Gravitational Field Strength

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 how to describe and quantify these fields is the foundation of this entire topic and underpins everything from satellite motion to astrophysics.

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Defining Gravitational Field Strength

Gravitational field strength, denoted g, is defined as the force per unit mass experienced by a small test mass placed at a point in a gravitational field:

$g = \frac{F}{m}$g=mF​

where F is the gravitational force (in newtons) and m is the mass of the test object (in kilograms). The units of g are N kg⁻¹, which are equivalent to m s⁻².

The word "small" in the definition of the test mass is important. We use a small test mass so that it does not significantly alter the gravitational field it is measuring. If you placed a massive planet next to Earth to "test" Earth's field, the second planet's own field would distort everything. In practice, any mass can be used in calculations, but the definition requires this caveat.

Gravitational field strength is a vector quantity — it has both magnitude and direction. The direction of g at any point is the direction a free mass would accelerate if placed there, which is always towards the centre of the mass creating the field.

Common exam mistake: Students sometimes write that g is the "acceleration due to gravity." While numerically correct (9.81 m s⁻² at Earth's surface), the definition required by examiners is force per unit mass. Writing "acceleration due to gravity" in a definition question will lose marks.

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Uniform Fields Near Earth's Surface

Close to the Earth's surface, the gravitational field is approximately uniform. This means that the field strength has the same magnitude and direction at every point within the region. The value of g at the Earth's surface is approximately 9.81 N kg⁻¹ (or 9.81 m s⁻²).

In a uniform field, the field lines are parallel, equally spaced, and all point in the same direction — downwards towards the centre of the Earth. This approximation works well provided the region being considered is small compared to the radius of the Earth. Over a few kilometres of altitude near the surface, the variation in g is negligible for most calculations.

This uniform field approximation is the one you have been using since GCSE when you wrote W = mg. The weight of an object is simply the gravitational force on it, and since g is approximately constant near the surface, weight is proportional to mass.

How g Varies with Altitude Near the Surface

Even near the surface, g does change slightly with altitude. At height h above the surface (where h << R):

$g_h \approx g_0 \left(1 - \frac{2h}{R}\right)$gh​≈g0​(1−R2h​)

For example, at the top of Mount Everest (h ≈ 8,849 m, R = 6.37 × 10⁶ m):

$g \approx 9.81 \times \left(1 - \frac{2 \times 8849}{6.37 \times 10^6}\right) = 9.81 \times (1 - 0.00278) = 9.78 \text{ N kg}^{-1}$g≈9.81×(1−6.37×1062×8849​)=9.81×(1−0.00278)=9.78 N kg−1

This is only a 0.3% reduction — justifying the uniform field approximation for everyday calculations.

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Radial Fields of Point and Spherical Masses

For a point mass or any spherically symmetric mass, the gravitational field is radial. Field lines point inward towards the centre of the mass from all directions. The field lines are not parallel — they converge at the centre.

The field strength at a distance r from the centre of a spherical mass M is given by:

$g = \frac{GM}{r^2}$g=r2GM​

where G is the gravitational constant (6.674 × 10⁻¹¹ N m² kg⁻²). This equation tells us that g follows an inverse square law: if you double the distance from the centre, the field strength drops to one quarter of its original value.

Key points about radial fields:

• Outside a uniform sphere, the mass behaves as if it were all concentrated at the centre. This remarkable result (proven by Newton) means you can treat planets and stars as point masses when calculating field strength at their surfaces or beyond.
• At the surface of a planet of mass M and radius R, the field strength is g = GM/R².
• Inside a uniform sphere, only the mass enclosed within radius r contributes to the field strength. The mass further out has no net gravitational effect (the shell theorem).

The field lines in a radial field get further apart as distance increases, which visually represents the decreasing field strength.

