Gravitational Potential and Orbits

Gravitational potential extends the concept of gravitational field strength by introducing an energy-based description of the field. Combined with orbital mechanics, this topic forms the heart of AQA specification 3.7.2 and is tested extensively in Paper 2.

Spec mapping (AQA 7408 §3.7.2 — Gravitational potential and orbits): Required content for this lesson includes the definition of gravitational potential as work done per unit mass against the field from infinity, the radial form V = −GM/r, the relationship g = −dV/dr (potential gradient), equipotential surfaces and their orthogonality to field lines, the derivation of escape velocity from energy conservation, circular-orbit dynamics with gravity as the centripetal force, Kepler's third law T² ∝ r³, the geostationary orbit conditions, and total orbital energy E = −GMm/(2r). Refer to the official AQA 7408 specification document for the precise wording.

Synoptic links — where this resurfaces in 7408:

• §3.6.1 Circular motion — the centripetal-force balance F = mv²/r = GMm/r² is the engine of every orbital-velocity question; the relationship is examined directly in Paper 2.
• §3.4 Mechanics — conservation of energy underlies the escape-velocity derivation; the same energy bookkeeping (KE + PE = constant) is used in projectile and spring problems.
• §3.7.3 Electric fields (Lesson 3 here) — V_grav = −GM/r and V_elec = kQ/r are mathematically parallel; the sign and "always-negative-for-grav" feature is the major synoptic discriminator.
• §3.9 Astrophysics — orbital mechanics generalises to binary-star systems and exoplanet detection via radial-velocity and transit methods (the Kepler-mission is named after Kepler's laws covered here).
• §3.7.5 Magnetic fields — charged-particle orbits in magnetic fields use the same "centripetal-force-from-something" template, but with F = BQv replacing gravitational attraction.
• §3.5 Materials / §3.4.1.7 Energy — work done against a conservative force = change in potential energy; the same framework spans gravity, springs and electricity.

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Gravitational Potential

Key Definition: The gravitational potential (V) at a point in a gravitational field is the work done per unit mass in bringing a small test mass from infinity to that point.

V = −GM/r

where V is the gravitational potential (J kg⁻¹), G = 6.67 × 10⁻¹¹ N m² kg⁻², M is the mass creating the field (kg), and r is the distance from the centre of the mass (m).

Key features:

• Gravitational potential is a scalar quantity (it has magnitude but no direction).
• It is always negative because gravity is always attractive. Work must be done against the gravitational field to move a mass from the surface to infinity, so the potential at infinity is defined as zero, and at any finite distance from a mass, the potential is less than zero (i.e., negative).
• As r → ∞, V → 0 (the potential at infinity is zero by definition).
• As r decreases (closer to the mass), V becomes more negative (the potential well deepens).

Common Misconception: Students sometimes think that a more negative potential means "less energy." In fact, a more negative potential means that more work must be done to move the test mass to infinity. The mass is more tightly bound in the gravitational field.

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Equipotential Surfaces

Equipotential surfaces connect all points at the same gravitational potential. For a spherical mass, equipotentials are concentric spheres.

Described diagram — Equipotential surfaces around a planet: Concentric circles (representing spherical surfaces) are drawn around the planet, each labelled with a potential value that becomes less negative (closer to zero) further from the planet. The field lines are radial, pointing inward, and are everywhere perpendicular to the equipotential surfaces.

Key properties:

• No work is done when a mass moves along an equipotential surface (since ΔV = 0).
• Equipotential surfaces are always perpendicular to gravitational field lines.
• Closer spacing of equipotentials indicates a stronger field (steeper potential gradient).

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Gravitational Potential Energy

The gravitational potential energy (E_p) of a mass m at a point where the gravitational potential is V is:

E_p = mV = −GMm/r

This is the work done in bringing the mass m from infinity to the distance r from mass M.

Work Done Moving Between Two Points

The work done in moving a mass m from one point to another in a gravitational field is:

W = mΔV = m(V₂ − V₁)

If V₂ > V₁ (moving further from the mass, to a less negative potential), W is positive — you must do work against the field. If V₂ < V₁ (moving closer), W is negative — the field does work on the mass.

