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Circular Orbits & Geostationary Satellites

Key idea: Analyse circular orbits using gravity as centripetal force, use v = √(GM/r) and T² ∝ r³, and solve geostationary satellite problems (A Level Physics).

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Analyse circular gravitational orbits and geostationary satellite conditions.

1. Definitions (Must Know)

A. Circular orbit

A circular orbit has constant orbital radius r (distance from the centre of the planet/star).

Gravity acts towards the centre and provides the centripetal force.

B. Orbital speed, v

For a mass m in a circular orbit around a mass M:

v = square root of (GM/r)

C. Orbital period, T

The orbital period, T, is the time for one complete orbit:

T = (2π r)/v

Combining with v = square root of (GM/r) gives:

T = 2π square root of (r³/GM)

D. Geostationary orbit

A geostationary satellite:

  • orbits above the equator,
  • moves west to east (same direction as Earth’s rotation),
  • has the same period as Earth’s rotation (approximately 24 h),
  • appears fixed above one point on Earth.
Circular orbit dynamics and geostationary conditionsA satellite in circular orbit has tangential velocity and inward gravity providing centripetal acceleration. A geostationary satellite lies in the equatorial plane, travels in Earth's rotational direction and matches Earth's rotation period.Circular orbit modelgravity, FvGMm/r² = mv²/rGeostationary orbitequatorial planeT ≈ 24 hsame direction as Earth rotates
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.

2. Key Ideas (What Earns Marks)

  • Use “gravity provides centripetal”: GMm/r² = mv²/r
  • Then: v = square root of (GM/r)
  • Period–radius relationship (very exam-useful): T = 2π square root of (r³/GM) ⇒ T² ∝ r³
  • For a circular orbit only: K = GMm/2r, U = -GMm/r and E = K + U = -GMm/2r.
  • For geostationary orbit, use T = 24 × 3600 s and solve for r, then altitude h = r-R_E.
Prerequisites

You need:

3. Detailed Explanations

A. Deriving orbital speed for a circular orbit

Gravitational force provides the centripetal force:

GMm/r² = mv²/r; GM/r = v²; v = square root of (GM/r)

B. Deriving the period formula

Distance for one orbit is 2π r, so:

v = (2π r)/T

Substitute v² = GM/r:

((2π r)/T)² = GM/r; T² = (4π²/GM)r³

Kepler’s 3rd law (linear form): T² vs r³

For circular orbits around the same central mass, T squared is proportional to r cubed.

Scroll across the graph to read all labels.

For circular orbits around the same central mass, T squared is proportional to r cubed.For circular orbits around the same central mass, T squared is proportional to r cubed.
For Earth, T² = (4π²/GM_E)r³. The point near (75,74) corresponds roughly to geostationary radius r ≈ 4.22 × 10⁷ m.
Open full-size graph
View figure data
Values and uncertainty for Kepler’s 3rd law (linear form): T² vs r³
Seriesr³ (10²¹ m³)r³ uncertaintyT² (10⁸ s²)T² uncertainty
Example points (Earth)0.3430.339
Example points (Earth)10.989
Example points (Earth)87.91
Example points (Earth)75.274.4
Straight-line trend00
Straight-line trend8079.1

C. Energy in a circular orbit

From the radial force equation,

GMm/r² = mv²/r ⇒ mv² = GMm/r.

Therefore

K = (1/2)mv² = GMm/2r = -(1/2)U,

and the total mechanical energy is

E = K + U = -GMm/2r.

The negative total energy identifies a bound orbit. These energy expressions are for a circular orbit at radius r; do not apply them unchanged to an arbitrary point on an elliptical path.

D. Geostationary satellites (what makes them special)

Geostationary satellites must match Earth’s angular speed. This forces a unique orbital radius r for Earth.

Once you find r, you can find:

  • orbital speed: v = square root of (GM/r),
  • altitude above Earth: h = r-R_E,
  • applications: communications links and continuous weather observation of the same region.

4. Common Mistakes

  • Using altitude h as r (use r = R_E + h).
  • Mixing up orbital speed v = square root of (GM/r) with escape speed v_esc = square root of (2GM/r).
  • Forgetting to convert 24 h to seconds.
  • Claiming a polar satellite can be geostationary (it cannot).
  • Using the circular-orbit energy result E = -GMm/(2r) as though it were the gravitational potential energy; U = -GMm/r.

5. Exam Tips

  • Start by writing GMm/r² = mv²/r before substituting numbers.
  • Use T² ∝ r³ for ratio questions: T₁/T₂ = (r₁/r₂)^(3/2)
  • Keep r as a distance from the centre until the very end.

6. Worked Examples

Modelled example 1

Orbital speed around Earth

Core

Problem

Find the speed of a satellite in a circular orbit of radius 7.00 × 10⁶ m around Earth. Take GM_E = 3.99 × 10¹⁴ m³ s⁻².
Study the worked solution

  1. Identify the inward force

    Method

    Set gravitational force equal to the required centripetal force.

    Reason

    Gravity is the resultant radial force maintaining the circular orbit.

    Working

    GMm/r² = mv²/r

  2. Derive orbital speed

    Method

    Cancel m and one factor of r to get v = square root of (GM/r).

    Reason

    The orbiting mass does not determine the speed at a given radius around the same source.

    Working

    v² = GM/r, v = square root of (GM/r)

  3. Evaluate

    Method

    Obtain 7.55 × 10³ m s⁻¹.