Worked Example: How g Varies with Distance

Calculate the gravitational field strength at 1R, 2R, 3R, and 4R from the centre of a planet with surface gravity g₀ = 9.81 N kg⁻¹:

Distance from centre	g (N kg⁻¹)	Fraction of surface value
1R (surface)	9.81	1
2R	2.45	1/4
3R	1.09	1/9
4R	0.613	1/16
10R	0.0981	1/100

The inverse square law means g drops off rapidly at first, then more gradually. At 10 Earth radii (about 64,000 km), the field is just 1% of the surface value — but it is definitely not zero.

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Gravitational Field Lines

Field lines are a visual tool for representing gravitational fields. The rules for drawing gravitational field lines are:

• They always point in the direction of the gravitational force (towards the mass).
• They never cross each other.
• The closer together the lines, the stronger the field.
• For a point mass, the lines are radial, pointing inward.
• For a uniform field, the lines are parallel and equally spaced.

flowchart TD
    subgraph Uniform["Uniform Field (near surface)"]
        direction TB
        U1["↓  ↓  ↓  ↓  ↓  ↓"]
        U2["Parallel, equally spaced"]
        U3["g constant everywhere"]
    end
    subgraph Radial["Radial Field (point/sphere)"]
        direction TB
        R1["Lines converge to centre"]
        R2["Spacing increases with r"]
        R3["g ∝ 1/r²"]
    end
    Uniform --- Radial

When two or more masses are present, the field at any point is the vector sum of the individual fields (the principle of superposition). Between the Earth and Moon, for example, there is a point where the fields cancel and g = 0. This is the neutral point — closer to the Moon because the Moon has less mass.

Worked Example: Finding the Neutral Point

Find the neutral point between the Earth (M_E = 5.97 × 10²⁴ kg) and Moon (M_M = 7.35 × 10²² kg) separated by d = 3.84 × 10⁸ m.

At the neutral point, distance x from Earth's centre:

$\frac{GM_E}{x^2} = \frac{GM_M}{(d - x)^2}$x2GME​​=(d−x)2GMM​​

$\frac{M_E}{x^2} = \frac{M_M}{(d - x)^2}$x2ME​​=(d−x)2MM​​

$\frac{d - x}{x} = \sqrt{\frac{M_M}{M_E}} = \sqrt{\frac{7.35 \times 10^{22}}{5.97 \times 10^{24}}} = 0.1109$xd−x​=ME​MM​​​=5.97×10247.35×1022​​=0.1109

$d - x = 0.1109x$d−x=0.1109x

$d = 1.1109x$d=1.1109x

$x = \frac{3.84 \times 10^8}{1.1109} = 3.46 \times 10^8 \text{ m from Earth}$x=1.11093.84×108​=3.46×108 m from Earth

This is about 90% of the way from Earth to the Moon, confirming the neutral point is much closer to the less massive body.

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Comparing Gravitational and Electric Fields

Gravitational and electric fields share many mathematical similarities, but there are important differences:

Property	Gravitational Field	Electric Field
Source	Mass	Charge
Force law	F = GMm/r²	F = kQq/r²
Field strength	g = F/m (N kg⁻¹)	E = F/q (N C⁻¹)
Potential	V = −GM/r	V = kQ/r
Nature	Always attractive	Attractive or repulsive
Shielding	Cannot be shielded	Can be shielded
Relative strength	Very weak	Much stronger
Field lines	Always point inward	Point inward (−) or outward (+)
Mediating particle	Graviton (hypothetical)	Photon (virtual)

The most critical difference is that gravitational fields are always attractive, whereas electric fields can be either attractive or repulsive. There is no "negative mass" equivalent to negative charge. This means gravitational field lines always point towards the source mass, and the force between any two masses always pulls them together.

Despite gravity being far weaker than the electromagnetic force (by a factor of roughly 10³⁶ for fundamental particles), gravity dominates on astronomical scales because it is always attractive and cannot be cancelled out. Electric charges tend to neutralise each other, but masses always add.

