Circular and geostationary orbits

Key idea: A satellite in circular orbit is continually falling around the planet. Gravity supplies the inward acceleration; no forward thrust is required in the ideal model.

GCE A-Level H1 Physics 2027

H1 Physics 8867 · Lesson 3 of 3

Check your understanding

By the end of this lesson, you should be able to

Analyse a circular orbit by equating gravity with centripetal force.
Explain how orbital speed and period depend on radius.
State every geostationary condition and give appropriate applications.

Learn the idea

Big question: How can a satellite be continuously falling yet never get closer to the planet?

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.

Scroll diagram horizontally to read all labels.

Gravity is the inward resultant in every circular orbit. A geostationary orbit adds three constraints: equatorial plane, same rotational direction, and the same period as Earth.

Key ideas

Gravity changes velocity direction even when speed is constant.
A 24-hour period alone does not make an orbit geostationary.
An ideal orbit needs no continuous tangential thrust.

Relationships to know

GMm/r² = mv²/r
v = √(GM/r)
T² = 4π²r³/(GM)

Follow 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.

Now try it with support

Practise with support

A satellite moves to a circular orbit with four times the original radius around the same planet. Compare its speed and period with the originals.

Hints

v ∝ r⁻¹/².
T ∝ r³/².

View the guided answer

The speed becomes 4⁻¹/² = 1/2 of the original. The period becomes 4³/² = 8 times the original.

Your turn

Practise independently

Explain why a satellite in a 24-hour polar orbit is not geostationary even if its angular speed matches Earth’s rotation.

Check your answer

A polar satellite crosses different latitudes and longitudes, so it does not remain above one point on the equator. Matching Earth’s angular speed is insufficient: the orbit must also be circular, equatorial and eastward.

Common mistakes and exam guidance

Watch out for

Saying the gravitational force is balanced in a circular orbit.
Calling every satellite with a one-day period geostationary.

In an exam

Begin an orbit calculation with ‘gravity provides the centripetal resultant’ and write the equality.
For geostationary questions, list all four conditions before discussing applications.

Put the ideas together

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.

View the marking points and model answer

Marking points

Finds vouter/vinner = 1/3.
Finds Touter/Tinner = 27.
Finds ωouter/ωinner = 1/27.
States circular orbit.
States equatorial plane.
States eastward/same direction as Earth rotation.
States period equal to Earth's rotational period.

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.

Finish from memory

Three-question recap

What supplies centripetal acceleration in an ideal satellite orbit?
Check
Gravitational force.
How does circular-orbit speed vary with radius?
Check
v ∝ r⁻¹/² for the same central mass.
Why is a geostationary satellite useful?
Check
It stays above one longitude, allowing continuous coverage of the same region.

Try this next

Compare low-Earth and geostationary orbits using speed, period and coverage rather than memorised labels.

Check the whole topic Return to Circular Motion and Gravitation Review Inverse-square gravitation and near-Earth field

Continue with the next resource in this course.

Course and syllabus information

Course: GCE A-Level H1 Physics
Edition: GCE A-Level H1 Physics 2027