Worked Example 1 — Work Done Launching a Satellite

Question: Calculate the minimum energy required to move a 500 kg satellite from the Earth's surface to an orbit at an altitude of 600 km. (Earth mass = 5.97 × 10²⁴ kg, Earth radius = 6.37 × 10⁶ m.)

Solution:

At the surface: r₁ = 6.37 × 10⁶ m At the orbit: r₂ = 6.37 × 10⁶ + 600 × 10³ = 6.97 × 10⁶ m

V₁ = −GM/r₁ = −(6.67 × 10⁻¹¹ × 5.97 × 10²⁴) / (6.37 × 10⁶) = −3.98 × 10¹⁴ / 6.37 × 10⁶ = −6.25 × 10⁷ J kg⁻¹

V₂ = −GM/r₂ = −3.98 × 10¹⁴ / 6.97 × 10⁶ = −5.71 × 10⁷ J kg⁻¹

ΔV = V₂ − V₁ = (−5.71 × 10⁷) − (−6.25 × 10⁷) = 0.54 × 10⁷ = 5.4 × 10⁶ J kg⁻¹

W = mΔV = 500 × 5.4 × 10⁶ = 2.7 × 10⁹ J = 2.7 GJ

Note: This is only the energy needed to raise the satellite against gravity. Additional energy is needed to give it orbital speed.

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Relationship Between g and V

The gravitational field strength is the negative of the potential gradient:

g = −dV/dr

This means:

• The field points in the direction of decreasing potential (from less negative to more negative).
• The magnitude of g equals the rate at which V changes with distance.
• On a V-r graph, the gradient at any point equals −g at that distance.

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Escape Velocity

Key Definition: The escape velocity is the minimum speed an object must have at the surface of a planet to escape the gravitational field completely (i.e., to reach infinity with zero kinetic energy remaining).

To derive the escape velocity, set the total energy (kinetic + potential) equal to zero (the value at infinity):

½mv² + (−GMm/r) = 0

½mv² = GMm/r

v² = 2GM/r

v_escape = √(2GM/r)

Worked Example 2 — Escape Velocity from Earth

Question: Calculate the escape velocity from the Earth's surface. (Earth mass = 5.97 × 10²⁴ kg, Earth radius = 6.37 × 10⁶ m.)

Solution:

v = √(2GM/R) = √(2 × 6.67 × 10⁻¹¹ × 5.97 × 10²⁴ / 6.37 × 10⁶)

v = √(2 × 3.98 × 10¹⁴ / 6.37 × 10⁶)

v = √(7.96 × 10¹⁴ / 6.37 × 10⁶)

v = √(1.25 × 10⁸)

v = 1.12 × 10⁴ m s⁻¹ ≈ 11.2 km s⁻¹

Exam Tip: Note that escape velocity does not depend on the mass of the escaping object — only on M and r. This is a consequence of the equivalence of inertial and gravitational mass.

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Orbital Mechanics

For an object in a circular orbit, gravity provides the centripetal force:

GMm/r² = mv²/r

Simplifying:

v_orbital = √(GM/r)

Note that the orbital speed depends only on M (the central mass) and r (the orbital radius), not on the mass of the orbiting object.

Orbital Period and Kepler's Third Law

The period T of a circular orbit is the circumference divided by the orbital speed:

T = 2πr/v = 2πr/√(GM/r) = 2πr × √(r/GM) = 2π√(r³/GM)

Squaring both sides:

T² = (4π²/GM) × r³

This is Kepler's third law: T² is proportional to r³, with the constant of proportionality depending only on the central mass M.

T² ∝ r³

Worked Example 3 — Orbital Period of the ISS

Question: The ISS orbits at an altitude of 408 km above the Earth's surface. Calculate its orbital speed and period. (Earth mass = 5.97 × 10²⁴ kg, Earth radius = 6.37 × 10⁶ m.)

Solution:

Orbital radius: r = 6.37 × 10⁶ + 408 × 10³ = 6.778 × 10⁶ m

Orbital speed: v = √(GM/r) = √(6.67 × 10⁻¹¹ × 5.97 × 10²⁴ / 6.778 × 10⁶) v = √(3.98 × 10¹⁴ / 6.778 × 10⁶) v = √(5.87 × 10⁷) v = 7.66 × 10³ m s⁻¹ ≈ 7.66 km s⁻¹

Period: T = 2πr/v = 2π × 6.778 × 10⁶ / 7.66 × 10³ T = 4.259 × 10⁷ / 7.66 × 10³ T = 5560 s ≈ 92.7 minutes

This matches the well-known ISS orbital period of approximately 93 minutes.