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Calculating g at Planetary Surfaces

Using the equation g = GM/R², you can calculate the surface value of g for any body in the solar system:

Body	Mass (kg)	Radius (m)	g (N kg⁻¹)
Mercury	3.30 × 10²³	2.44 × 10⁶	3.70
Venus	4.87 × 10²⁴	6.05 × 10⁶	8.87
Earth	5.97 × 10²⁴	6.37 × 10⁶	9.81
Moon	7.35 × 10²²	1.74 × 10⁶	1.62
Mars	6.42 × 10²³	3.39 × 10⁶	3.72
Jupiter	1.90 × 10²⁷	6.99 × 10⁷	25.9
Saturn	5.68 × 10²⁶	5.82 × 10⁷	11.2

Verification for Earth:

$g = \frac{GM}{R^2} = \frac{6.674 \times 10^{-11} \times 5.97 \times 10^{24}}{(6.37 \times 10^6)^2} = \frac{3.984 \times 10^{14}}{4.058 \times 10^{13}} = 9.81 \text{ N kg}^{-1}$g=R2GM​=(6.37×106)26.674×10−11×5.97×1024​=4.058×10133.984×1014​=9.81 N kg−1

Notice that Jupiter, despite being 318 times more massive than Earth, has only about 2.6 times the surface gravity. This is because Jupiter's radius is 11 times larger, and the R² in the denominator partially offsets the much larger mass.

Exam tip: When comparing g values between planets, always express the comparison as a ratio. For example: g_Mars/g_Earth = (M_Mars/M_Earth) × (R_Earth/R_Mars)² = (0.107) × (1.878)² = 0.107 × 3.53 = 0.379. So g on Mars is about 38% of Earth's value.

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Summary

• Gravitational field strength g = F/m is the force per unit mass at a point in a field (not "acceleration due to gravity" in definitions)
• Near Earth's surface the field is approximately uniform with g ≈ 9.81 N kg⁻¹
• For radial fields around spheres: g = GM/r², an inverse square law
• Field lines show direction and strength; they never cross
• The shell theorem: no net field inside a hollow spherical shell
• Superposition: the net field is the vector sum of individual fields
• Gravitational fields are always attractive, unlike electric fields

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A-Level Deep Dive: Gravitational Field Strength

Spec mapping

Edexcel 9PH0 specification Topic 12 — Gravitational Fields introduces field strength as force per unit mass, develops the radial-field formula $g = GM/r^2$g=GM/r2, and connects field strength to potential and to orbital motion (refer to the official specification document for exact wording). Although Topic 12 sits in Year 2 of the linear A-Level, the underpinning ideas — Newton's second law, vector addition, inverse-square thinking — are bridged from Topic 2 (Mechanics) and Topic 3 (Electric Circuits) language about analogies between fields. Gravitational field strength is examined in Paper 2 (9PH0/02) alongside electric and magnetic fields, and routinely appears in Paper 3 (9PH0/03) synoptic questions linking orbital mechanics, energy conservation, and the deformation of stars in binary systems. The Edexcel formula booklet provides $g = -GM/r^2$g=−GM/r2 (radial) and $F = Gm_1 m_2 / r^2$F=Gm1​m2​/r2, but candidates must remember that $g = F/m$g=F/m is the defining equation and that the negative sign denotes the attractive direction toward the source mass.

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Worked example with full mark scheme

Question (8 marks):

(a) Show that the gravitational field strength at the surface of the Earth is approximately $9.8\ \text{N kg}^{-1}$9.8 N kg−1, given $M_E = 5.97 \times 10^{24}\ \text{kg}$ME​=5.97×1024 kg, $R_E = 6.37 \times 10^{6}\ \text{m}$RE​=6.37×106 m and $G = 6.67 \times 10^{-11}\ \text{N m}^2 \text{kg}^{-2}$G=6.67×10−11 N m2kg−2. (3)

(b) A weather satellite orbits at altitude $h = 800\ \text{km}$h=800 km above Earth's surface. Calculate the gravitational field strength experienced by the satellite. (3)

(c) Hence find the percentage reduction in $g$g between the surface and the satellite's orbit. (2)

Solution with mark scheme:

(a) Step 1 — apply the radial-field equation.