    Reason

    The speed is tangential while acceleration and gravity point inward.

    Working

    v = square root of ((3.99 × 10¹⁴)/(7.00 × 10⁶)) ≈ 7.55 × 10³ m s⁻¹

Guided practice 2

Radius and altitude of a geostationary orbit

About 6 min

Problem

Find the radius and altitude of a geostationary orbit. Take T = 24 × 3600 s, GM_E = 3.99 × 10¹⁴ m³ s⁻², and R_E = 6.37 × 10⁶ m.

Try this before viewing the solution

Hints

Hint 1: solve for centre distance first
Rearrange T² = 4π²r³/(GM), then take the cube root. Altitude is r-R_E.
View solution step by step

  1. Find orbital radius

    Method

    Obtain r ≈ 4.22 × 10⁷ m.

    Reason

    The period equation returns distance from Earth’s centre.

    Working

    r = (GMT²/4π²)^(1/3) = (((3.99 × 10¹⁴)(86400)²)/4π²)^(1/3) ≈ 4.22 × 10⁷ m

  2. Convert radius to altitude

    Method

    Obtain h ≈ 3.58 × 10⁷ m = 3.58 × 10⁴ km.

    Reason

    Altitude is measured from Earth’s surface.

    Working

    h = r-R_E = (4.22-0.637) × 10⁷ ≈ 3.58 × 10⁷ m

Common misconception 3

Ratio of orbital periods (Mars satellites)

Find and correct the mistake

Learner claim

Phobos and Deimos orbit Mars at radii 9.5 × 10⁶ m and 24.5 × 10⁶ m. A learner uses T_P/T_D = (r_P/r_D)³. Diagnose the exponent and calculate the ratio.

Try this before viewing the solution

View solution step by step

  1. Take the square root of the law

    Method

    Use T ∝ r^(3/2).

    Reason

    T² ∝ r³ does not mean T ∝ r³.

    Working

    T ∝ square root of r³ = r^(3/2)

  2. Evaluate the ratio

    Method

    Obtain T_P/T_D ≈ 0.241.

    Reason

    Both satellites orbit the same central mass, so the proportionality constant cancels.

    Working

    T_P/T_D = (9.5/24.5)^(3/2) ≈ 0.241

Examiner practice 4

Find orbital period from orbital radius

3 marks

Examination question

A satellite has a circular Earth orbit of radius 7.00 × 10⁶ m. Take GM_E = 3.99 × 10¹⁴ m³ s⁻². Find its orbital period in seconds and minutes. [3 marks]

Try this before viewing the solution

View solution step by step

  1. Select the period relation

    1 mark

    Method

    Use T = 2π square root of (r³/(GM)).

    Reason

    This combines circular path length with gravity providing centripetal acceleration.

    Working

    T = 2π square root of (r³/GM)

  2. Calculate in seconds

    1 mark

    Method

    Obtain 5.83 × 10³ s.

    Reason

    The supplied quantities are in SI units.

    Working

    T = 2π square root of (((7.00 × 10⁶)³)/(3.99 × 10¹⁴)) ≈ 5.83 × 10³ s

  3. Convert to minutes

    1 mark

    Method

    Obtain about 97 min.

    Reason

    Divide seconds by 60 s min⁻¹.

    Working

    T ≈ 5830/60 ≈ 97 min

Challenge 5

Find the mass of the central body from an orbit

Minimal support

Independent transfer

A moon moves in a circular orbit of radius 4.50 × 10⁸ m around a planet, with period 2.00 × 10⁶ s. Infer the planet’s mass.

Try this before viewing the solution

Hints

Hint 1: make M the subject
Start with T² = 4π²r³/(GM) and multiply through by GM.
View solution step by step

  1. Rearrange the period law

    Method

    Use M = 4π²r³/(GT²).

    Reason

    The orbit supplies a gravitational measurement of the central mass.

    Working

    M = 4π²r³/GT²

  2. Evaluate

    Method

    Obtain 1.35 × 10²⁵ kg.

    Reason

    Cube the radius and square the period before applying G.

    Working

    M = (4π²(4.50 × 10⁸)³)/((6.67 × 10⁻¹¹)(2.00 × 10⁶)²) ≈ 1.35 × 10²⁵ kg

7. Mind Stretchers

Mind stretcher 1: Why a polar satellite cannot be geostationaryExtension

Explain why a satellite cannot be geostationary above the North Pole.

Show answer

To appear fixed above a point on Earth, the satellite must rotate with Earth and remain above the same latitude. The only circular orbit that stays above a fixed point is an equatorial orbit (the orbit plane must match Earth’s equatorial plane). A satellite above the pole would require an orbit plane through the pole, which cannot keep the satellite above the same point while Earth rotates.

Mind stretcher 2: How v and T change with larger orbital radiusExtension

A satellite is moved to a circular orbit with larger radius. State what happens to its orbital speed and period.

Show answer

From v = square root of (GM/r), increasing r decreases v.

From T = 2π square root of (r³/(GM)), increasing r increases T.

Mind stretcher 3: Optional (Enrichment)Extension

A. Elliptical orbits (beyond syllabus)

Most real orbits are slightly elliptical. A common extension is that the period depends on the semi-major axis a via T² ∝ a³ (Kepler’s third law). Detailed ellipse geometry and energy methods are not required for the A Level gravitation learning outcomes.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027