Gravitational Fields

A gravitational field is a region of space in which a mass experiences a force due to the presence of another mass. Every object with mass creates a gravitational field around it. Understanding gravitational fields is essential for A-Level Physics (AQA specification 3.7.1) and forms the foundation for orbital mechanics, satellite motion, and astrophysics.

Spec mapping (AQA 7408 §3.7.2 — Gravitational fields): This lesson covers the foundational subsection on Newton's law of gravitation, gravitational field strength as force per unit mass, uniform vs radial fields, field lines and their density as a representation of g, and the inverse-square dependence g = GM/r² for a point or spherical mass. The AQA data and formulae booklet supplies G and the inverse-square form; you must be able to apply it, derive g at any radius from a uniform sphere, and treat the sphere as a point mass concentrated at its centre. Refer to the official AQA 7408 specification document for the precise wording of each statement and any subject-content limitations.

Synoptic links — where this resurfaces in 7408:

• §3.4.1 Force, energy and momentum — F = mg as a special case of F = ma when the field is uniform; weight and free-fall acceleration in projectile problems.
• §3.7.3 Electric fields — Coulomb's law F = kQ₁Q₂/r² shares the inverse-square mathematical form; the comparison of gravitational and electric radial fields is a frequent synoptic prompt and is examined in Lesson 2 of this course.
• §3.4.1.4 Newton's laws of motion — Newton's third law applied to gravitational pairs (Earth pulls satellite, satellite pulls Earth with equal magnitude); reaction-force confusion is a common mark-loss point.
• §3.7.2 (cont.) Gravitational potential and orbits — g = −dV/dr links field strength to potential gradient and underpins escape velocity and Kepler's third law (Lesson 1 here).
• §3.9 Astrophysics (optional unit) — radial g profiles inside stars and around neutron stars rely on this lesson's machinery extended via Gauss-style shell theorems.
• §3.6.1 Circular motion — the centripetal-force requirement F = mv²/r equated to GMm/r² generates orbital velocity (Lesson 1) and the geostationary-radius derivation.

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Field Lines and Their Properties

Gravitational field lines are used to represent the direction and relative strength of a gravitational field. They are drawn as arrows that point in the direction a small test mass would accelerate if placed at that point.

Described diagram — Field lines around a spherical mass: Lines are drawn radially inward towards the centre of the mass, uniformly distributed around its surface. The lines are closer together near the surface (indicating a stronger field) and spread further apart at greater distances (indicating a weaker field). The arrows on every line point towards the centre of the mass, because gravity is always attractive.

Key properties of gravitational field lines:

• They always point towards the mass creating the field (gravity is always attractive).
• They never cross each other.
• The density of field lines (number per unit area perpendicular to the lines) indicates the strength of the field.
• For a spherically symmetric mass, the field lines are radial and converge at the centre.

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Uniform Gravitational Fields

Near the surface of a large body such as the Earth, over small distances, the gravitational field is approximately uniform. This means the field strength has the same magnitude and direction at every point in the region.

Described diagram — Uniform gravitational field: Equally spaced, parallel, vertical lines pointing downward. The uniform spacing indicates that the field strength is the same everywhere in the region.

In a uniform field:

• The acceleration due to gravity, g, is constant.
• The force on a mass m is F = mg, directed downward.
• Field lines are parallel and equally spaced.

Near the Earth's surface, g ≈ 9.81 N kg⁻¹ (equivalently, 9.81 m s⁻²). This approximation is valid for heights much smaller than the Earth's radius (6.37 × 10⁶ m).

Exam Tip: The units of gravitational field strength are N kg⁻¹, which are dimensionally equivalent to m s⁻². Both are acceptable, but N kg⁻¹ is preferred when discussing field strength, while m s⁻² is preferred when discussing acceleration. The conceptual distinction matters: g as field strength is a property of the field at a point; g as acceleration is the response of a mass placed in that field.

