect. For example, a ruler has a resolution of 1 mm. photography, weather monitoring, and military applications.
- LEO satellites have shorter periods and faster speeds than geostationary satellites.
Worked Example
A LEO satellite orbits at 250 km altitude with a periodThe time taken for one complete oscillation or wave cycle. Measured in seconds (s). of 89 minutes. Calculate the height of a geostationary orbitAn orbit with a periodThe time taken for one complete oscillation or wave cycle. Measured in seconds (s). of exactly 24 hours, directly above the equator, so the satellite remains above the same point on Earth's surface. above the Earth's surface. Radius of Earth = 6370 km.
Show Solution
1
List known values
- LEO periodThe time taken for one complete oscillation or wave cycle. Measured in seconds (s).: $T_L = 89$ min
- Geostationary period: $T_G = 24 \times 60 = 1440$ min
- LEO altitude: $h_L = 250$ km
- $R_E = 6370$ km
2
Calculate LEO orbital radius
$$r_L = R_E + h_L = 6370 + 250 = 6620 \text{ km} = 6.62 \times 10^6 \text{ m}$$
3
Apply Kepler's third law as a ratio
$$\frac{T_G^2}{T_L^2} = \frac{r_G^3}{r_L^3}$$
$$r_G = r_L \left(\frac{T_G}{T_L}\right)^{2/3}$$4
Substitute and evaluate
$$r_G = 6.62 \times 10^6 \times \left(\frac{1440}{89}\right)^{2/3} = 4.24 \times 10^7 \text{ m}$$
$$Y = r_G - R_E = 4.24 \times 10^7 - 6.37 \times 10^6 = 3.6 \times 10^7 \text{ m}$$Answer
$Y = 36\,000$ km above the Earth's surface
Examiner Tips and Tricks
- Memorise the three key features of a geostationary orbitAn orbit with a period of exactly 24 hours, directly above the equator, so the satellite remains above the same point on Earth's surface.: equatorial plane, west to east, period of 24 hours.
- This comes up frequently as a 'state' question worth 2-3 marks.