Gravity as centripetal force; geostationary and low-Earth orbits; uses of artificial satellites
AQA GCSE Physics 4.8 ยท Year 11 Higher Tier
๐ Explain how gravity provides the centripetal force that keeps objects in orbit
๐ Use the relationship between orbital speed, radius, and period
๐ฐ๏ธ Describe the properties of low-Earth orbits (LEO) and geostationary orbits (GEO)
๐ก Explain why geostationary satellites are used for communications and weather monitoring
๐ญ Explain why LEO satellites are used for imaging, GPS, and scientific research
โ๏ธ Analyse how orbital speed and period change with orbital radius
๐ Gravity as Centripetal Force
Any object moving in a circle must experience a centripetal force โ a force directed towards the centre of the circle. For satellites and planets orbiting in space, this centripetal force is provided entirely by gravity.
Centripetal force is the resultant force acting towards the centre of a circular path, causing an object to continuously change direction while maintaining (approximately) constant speed.
Without gravity, a satellite would travel in a straight line at constant velocity (Newton's First Law). Gravity continuously "pulls" the satellite inward, bending its path into a curve. At exactly the right speed for a given height, this curve matches the curvature of the Earth โ producing a stable circular orbit.
Centripetal force = (mass ร orbital speedยฒ) รท orbital radius F = mvยฒ / r
where F = centripetal force (N), m = mass of satellite (kg),
v = orbital speed (m/s), r = orbital radius (m)
๐ก The gravitational force does NOT do work on the satellite in a circular orbit because it acts perpendicular to the direction of motion. The satellite's speed stays constant but its direction changes continuously.
The orbital radius r is measured from the centre of the Earth, not from the surface. So if a satellite orbits at height h above the surface, and Earth's radius is RE โ 6.4 ร 10โถ m, then r = RE + h.
โฑ๏ธ Orbital Speed and Period
The orbital period T is the time taken for one complete orbit. The orbital speed v can be calculated from the circumference of the orbit divided by the period:
v = 2ฯr / T
where v = orbital speed (m/s), r = orbital radius (m), T = orbital period (s)
By combining this with the centripetal force equation, we can see a fundamental relationship: satellites in larger orbits travel more slowly and have longer orbital periods.
โฌ๏ธ Larger orbital radius โ Slower orbital speed โ Longer orbital period
โฌ๏ธ Smaller orbital radius โ Faster orbital speed โ Shorter orbital period
This makes intuitive sense: the further from Earth, the weaker gravity is, so less centripetal force is needed, meaning a slower speed is sufficient to maintain the orbit. However, the orbit path is also much longer, so the period increases significantly.
Symbol
Quantity
SI Unit
v
Orbital speed
m/s
r
Orbital radius (from Earth's centre)
m
T
Orbital period
s
m
Mass of satellite
kg
F
Centripetal (gravitational) force
N
๐ฐ๏ธ Low-Earth Orbit (LEO)
Low-Earth orbit satellites orbit at altitudes typically between 200 km and 2,000 km above the Earth's surface. This gives an orbital radius of roughly 6,600 km to 8,400 km from Earth's centre.
Low-Earth Orbit (LEO): An orbit with an altitude of approximately 200โ2,000 km above Earth's surface, with an orbital period of roughly 90โ130 minutes.
Properties of LEO satellites:
Fast orbital period: typically around 90โ120 minutes per orbit
High orbital speed: approximately 7,000โ8,000 m/s (~28,000 km/h)
Not fixed above one location: the satellite moves across the sky as seen from the ground
Closer to Earth: stronger gravitational field, so more centripetal force is needed
Lower launch energy: cheaper to launch than higher-orbit satellites
Uses of LEO satellites:
๐ Earth observation and imaging โ high resolution images of the surface
๐ฌ Scientific research โ International Space Station is in LEO (~400 km)
๐ GPS navigation โ constellations of LEO/MEO satellites
๐ก Broadband internet โ Starlink-type constellations use LEO for low latency
โ ๏ธ Because LEO satellites are not stationary above one point, a network (constellation) of many satellites is needed for continuous global coverage.