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Gravitational Field Strength

Key Definition: The gravitational field strength (g) at a point is the force per unit mass acting on a small test mass placed at that point.

g = F/m

where g is the gravitational field strength (N kg⁻¹), F is the gravitational force (N), and m is the mass of the test mass (kg).

The definition uses a "small test mass" to ensure that the test mass does not significantly disturb the field it is measuring. If the test mass were very large, its own gravitational field would alter the field being measured.

Gravitational field strength is a vector quantity — it has both magnitude and direction. The direction is always towards the mass creating the field.

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Newton's Law of Gravitation

Key Definition: Newton's law of universal gravitation states that every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.

F = GMm/r²

where:

• F is the gravitational force between the two masses (N)
• G is the gravitational constant, G = 6.67 × 10⁻¹¹ N m² kg⁻² (given on the data sheet)
• M and m are the two masses (kg)
• r is the distance between the centres of the two masses (m)

Key features:

• The force is always attractive (there is no gravitational repulsion).
• The force obeys an inverse square law: doubling the distance reduces the force by a factor of four.
• The force acts along the line joining the centres of the two masses.
• Newton's third law applies: M exerts a force on m equal in magnitude but opposite in direction to the force m exerts on M.

Gravitational Field Strength for a Point or Spherical Mass

Combining g = F/m with F = GMm/r²:

g = F/m = (GMm/r²)/m = GM/r²

g = GM/r²

This gives the gravitational field strength at a distance r from the centre of a mass M. For a uniform sphere, this formula applies outside the sphere (r ≥ R, where R is the radius of the sphere), and the mass behaves as if concentrated at the centre.

Worked Example 1 — Gravitational Field Strength at the Earth's Surface

Question: The Earth has a mass of 5.97 × 10²⁴ kg and a radius of 6.37 × 10⁶ m. Calculate the gravitational field strength at the Earth's surface.

Solution:

Using g = GM/r²:

g = (6.67 × 10⁻¹¹ × 5.97 × 10²⁴) / (6.37 × 10⁶)²

g = (3.98 × 10¹⁴) / (4.06 × 10¹³)

g = 9.81 N kg⁻¹

This confirms the familiar value of g at the Earth's surface.

Worked Example 2 — Force Between Two Masses

Question: Calculate the gravitational force between the Earth (mass 5.97 × 10²⁴ kg) and the Moon (mass 7.35 × 10²² kg). The mean Earth–Moon distance is 3.84 × 10⁸ m.

Solution:

Using F = GMm/r²:

F = (6.67 × 10⁻¹¹ × 5.97 × 10²⁴ × 7.35 × 10²²) / (3.84 × 10⁸)²

Numerator: 6.67 × 10⁻¹¹ × 5.97 × 10²⁴ = 3.98 × 10¹⁴ 3.98 × 10¹⁴ × 7.35 × 10²² = 2.93 × 10³⁷

Denominator: (3.84 × 10⁸)² = 1.47 × 10¹⁷

F = 2.93 × 10³⁷ / 1.47 × 10¹⁷ = 1.99 × 10²⁰ N

This enormous force (approximately 2 × 10²⁰ N) is what keeps the Moon in orbit around the Earth.

Worked Example 3 — Variation of g with Altitude

Question: A satellite orbits at an altitude of 400 km above the Earth's surface. Calculate the gravitational field strength at the satellite's orbital altitude. (Earth mass = 5.97 × 10²⁴ kg, Earth radius = 6.37 × 10⁶ m.)

Solution:

The distance from the centre of the Earth is: r = 6.37 × 10⁶ + 400 × 10³ = 6.37 × 10⁶ + 0.40 × 10⁶ = 6.77 × 10⁶ m

g = GM/r² = (6.67 × 10⁻¹¹ × 5.97 × 10²⁴) / (6.77 × 10⁶)²

g = 3.98 × 10¹⁴ / 4.58 × 10¹³ = 8.69 N kg⁻¹

Note that g has only decreased from 9.81 to 8.69 N kg⁻¹ — a reduction of about 11%. Astronauts aboard the International Space Station (which orbits at approximately this altitude) experience "weightlessness" not because gravity is zero, but because both the station and the astronauts are in free fall together.