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Geostationary Orbits

A geostationary orbit (also called a geosynchronous equatorial orbit) has specific characteristics:

• Period T = 24 hours = 86 400 s (same as Earth's rotation)
• Orbits directly above the equator
• Direction of orbit is the same as Earth's rotation (west to east)
• The satellite appears stationary when viewed from the ground

Worked Example 4 — Geostationary Orbit Radius

Question: Calculate the radius and altitude of a geostationary orbit. (Earth mass = 5.97 × 10²⁴ kg, Earth radius = 6.37 × 10⁶ m.)

Solution:

Using T² = (4π²/GM) × r³:

r³ = GMT²/(4π²)

r³ = (6.67 × 10⁻¹¹ × 5.97 × 10²⁴ × (86 400)²) / (4π²)

r³ = (3.98 × 10¹⁴ × 7.46 × 10⁹) / 39.48

r³ = 2.97 × 10²⁴ / 39.48

r³ = 7.52 × 10²² m³

r = (7.52 × 10²²)^(1/3) = 4.22 × 10⁷ m ≈ 42 200 km from the Earth's centre.

Altitude = r − R_E = 4.22 × 10⁷ − 6.37 × 10⁶ = 3.58 × 10⁷ m ≈ 35 800 km

Geostationary satellites are used for communications and weather monitoring because their fixed position relative to the ground means ground-based antennas can point at a fixed direction.

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Total Energy of an Orbiting Satellite

For a satellite in a circular orbit:

Kinetic energy: E_k = ½mv² = ½m(GM/r) = GMm/(2r)

Potential energy: E_p = −GMm/r

Total energy: E_total = E_k + E_p = GMm/(2r) − GMm/r = −GMm/(2r)

E_total = −GMm/(2r)

Key relationships:

• E_k = −E_total (the kinetic energy is the negative of the total energy)
• E_p = 2 × E_total (the potential energy is twice the total energy)
• |E_k| = ½|E_p| (the kinetic energy is half the magnitude of the potential energy)
• The total energy is always negative — the satellite is bound.

Exam Tip: If a satellite moves to a higher orbit: r increases, E_p increases (less negative), E_k decreases, v decreases, but the total energy increases (less negative). Energy must be supplied to move to a higher orbit — this seems counterintuitive because the satellite slows down, but the increase in potential energy exceeds the decrease in kinetic energy.

Worked Example 5 — Energy to Reach a Higher Orbit

Question: A satellite of mass 800 kg is initially in a circular orbit at 500 km altitude around Earth. Calculate the additional energy required to lift it to a higher circular orbit at 1500 km altitude. (Earth mass = 5.97 × 10²⁴ kg, Earth radius = 6.37 × 10⁶ m.)

Solution:

Initial orbital radius: r₁ = 6.37 × 10⁶ + 500 × 10³ = 6.87 × 10⁶ m Final orbital radius: r₂ = 6.37 × 10⁶ + 1500 × 10³ = 7.87 × 10⁶ m

Total energy at each orbit: E₁ = −GMm/(2r₁) = −(6.67 × 10⁻¹¹ × 5.97 × 10²⁴ × 800) / (2 × 6.87 × 10⁶) E₁ = −(3.18 × 10¹⁷) / (1.374 × 10⁷) E₁ = −2.32 × 10¹⁰ J

E₂ = −GMm/(2r₂) = −(3.18 × 10¹⁷) / (1.574 × 10⁷) E₂ = −2.02 × 10¹⁰ J

Additional energy required: ΔE = E₂ − E₁ = −2.02 × 10¹⁰ − (−2.32 × 10¹⁰) = +3.0 × 10⁹ J = 3.0 GJ

Note that although the satellite slows down in the higher orbit (v ∝ 1/√r), the total mechanical energy has increased — the gain in potential energy (less negative) exceeds the loss in kinetic energy. This is one of the most counter-intuitive results in orbital mechanics and is heavily examined.