$g = \dfrac{GM_E}{R_E^2} = \dfrac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{(6.37 \times 10^{6})^2}$g=RE2​GME​​=(6.37×106)2(6.67×10−11)(5.97×1024)​

M1 — correct substitution into $g = GM/r^2$g=GM/r2 with $r = R_E$r=RE​ (the surface).

Step 2 — evaluate the denominator.

$R_E^2 = (6.37 \times 10^{6})^2 = 4.058 \times 10^{13}\ \text{m}^2$RE2​=(6.37×106)2=4.058×1013 m2.

Step 3 — evaluate the field strength.

$g = \dfrac{3.982 \times 10^{14}}{4.058 \times 10^{13}} = 9.81\ \text{N kg}^{-1}$g=4.058×10133.982×1014​=9.81 N kg−1

M1 — correct numerical evaluation. Common slip: forgetting to square $R_E$RE​, or inputting $R_E$RE​ in km rather than m.

A1 — answer to 3 s.f. with units $\text{N kg}^{-1}$N kg−1 (or equivalently $\text{m s}^{-2}$m s−2).

(b) Step 1 — find the orbital radius.

$r = R_E + h = 6.37 \times 10^{6} + 8.00 \times 10^{5} = 7.17 \times 10^{6}\ \text{m}$r=RE​+h=6.37×106+8.00×105=7.17×106 m.

M1 — adding altitude to Earth's radius. Common slip: using $h$h alone as the orbital radius — this is the single most penalised error on Topic 12 questions.

Step 2 — apply the radial-field equation.

$g = \dfrac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{(7.17 \times 10^{6})^2} = \dfrac{3.982 \times 10^{14}}{5.141 \times 10^{13}}$g=(7.17×106)2(6.67×10−11)(5.97×1024)​=5.141×10133.982×1014​

M1 — correct substitution.

A1 — $g = 7.74\ \text{N kg}^{-1}$g=7.74 N kg−1 (accept 7.7 to 7.8).

(c) Step 1 — compute the fractional change.

$\dfrac{\Delta g}{g_{\text{surface}}} = \dfrac{9.81 - 7.74}{9.81} = 0.211$gsurface​Δg​=9.819.81−7.74​=0.211

M1 — correct ratio.

A1 — 21.1% reduction (accept 21% to 22%).

Total: 8 marks (M5 A3, split as shown).

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Specimen question modelled on the Edexcel 9PH0 Paper 2 format

Question (6 marks): The Moon has mass $7.35 \times 10^{22}\ \text{kg}$7.35×1022 kg and radius $1.74 \times 10^{6}\ \text{m}$1.74×106 m.

(a) Calculate the gravitational field strength at the Moon's surface. (2)

(b) Express the surface field strength on the Moon as a fraction of that on Earth. (1)

(c) An astronaut of mass $70\ \text{kg}$70 kg stands on the Moon. State the value of (i) her mass, (ii) her weight on the Moon, and (iii) explain why these answers do not contradict each other. (3)

Mark scheme decomposition by AO:

(a)

• M1 (AO2.1) — substituting into $g = GM/r^2$g=GM/r2 with $M = 7.35 \times 10^{22}$M=7.35×1022 and $r = 1.74 \times 10^{6}$r=1.74×106.
• A1 (AO2.1) — $g_{\text{Moon}} = 1.62\ \text{N kg}^{-1}$gMoon​=1.62 N kg−1 (accept 1.6 to 1.7).

(b)

• B1 (AO2.2) — $g_{\text{Moon}}/g_{\text{Earth}} = 1.62/9.81 \approx 0.165$gMoon​/gEarth​=1.62/9.81≈0.165, or "about one-sixth".