Common Misconception: Students often believe that astronauts float in the ISS because there is no gravity in space. In fact, gravity at the ISS orbit is about 89% as strong as at the surface. The apparent weightlessness occurs because the ISS and everything inside it are in continuous free fall — they are all accelerating towards the Earth at the same rate.

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Superposition of Gravitational Fields

When more than one mass is present, the total gravitational field at any point is the vector sum of the individual fields due to each mass. This is the principle of superposition.

Worked Example 4 — Neutral Point Between Two Masses

Question: The Earth has mass M_E = 5.97 × 10²⁴ kg and the Moon has mass M_M = 7.35 × 10²² kg. The distance between their centres is d = 3.84 × 10⁸ m. Find the distance from the Earth's centre to the point where the gravitational field strength is zero (the neutral point).

Solution:

At the neutral point, the field due to the Earth equals the field due to the Moon (in magnitude but opposite in direction).

GM_E/x² = GM_M/(d − x)²

Cancelling G:

M_E/x² = M_M/(d − x)²

Taking square roots:

√M_E / x = √M_M / (d − x)

(d − x)/x = √(M_M/M_E) = √(7.35 × 10²² / 5.97 × 10²⁴) = √(0.01231) = 0.1109

d − x = 0.1109x

d = 1.1109x

x = d/1.1109 = 3.84 × 10⁸ / 1.1109 = 3.46 × 10⁸ m from the Earth's centre.

This is about 90% of the way to the Moon, confirming that the Earth's much larger mass dominates most of the space between them.

Worked Example 5 — Field Strength at the Surface of Jupiter

Question: Jupiter has a mass of 1.90 × 10²⁷ kg and an equatorial radius of 6.99 × 10⁷ m. (a) Calculate the gravitational field strength at Jupiter's cloud-top surface. (b) Compare your answer to Earth's surface gravity and comment on the implications for a probe descending into Jupiter's atmosphere.

Solution:

(a) Using g = GM/r²:

g = (6.67 × 10⁻¹¹ × 1.90 × 10²⁷) / (6.99 × 10⁷)²

g = (1.267 × 10¹⁷) / (4.886 × 10¹⁵)

g = 25.9 N kg⁻¹

(b) The ratio g_Jupiter / g_Earth = 25.9 / 9.81 = 2.64. A 70 kg human at Jupiter's cloud tops would have an apparent weight of 70 × 25.9 ≈ 1810 N — over two-and-a-half times their Earth weight. A descending probe must be engineered for substantially higher structural loads, faster terminal velocity (the same drag formula gives a higher v_terminal because mg has increased), and rapid heating from atmospheric friction. This is why the Galileo entry probe (which entered Jupiter's atmosphere in 1995) was packaged behind a heatshield far more substantial than any Earth re-entry shield.

Common Misconception: Students often assume Jupiter's surface gravity is enormous because Jupiter is so massive. In fact, it is "only" 2.6 × Earth's — the planet's huge radius (~11× Earth's) partly cancels its huge mass (~318× Earth's) via the 1/r² factor. The lesson: g depends on both M and r — never on M alone.

Worked Example 6 — Field Strength on a Line Between Two Masses

Question: Two identical stars, each of mass 2.0 × 10³⁰ kg, are separated by a distance of 4.0 × 10¹¹ m (centre to centre). Calculate the magnitude of the gravitational field strength at a point on the line joining them, 1.0 × 10¹¹ m from the first star.

Solution:

Distance from second star: 4.0 × 10¹¹ − 1.0 × 10¹¹ = 3.0 × 10¹¹ m.