๐ก Geostationary Orbit (GEO)
A geostationary satellite has a very specific orbit: it orbits directly above the Earth's equator at an altitude of approximately 35,786 km (โ 36,000 km), with an orbital period of exactly 24 hours. This means it rotates at the same rate as the Earth rotates, so it appears stationary over the same point on the equator.
Geostationary orbit: An equatorial orbit at ~36,000 km altitude where the satellite's orbital period equals Earth's rotation period (24 hours), keeping the satellite fixed above the same point on the equator.
Properties of GEO satellites:
Orbital period: exactly 24 hours
Orbital radius: โ 4.2 ร 10โท m from Earth's centre
Orbital speed: โ 3,070 m/s (~11,000 km/h)
Fixed position in sky: appears stationary from the ground
Orbit must be equatorial: directly above the equator
High altitude: can "see" about 40% of Earth's surface at once
Uses of geostationary satellites:
๐บ Television broadcasting โ dish always points same direction
๐ฑ Telecommunications โ phone and data relay
๐ฆ๏ธ Weather forecasting โ continuous monitoring of weather systems
โก Disadvantage: The large distance (~36,000 km) causes a signal delay of about 0.24 seconds each way โ significant for real-time communications like phone calls or video conferencing.
โ๏ธ Comparing LEO and GEO
Feature
LEO
GEO
Altitude
200โ2,000 km
~35,786 km
Orbital period
~90โ130 minutes
24 hours
Orbital speed
~7,000โ8,000 m/s
~3,070 m/s
Appears stationary?
No
Yes
Signal delay
Very small (~ms)
~0.24 s each way
Coverage
Small strip per pass
~40% of Earth
Image resolution
High (closer)
Lower (further)
Launch cost
Lower
Higher
Satellites needed for global coverage
Many (~20+)
3โ4
๐ Only 3โ4 geostationary satellites are needed for near-global coverage (excluding polar regions), whereas dozens of LEO satellites are required for continuous coverage.
Note that geostationary satellites cannot cover polar regions because they sit above the equator and the viewing angle becomes too shallow near the poles. For polar coverage, scientists use polar-orbiting LEO satellites that pass over both poles in each orbit.
Example 1: The International Space Station (ISS) orbits Earth at an altitude of 400 km. Earth's radius is 6.4 ร 10โถ m. Calculate the orbital speed of the ISS if its orbital period is 92 minutes.
1Identify the orbital radius:
r = REarth + altitude = 6.4 ร 10โถ m + 400 ร 10ยณ m
r = 6.4 ร 10โถ + 0.4 ร 10โถ = 6.8 ร 10โถ m
2Convert the period to seconds:
T = 92 minutes ร 60 s/min = 5,520 s
3Apply the orbital speed formula:
v = 2ฯr / T
v = (2 ร ฯ ร 6.8 ร 10โถ) / 5,520
v = (4.272 ร 10โท) / 5,520
4Calculate:
v = 7,739 m/s โ 7,740 m/s
Orbital speed of ISS โ 7,740 m/s (โ 7.7 km/s)
Example 2: A geostationary satellite orbits at a radius of 4.2 ร 10โท m from Earth's centre. Calculate its orbital speed and show that the orbital period is approximately 24 hours.
1Write down known values:
r = 4.2 ร 10โท m
T = 24 hours = 24 ร 3,600 = 86,400 s
2Calculate the circumference of the orbit:
Circumference = 2ฯr = 2 ร ฯ ร 4.2 ร 10โท
= 2.639 ร 10โธ m
3Calculate orbital speed:
v = 2ฯr / T = 2.639 ร 10โธ / 86,400
v = 3,054 m/s โ 3,100 m/s
4Verify: rearrange to find T:
T = 2ฯr / v = 2.639 ร 10โธ / 3,054 = 86,415 s รท 3,600 โ 24 hours โ
Geostationary orbital speed โ 3,100 m/s; period โ 24 hours โ
Example 3: A satellite of mass 500 kg orbits Earth at an altitude of 800 km. Earth's radius is 6.4 ร 10โถ m and the satellite's orbital speed is 7,450 m/s. Calculate the centripetal force acting on the satellite.