(c)

• B1 (AO1.1) — mass on the Moon is $70\ \text{kg}$70 kg (mass is invariant).
• B1 (AO1.2) — weight on the Moon is $W = mg = 70 \times 1.62 = 113\ \text{N}$W=mg=70×1.62=113 N (accept 110 to 115).
• B1 (AO3.1) — explanation that mass measures quantity of matter (a scalar property of the body) while weight is the gravitational force on that mass and depends on the local field strength. The two quantities are physically distinct.

Total: 6 marks split AO1 = 2, AO2 = 3, AO3 = 1. Notice how Edexcel uses the mass/weight distinction repeatedly as a low-tariff AO3 conceptual gate. Rote substitution earns the AO1/AO2 marks; the AO3 mark requires articulating why.

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Synoptic links

Connects to:

1. Topic 11 — Newton's Law of Gravitation: the radial field equation $g = GM/r^2$g=GM/r2 is a direct corollary of $F = Gm_1 m_2/r^2$F=Gm1​m2​/r2 divided by the test mass. Field strength is the per-unit-mass abstraction of force — the conceptual bridge from "force between masses" to "field permeating space".

2. Topic 7 — Electric Fields: the structural analogy between gravitational $g = GM/r^2$g=GM/r2 and electric $E = kQ/r^2$E=kQ/r2 is exact, with the crucial difference that gravity is always attractive while electric forces can repel. Examiners frequently ask candidates to compare or contrast the two field formulae — the Coulomb constant $k = 1/(4\pi\varepsilon_0)$k=1/(4πε0​) plays the role of $G$G, and charge $Q$Q plays the role of mass $M$M.

3. Topic 4 — Mechanics (circular motion): for a satellite in circular orbit, the gravitational field provides the centripetal acceleration: $g = v^2/r$g=v2/r at orbital radius $r$r. Equating $GM/r^2 = v^2/r$GM/r2=v2/r yields the orbital speed $v = \sqrt{GM/r}$v=GM/r​. This is the algebraic backbone of every satellite-orbit question on Paper 2.

4. Topic 12 — Gravitational Potential: field strength and potential are linked by $g = -dV/dr$g=−dV/dr. Where $V$V is steepest, $g$g is largest. A flat region of the potential corresponds to zero net field — the saddle point between two masses (Lagrange point L1 in idealised form).

5. Topic 13 — Astrophysics and Cosmology: stellar structure depends critically on the surface gravity $g = GM/R^2$g=GM/R2. White dwarfs and neutron stars have radii thousands of times smaller than the Sun's despite comparable masses, giving surface field strengths that exceed Earth's by factors of $10^6$106 to $10^{12}$1012.

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Mark-scheme literacy

Gravitational field strength questions on 9PH0/02 split AO marks as follows:

AO	Typical share	Earned by
AO1 (knowledge)	25–35%	Stating $g = F/m$g=F/m, naming units, recalling $g \approx 9.81\ \text{N kg}^{-1}$g≈9.81 N kg−1, defining "field"
AO2 (application)	45–55%	Substituting into $g = GM/r^2$g=GM/r2, computing field strength at altitude, comparing planets
AO3 (analysis / evaluation)	15–25%	Resolving conceptual mass-vs-weight questions, evaluating limits of the point-mass model, justifying the negative sign

Examiner-rewarded phrasing: "field strength is the force per unit mass at a point in the field"; "for an extended spherically symmetric body, the field outside is identical to that of a point mass at the centre (shell theorem)"; "at radius $r$r from the centre, where $r > R$r>R (the body's radius)". Phrases that lose marks: "g is the acceleration due to gravity" (this is equivalent in free fall but is not the definition — Edexcel mark schemes consistently reward the force-per-unit-mass formulation); omitting units $\text{N kg}^{-1}$N kg−1; treating $g$g as a scalar in vector-addition contexts.

A specific Edexcel pattern: questions about field strength inside a spherical shell expect the explicit answer "$g = 0$g=0 inside the shell, by the shell theorem" rather than calculation. Going through the integral wastes time and gains nothing.

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Grade-band model answers

3-mark question

Question: Calculate the gravitational field strength at a height equal to the Earth's radius above the surface.