Field due to first star (pointing towards it): g₁ = GM/r₁² = (6.67 × 10⁻¹¹ × 2.0 × 10³⁰) / (1.0 × 10¹¹)² = 1.334 × 10²⁰ / 1.0 × 10²² = 1.33 × 10⁻² N kg⁻¹ (towards star 1)

Field due to second star (pointing towards it, i.e. in the opposite direction): g₂ = GM/r₂² = (6.67 × 10⁻¹¹ × 2.0 × 10³⁰) / (3.0 × 10¹¹)² = 1.334 × 10²⁰ / 9.0 × 10²² = 1.48 × 10⁻³ N kg⁻¹ (towards star 2)

Because the two fields point in opposite directions along the line, the net field is the difference:

g_net = g₁ − g₂ = 1.33 × 10⁻² − 1.48 × 10⁻³ = 1.19 × 10⁻² N kg⁻¹, directed towards the nearer star.

This demonstrates the vector character of gravitational fields: at any point off the midpoint, the field of the nearer mass dominates.

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Specimen Exam Question (9-mark)

Specimen question modelled on the AQA 7408 Paper 2 format.

A newly discovered exoplanet, Kepler-X, is approximately spherical and has a mean radius of 4.20 × 10⁶ m. A small probe of mass 250 kg sitting on its surface registers a weight of 1.86 × 10³ N on a calibrated spring balance.

(a) State what is meant by the gravitational field strength at a point. (1 mark) (b) Calculate the gravitational field strength at the surface of Kepler-X. (2 marks) (c) Show that the mass of Kepler-X is approximately 2.0 × 10²⁴ kg. (3 marks) (d) The probe is moved to a position 1.20 × 10⁷ m above the surface of Kepler-X. Calculate the new gravitational field strength experienced by the probe and comment on how this value compares with the surface value. (3 marks)

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AO Breakdown

Part	Marks	Assessment Objective	What is being tested
(a)	1	AO1 (knowledge)	Verbatim-quality definition of g as force per unit mass on a small test mass
(b)	2	AO2 (application)	Use of g = F/m with correct unit (N kg⁻¹)
(c)	3	AO2 (application)	Rearrangement of g = GM/r² to find M; correct substitution and significant figures
(d)	3	AO2 (application) + AO3 (analysis)	New radius r = R + h; recompute g; comment using the inverse-square ratio

Total: 9 marks. AQA Paper 2 typically awards roughly 6–7 marks for execution (AO2) and 2–3 marks for definitions and qualitative comment (AO1 + AO3).

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Grade-band Model Answers

Grade C response (~210 words)

(a) "Gravitational field strength is the force on a mass divided by the mass." (1 mark — definition acceptable but lacks the explicit "per unit mass on a small test mass" phrasing.)

(b) g = F/m = 1860 / 250 = 7.44 N kg⁻¹. (2 marks — correct value with unit.)

(c) Using g = GM/r²: M = gr² / G = (7.44 × (4.20 × 10⁶)²) / (6.67 × 10⁻¹¹) M = (7.44 × 1.764 × 10¹³) / 6.67 × 10⁻¹¹ M = 1.31 × 10¹⁴ / 6.67 × 10⁻¹¹ = 1.97 × 10²⁴ kg ≈ 2.0 × 10²⁴ kg. (3 marks.)

(d) New radius r = 4.20 × 10⁶ + 1.20 × 10⁷ = 1.62 × 10⁷ m. g_new = GM/r² = (6.67 × 10⁻¹¹ × 2.0 × 10²⁴) / (1.62 × 10⁷)² g_new = 1.33 × 10¹⁴ / 2.62 × 10¹⁴ = 0.51 N kg⁻¹. "The gravitational field is much weaker far away from the planet." (2 marks of 3 — execution correct, but the comment is too vague to access the AO3 mark.)