1Calculate the orbital radius:
r = 6.4 ร 10โถ + 800 ร 10ยณ = 6.4 ร 10โถ + 0.8 ร 10โถ = 7.2 ร 10โถ m
2Write down the centripetal force formula:
F = mvยฒ / r
Centripetal force โ 3,850 N (provided by gravity)
Example 4: A communications satellite orbits at 36,000 km altitude. A spy satellite orbits at 300 km altitude. Compare their orbital periods and explain why the communications satellite is more suitable for TV broadcasting.
1Communications satellite period (GEO):
r = 6.4 ร 10โถ + 3.6 ร 10โท = 4.24 ร 10โท m
T = 24 hours (by definition for geostationary orbit)
2Spy satellite period (LEO, estimate using v โ 7,700 m/s):
r = 6.4 ร 10โถ + 3.0 ร 10โต = 6.7 ร 10โถ m
T = 2ฯr / v = (2 ร ฯ ร 6.7 ร 10โถ) / 7,700
T = 4.21 ร 10โท / 7,700 โ 5,466 s โ 91 minutes
3Comparison:
GEO: T = 24 hours โ appears stationary above equator
LEO spy satellite: T โ 91 min โ moves across sky rapidly
4Why GEO is better for TV broadcasting:
โข Appears stationary โ satellite dish can point at fixed position
โข No tracking required โ cheaper, simpler receivers
โข Continuous coverage of same area โ uninterrupted signal
โข LEO would need complex tracking and signal would drop out regularly
GEO period = 24 h vs LEO โ 91 min. GEO is better for TV: fixed position means dishes don't need to track and signal is continuous.
Question 1: What provides the centripetal force that keeps a satellite in orbit around the Earth?
Question 2: A geostationary satellite has an orbital period of:
Question 3: A satellite orbits Earth at a radius of 8.0 ร 10โถ m with an orbital period of 7,100 s. Calculate the orbital speed. Give your answer in m/s to 3 significant figures.
Question 4: Which of the following is NOT an advantage of a low-Earth orbit satellite compared to a geostationary satellite?
Question 5: A satellite of mass 800 kg orbits at a speed of 6,500 m/s at an orbital radius of 7.5 ร 10โถ m. Calculate the centripetal force on the satellite in Newtons. Give your answer to 3 significant figures.
Challenge 1 (6 marks): A weather satellite orbits Earth at an altitude of 1,200 km. Earth's radius = 6.4 ร 10โถ m. The satellite orbits at a speed of 7,120 m/s.
(a) Calculate the orbital radius from Earth's centre. [1 mark]
(b) Calculate the orbital period in minutes. [3 marks]
(c) Explain why this satellite is more suitable for monitoring polar weather than a geostationary satellite. [2 marks]
(a) r = 6.4 ร 10โถ + 1.2 ร 10โถ = 7.6 ร 10โถ m โ
(b) T = 2ฯr / v = (2 ร ฯ ร 7.6 ร 10โถ) / 7,120 โ
T = 4.775 ร 10โท / 7,120 = 6,707 s โ
T = 6,707 รท 60 โ 112 minutes โ
(c) Geostationary satellites orbit above the equator so they have a very shallow viewing angle to polar regions and cannot effectively observe them. [1 mark]
The LEO satellite is in a polar orbit, passing over both poles, so it can scan the entire Earth including polar regions within each orbit. [1 mark]
Challenge 2 (5 marks): A student claims: "The further a satellite is from Earth, the faster it must travel to maintain its orbit because the orbit circumference is larger."