Grade C response (~150 words):

The orbital radius is $2 R_E$2RE​, so $r = 2 \times 6.37 \times 10^{6} = 1.274 \times 10^{7}\ \text{m}$r=2×6.37×106=1.274×107 m.

$g = GM/r^2 = (6.67 \times 10^{-11})(5.97 \times 10^{24})/(1.274 \times 10^{7})^2 = 3.982 \times 10^{14}/1.623 \times 10^{14} = 2.45\ \text{N kg}^{-1}$g=GM/r2=(6.67×10−11)(5.97×1024)/(1.274×107)2=3.982×1014/1.623×1014=2.45 N kg−1.

So the field strength is about $2.45\ \text{N kg}^{-1}$2.45 N kg−1.

Examiner commentary: Full marks (3/3). The candidate correctly doubles the Earth's radius, applies the radial-field formula, and quotes the answer with units. A minor presentation point: the working would be cleaner if the doubling step used the inverse-square shortcut $g(2R)/g(R) = 1/4$g(2R)/g(R)=1/4, giving $g = 9.81/4 = 2.45\ \text{N kg}^{-1}$g=9.81/4=2.45 N kg−1 in a single line. Both methods earn full marks.

Grade A response (~180 words):*

At a height equal to one Earth radius above the surface, the distance from the Earth's centre is $r = 2 R_E$r=2RE​. Since $g \propto 1/r^2$g∝1/r2, doubling the distance reduces the field strength by a factor of $4$4:

$g(2R_E) = \dfrac{g_{\text{surface}}}{4} = \dfrac{9.81}{4} = 2.45\ \text{N kg}^{-1}$g(2RE​)=4gsurface​​=49.81​=2.45 N kg−1

To verify by direct substitution:

$g = \dfrac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{(1.274 \times 10^{7})^2} = 2.45\ \text{N kg}^{-1}$g=(1.274×107)2(6.67×10−11)(5.97×1024)​=2.45 N kg−1

So the field strength halves not at twice the radius but at $\sqrt{2} R_E$2​RE​ — a useful sanity check for "half the field" altitudes.

Examiner commentary: Full marks (3/3). The candidate exploits the inverse-square scaling to obtain the answer in one line, then verifies by direct substitution. The closing sanity check on "half the field" altitude shows the kind of structural fluency that signals an A* candidate. This question rewards understanding of the inverse-square law as much as substitution skill.

6-mark question

Question: A planet has half the radius of Earth and twice the mass. (a) Find the surface gravitational field strength as a multiple of $g_E$gE​. (b) Explain qualitatively why this planet would not retain a substantial atmosphere of light gases despite its strong surface gravity, given its small size.

Grade B response (~220 words):

(a) $g_{\text{planet}} = GM_p/R_p^2 = G(2M_E)/(R_E/2)^2 = (2GM_E)/(R_E^2/4) = 8GM_E/R_E^2 = 8 g_E$gplanet​=GMp​/Rp2​=G(2ME​)/(RE​/2)2=(2GME​)/(RE2​/4)=8GME​/RE2​=8gE​.

So the surface field strength is $8\ g_E \approx 78.5\ \text{N kg}^{-1}$8 gE​≈78.5 N kg−1.

(b) Even though gravity is strong, the planet is small so its escape velocity might not be high enough to retain hydrogen and helium, which move very fast at typical temperatures.

Examiner commentary: (a) is full marks (3/3) — the algebraic manipulation is clean and the numerical value correctly quoted. (b) is 1/3. The candidate gestures at the right concepts (escape velocity, light-gas thermal speed) but does not connect them to the radius of the planet. The escape velocity $v_{\text{esc}} = \sqrt{2GM/R}$vesc​=2GM/R​ scales as $\sqrt{M/R}$M/R​, so doubling $M$M and halving $R$R gives a factor-of-2 increase in $v_{\text{esc}}$vesc​ — actually better atmospheric retention. The candidate has confused field strength with escape velocity. Total: 4/6.