Examiner commentary: The candidate scores 8/9. The mark lost in (d) is the comparative analysis mark. A C-grade response handles the algebra competently but tends to under-evaluate qualitative comparisons.

Grade A* response (~340 words)

(a) "The gravitational field strength g at a point is the gravitational force per unit mass experienced by a small test mass placed at that point. It is a vector quantity directed towards the source mass." (1 mark — full, technically precise.)

(b) g = F/m = 1860 N / 250 kg = 7.44 N kg⁻¹ (vertically downward, towards the planet's centre). (2 marks.)

(c) Treating Kepler-X as a uniform sphere, the field outside obeys g = GM/r², with r measured from the planet's centre. Rearranging: M = gr²/G M = (7.44 × (4.20 × 10⁶)²) / (6.67 × 10⁻¹¹) M = (7.44 × 1.764 × 10¹³) / 6.67 × 10⁻¹¹ M = 1.31 × 10¹⁴ / 6.67 × 10⁻¹¹ M = 1.97 × 10²⁴ kg ≈ 2.0 × 10²⁴ kg (3 s.f., consistent with the data). (3 marks.)

(d) New distance from the centre: r = R + h = 4.20 × 10⁶ + 1.20 × 10⁷ = 1.62 × 10⁷ m. g_new = GM/r² = (6.67 × 10⁻¹¹ × 1.97 × 10²⁴) / (1.62 × 10⁷)² = 0.50 N kg⁻¹. Ratio: g_new / g_surface = 0.50 / 7.44 ≈ 1/15. Geometrically, r has increased by a factor of 1.62 × 10⁷ / 4.20 × 10⁶ ≈ 3.86, and an inverse-square law predicts the field falls by a factor of (3.86)² ≈ 14.9 — consistent. The result demonstrates that gravitational field strength is not uniform on the scale of distances comparable to the planet's radius, and treating g as a constant is only valid for h ≪ R. (3 marks: correct numerical value + explicit inverse-square comparison + limit-of-validity comment.)

Examiner commentary: Full 9/9. The A* moves are: (i) including the direction in (a) and (b); (ii) the explicit limit-of-validity comment in (d) — the inverse-square ratio is computed both numerically and geometrically and the two are reconciled. That structural cross-check distinguishes A* from A.

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Common Errors

Frequently observed mark-loss patterns on field-strength calculations:

• Using altitude h instead of r in g = GM/r². The formula needs the distance from the planet's centre, not the height above the surface. A common error in part (d) is to write r = 1.20 × 10⁷ m and lose 2 marks.
• Treating g as a constant beyond ~10% of the planet's radius. The "uniform field near Earth's surface" approximation collapses quickly. Once h ≳ 0.1R, the inverse-square correction matters.
• Wrong-direction field lines. Gravitational field lines always point towards the source mass (gravity is attractive). Candidates occasionally draw them outward by analogy with electric fields around a positive charge.
• Dropping units mid-calculation. AQA mark schemes accept either N kg⁻¹ or m s⁻² for g, but a numerical answer with no unit at all loses the final mark.
• Confusing "field strength" with "force". g is force per unit mass. A common misstatement is to define g as "the gravitational force on an object", which loses the AO1 mark.
• Sign-of-G confusion. G is a positive constant (6.67 × 10⁻¹¹ N m² kg⁻²). The attractive nature of gravity is conveyed by the direction of the field, not by giving G a negative sign.
• Forgetting that the sphere theorem only applies outside the body. Inside a uniform sphere (r < R), g varies linearly with r — it does not obey GM/r². AQA can examine this at the boundary.

