Circular and geostationary orbits
Key idea: H2 Physics lessons on inverse-square gravitation, field strength, gravitational potential, escape speed and satellite orbits.
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The core idea
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Big question: How does gravity set the speed and period of a circular orbit?
For a circular orbit, gravity alone supplies the radial resultant: GMm/r² = mv²/r. This gives v = √(GM/r) and T² = 4π²r³/(GM). A geostationary satellite must have a 24-hour period, orbit above the equator in the same direction as Earth's rotation and remain in one circular orbital plane.
Let gravity provide the inward resultant
A circular-orbit satellite has tangential velocity and inward gravitational acceleration. It continually falls away from its tangent as the planet's surface curves beneath it.
Equating GMm/r² to mv²/r gives v = √(GM/r). Satellite mass cancels: at the same radius, ideal circular-orbit speed does not depend on satellite mass.
Check your understanding: Is gravity balanced by a centripetal force in orbit?
No. Gravity is the unbalanced inward resultant that produces centripetal acceleration.
Connect radius, speed, period and purpose
Using v = 2πr/T gives T² = 4π²r³/(GM). Larger circular orbits are slower and take much longer to complete.
A geostationary satellite must be circular, equatorial, eastward and have Earth's rotation period. Remaining above one longitude makes continuous communication and weather observation possible.
Check your understanding: Why is a 24-hour polar orbit not geostationary?
It does not lie over the equator and therefore does not remain above one longitude.
Key ideas to keep
- A higher circular orbit has lower speed but a longer period.
- The satellite is continually falling, not beyond gravity.
- Geostationary is more restrictive than merely geosynchronous.
See the reasoning
Worked example
Find orbit speed and period
Question: A satellite orbits a planet of mass 6.0 × 10²⁴ kg at centre distance 7.0 × 10⁶ m. Find its circular speed and period. Use G = 6.67 × 10⁻¹¹ N m² kg⁻².
Step 1: Equate gravity and inward resultant
Why: Gravity is the real force producing circular acceleration.
Working: GMm/r² = mv²/r, so v = √(GM/r).
Step 2: Calculate speed
Why: Satellite mass cancels from the orbit condition.
Working: v = √[(6.67 × 10⁻¹¹)(6.0 × 10²⁴)/(7.0 × 10⁶)] = 7.56 × 10³ m s⁻¹.
Step 3: Use circumference over speed
Why: One orbit covers distance 2πr at constant speed.
Working: T = 2πr/v = 2π(7.0 × 10⁶)/(7.56 × 10³) = 5.82 × 10³ s.
Answer: Orbital speed is 7.56 km s⁻¹ and period is 5.82 × 10³ s, or about 97.0 min.
Check: The period is longer than the time to travel one radius at that speed by the expected factor 2π.
Another worked model
Question
A satellite orbits Earth at r = 4.22 × 10⁷ m. Find its speed and period, then state the additional conditions and one use needed for a geostationary interpretation.
Check the worked solution
v = √(GM/r) = 3.07 × 10³ m s⁻¹. T = 2πr/v = 8.64 × 10⁴ s = 24.0 h. It must orbit circularly in the equatorial plane, west-to-east. Remaining over one longitude supports continuous communications or weather monitoring.
Use a hint if needed
Practise with support
Try this
Find the speed and period of an Earth satellite in a circular orbit of radius 8.00 × 10⁶ m. Use GM = 3.99 × 10¹⁴ m³ s⁻².
Hint: Set gravity equal to mv²/r, then use T = 2πr/v.
Check your answer
v = √(GM/r) = 7.06 × 10³ m s⁻¹. T = 2π(8.00 × 10⁶)/(7.06 × 10³) = 7.12 × 10³ s = 119 min, so it is not geostationary.
Now work without the hint
Practise independently
Your turn
Show that the period of a circular gravitational orbit satisfies T² = 4π²r³/(GM), then state why a geostationary satellite is useful.
Check your answer
GMm/r² = m(4π²r/T²), so T² = 4π²r³/(GM). A geostationary satellite stays above one longitude, allowing a fixed ground antenna and continuous coverage of the same region.
Avoid these traps
Common mistakes
Common mistake
Any satellite with a 24-hour period is geostationary.
What is wrong with this reasoning?
Show better thinking
It must also have a circular equatorial orbit and move west-to-east with Earth's rotation. Then it stays above one longitude and can support continuous communication or weather observation.
Write for the examiner
Exam guidance
Use orbital radius from the centre and list every geostationary condition when the term appears.
Exam-style practice [7 marks]
Two satellites orbit the same planet in circular orbits of radii r and 9r. Compare their speeds, angular speeds and periods. Then state all conditions needed for the outer satellite to be geostationary.
Plan before you answer
- Use v ∝ r⁻¹/² and T ∝ r³/².
- Use ω = 2π/T.
- List geometry, direction and period separately.
Mark your answer and compare the model
Marking points
Tick each point only if your answer states it clearly.
Model answer
Since v ∝ r⁻¹/², the outer speed is 9⁻¹/² = 1/3 of the inner speed. Since T ∝ r³/², its period is 9³/² = 27 times as long, so its angular speed is 1/27 as large. To be geostationary it must have a circular equatorial orbit, travel eastward and have the same rotational period as Earth.
Come back in three days
Check what stayed with you
Recall question 1
What supplies centripetal acceleration in an ideal satellite orbit?
Check the answer
Gravitational force.
Recall question 2
How does circular-orbit speed vary with radius?
Check the answer
v ∝ r⁻¹/² for the same central mass.
Recall question 3
Why is a geostationary satellite useful?
Check the answer
It stays above one longitude, allowing continuous coverage of the same region.
Syllabus and review details
This lesson covers the listed H2 Physics 9478 outcomes. The official topic states no explicit exclusions. Models use centre-to-centre distance, zero potential at infinity, point masses or spherical bodies viewed externally, and ideal circular orbits unless a question states otherwise.
- GCE A-Level H2 PhysicsTopic 8(j) / Topic 8(k) · 2027Checked against the syllabus · partial topic coverageOfficial 9478 syllabus
Course and syllabus information
- Course
- GCE A-Level H2 Physics
- Edition
- GCE A-Level H2 Physics 2027