(a) State whether this claim is correct and explain what actually happens to orbital speed as orbital radius increases. [3 marks]
(b) Calculate the orbital speeds of two satellites: one at r = 7.0 ร 10โถ m (T = 5,800 s) and one at r = 2.0 ร 10โท m (T = 28,200 s). Use these to support your answer to (a). [2 marks]
(a) The claim is incorrect. [1 mark]
As orbital radius increases, orbital speed decreases. [1 mark]
The gravitational force (centripetal force) decreases with distance, so a smaller centripetal force is needed, meaning a lower speed can maintain the orbit. Although the circumference increases, the period increases even more rapidly, so speed decreases. [1 mark]
(b) Satellite 1: v = 2ฯ ร 7.0 ร 10โถ / 5,800 = 7,584 m/s โ 7,580 m/s โ
Satellite 2: v = 2ฯ ร 2.0 ร 10โท / 28,200 = 4,453 m/s โ 4,450 m/s โ
Larger radius โ smaller speed, confirming (a). โ
Challenge 3 (6 marks): A telecommunications company wants to provide satellite internet with minimal signal delay. An engineer proposes using a constellation of LEO satellites at 550 km altitude instead of a single geostationary satellite at 36,000 km altitude.
(a) The speed of electromagnetic waves is 3.0 ร 10โธ m/s. Calculate the signal travel time from Earth to a geostationary satellite and back. [2 marks]
(b) Calculate the signal travel time from Earth to an LEO satellite at 550 km and back. [2 marks]
(c) Explain why a constellation of many LEO satellites is needed for continuous internet coverage, whereas fewer geostationary satellites are needed for TV broadcasting. [2 marks]
(a) Distance to GEO = 36,000 km = 3.6 ร 10โท m
Round trip = 2 ร 3.6 ร 10โท = 7.2 ร 10โท m
Time = distance / speed = 7.2 ร 10โท / 3.0 ร 10โธ = 0.24 s โโ
(b) Distance to LEO = 550 km = 5.5 ร 10โต m
Round trip = 2 ร 5.5 ร 10โต = 1.1 ร 10โถ m
Time = 1.1 ร 10โถ / 3.0 ร 10โธ = 0.0037 s = 3.7 ms โโ
(c) LEO satellites move across the sky and are not fixed over one location, so any single satellite is only in view for minutes at a time. A large constellation ensures at least one satellite is always visible from any location. [1 mark]
TV broadcasting only requires signal to be sent in one direction (broadcast), and a dish can always point at the fixed GEO satellite โ no tracking needed, so fewer satellites give continuous coverage. [1 mark]
Challenge 4 โ Extended Response (6 marks): Describe and explain the key differences between a geostationary orbit and a low-Earth orbit satellite, including: altitude, orbital period, orbital speed, and the most suitable applications for each type. Include relevant physics principles in your answer.
Model Answer (6 marks):
GEO satellites orbit at approximately 36,000 km altitude above the equator, with a 24-hour period matching Earth's rotation, so they appear stationary from the ground. Their orbital speed is approximately 3,100 m/s. Because they are further from Earth, gravitational force is weaker, so they need less centripetal force โ achieved by a lower orbital speed. [2 marks]
LEO satellites orbit at 200โ2,000 km altitude. At this lower altitude, gravitational force is stronger, requiring a greater centripetal force โ provided by their higher orbital speed (~7,000โ8,000 m/s). The shorter orbital circumference combined with the high speed gives a period of only ~90โ130 minutes. [2 marks]
Applications: GEO is best for TV broadcasting and telecommunications because the fixed position means simple fixed-dish receivers can be used with no tracking, and coverage is continuous. LEO is best for Earth imaging (closer = higher resolution), scientific research (e.g. ISS), and polar weather monitoring, but many satellites are needed for global coverage as they are not stationary. [2 marks]