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Going Further

For students preparing for STEP, MAT or competitive physics interviews:

• Shell theorem proof: Newton showed that a uniform spherical shell exerts no net gravitational force on a mass inside it, and behaves like a point mass at its centre when viewed from outside. The proof uses an elegant geometric argument summing rings of mass on the shell — see Chapter 10 of Principia (Book I) for the original, or any undergraduate mechanics text (e.g. Kleppner & Kolenkow) for a modern integration-based proof.
• Tidal forces arise from the gradient of g across an extended body (dg/dr ≈ −2GM/r³). They explain ocean tides, Roche limits for tidally disrupted moons, and the spaghettification of objects approaching black-hole event horizons.
• General relativity replaces the Newtonian "force" with curvature of spacetime; for weak fields and slow motions, GR reduces to F = GMm/r². The pertinent admissions-interview question: at what point does the Newtonian approximation break down? (Answer: when GM/(rc²) becomes non-negligible — order 10⁻⁹ at Earth's surface, order 1 near a black hole's horizon.)
• Suggested reading: Gravity: An Introduction to Einstein's General Relativity (Hartle), Chapter 2; the Feynman Lectures on Physics Vol. 1 Chapter 7 ("The Theory of Gravitation").

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A-Level Depth Misconceptions

Subtle errors that separate A from A*:

• "g is the same everywhere on Earth" — false. g varies with latitude (centripetal effect at the equator) by ~0.5%, with altitude (g decreases by ~0.3 mN kg⁻¹ per 100 m), and with local geology (mountains and ore bodies produce measurable anomalies). AQA expects you to know g ≈ 9.81 N kg⁻¹ as a representative value, not a universal constant.
• "Gravitational mass and inertial mass are obviously the same" — they are experimentally indistinguishable to one part in 10¹³, but this equivalence is non-trivial. It is the empirical foundation of Einstein's principle of equivalence and underpins general relativity.
• "Field strength and force are interchangeable" — they are dimensionally distinct. g has units N kg⁻¹; F has units N. The conceptual move from F to g is exactly the move from "force on a particular mass" to "property of the field independent of any test mass" — this is the same conceptual leap students will repeat for E in §3.7.3.
• "A satellite in orbit has no weight" — false. Weight (the gravitational force on the satellite) is non-zero everywhere outside infinity. What the orbiting astronaut experiences is apparent weightlessness, because both the astronaut and the spacecraft accelerate at the same rate towards Earth — there is no normal contact force pushing the astronaut into the floor.

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Exam Tips

• Quote g = F/m as the definition before introducing g = GM/r². The definition is what scores the AO1 mark; the formula is what scores the AO2 marks. Both are needed.
• Always work from the centre of the planet/star. Even if the question gives altitude h, the very first algebra step should be r = R + h. State this explicitly on the page — it earns method marks even if the arithmetic stumbles.
• Use the data sheet. G, the masses of Earth/Moon/Sun and the radius of Earth are not provided in standard questions unless explicitly given — but G is on the AQA data and formulae booklet. Reach for it; don't try to remember 6.67 × 10⁻¹¹ to extra precision.
• State units in your final answer. N kg⁻¹ for field strength, N for force, m for distance. A unit-less number is a 0/1 mark.
• For "compare" or "comment" parts, write a sentence. AO3 evaluation marks reward prose explanations, not bare numbers. "The field has decreased by a factor of ~15, consistent with an inverse-square law" earns more than "g is smaller".
• Significant figures: AQA's tolerance is typically ±1 in the last significant figure of the published answer. Carry one extra s.f. through working; round only at the end.
• Distinguish "field strength" from "potential". g is a vector with units N kg⁻¹; V is a scalar with units J kg⁻¹. Lesson 1 builds on this distinction.

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Summary

Concept	Formula	Units
Gravitational field strength (definition)	g = F/m	N kg⁻¹
Newton's law of gravitation	F = GMm/r²	N
Field strength for a point/spherical mass	g = GM/r²	N kg⁻¹
Gravitational constant	G = 6.67 × 10⁻¹¹	N m² kg⁻²

Exam Tip: In multi-mark calculations, always show your substitution clearly and include units at every stage. State the formula, substitute, then evaluate. Even if you make an arithmetic error, you will gain method marks for correct setup and substitution.