Grade A response (~290 words):*

(a) Using $g = GM/R^2$g=GM/R2:

$\dfrac{g_{\text{planet}}}{g_E} = \dfrac{M_p/R_p^2}{M_E/R_E^2} = \dfrac{2M_E}{(R_E/2)^2} \cdot \dfrac{R_E^2}{M_E} = \dfrac{2 \cdot 4 \cdot R_E^2}{R_E^2} = 8$gE​gplanet​​=ME​/RE2​Mp​/Rp2​​=(RE​/2)22ME​​⋅ME​RE2​​=RE2​2⋅4⋅RE2​​=8

So the surface field strength is $8 g_E$8gE​, or about $78.5\ \text{N kg}^{-1}$78.5 N kg−1.

(b) Atmospheric retention depends on the escape velocity $v_{\text{esc}} = \sqrt{2GM/R}$vesc​=2GM/R​, not surface field strength. Substituting $M_p = 2M_E$Mp​=2ME​ and $R_p = R_E/2$Rp​=RE​/2:

$\dfrac{v_{\text{esc, planet}}}{v_{\text{esc, E}}} = \sqrt{\dfrac{2 \cdot 2}{1/2}} = \sqrt{8} \approx 2.83$vesc, E​vesc, planet​​=1/22⋅2​​=8​≈2.83

So the escape velocity is actually about 2.83 times Earth's. The premise of the question contains a subtle confusion: stronger surface gravity combined with smaller radius gives higher escape velocity, not lower. A planet meeting this description would retain even hydrogen and helium more effectively than Earth.

Examiner commentary: Full marks (6/6) — and crucially, the candidate refuses the question's framing. The A* mark in (b) goes to recognising that the question's premise is physically wrong: combining greater mass with smaller radius enhances atmospheric retention, not reduces it. Edexcel routinely tests the ability to challenge a question's framing — the AO3 mark explicitly rewards "appropriate critique of given assumptions". A weaker candidate writes what the question seems to want; an A* candidate writes what is true.

9-mark question (A* extended-prose)

Question: A binary star system consists of two stars of equal mass $M$M separated by a distance $2a$2a. Derive an expression for the net gravitational field strength at a point P on the perpendicular bisector of the line joining the two stars, at a distance $d$d from the midpoint. Discuss the limiting cases $d \ll a$d≪a and $d \gg a$d≫a.

Grade A response (~360 words):*

Place the stars at $(\pm a, 0)$(±a,0) and the point P at $(0, d)$(0,d). The distance from each star to P is $r = \sqrt{a^2 + d^2}$r=a2+d2​.

The magnitude of the field due to each star at P is:

$g_1 = g_2 = \dfrac{GM}{a^2 + d^2}$g1​=g2​=a2+d2GM​

By symmetry, the components along the line joining the stars cancel; the components perpendicular to that line (toward the midpoint, i.e. in the $-y$−y direction) add. The component perpendicular is $g_i \cos\theta$gi​cosθ where $\cos\theta = d/r = d/\sqrt{a^2 + d^2}$cosθ=d/r=d/a2+d2​.

So the net field at P, directed along the perpendicular bisector toward the midpoint, has magnitude:

$g_{\text{net}} = 2 g_1 \cos\theta = \dfrac{2GM}{a^2 + d^2} \cdot \dfrac{d}{\sqrt{a^2 + d^2}} = \dfrac{2GMd}{(a^2 + d^2)^{3/2}}$gnet​=2g1​cosθ=a2+d22GM​⋅a2+d2​d​=(a2+d2)3/22GMd​

Limit $d \ll a$d≪a: $(a^2 + d^2)^{3/2} \approx a^3$(a2+d2)3/2≈a3, so $g_{\text{net}} \approx 2GMd/a^3$gnet​≈2GMd/a3 — linear in $d$d. Close to the midpoint, the field strength rises proportionally with displacement, characteristic of a restoring force (the system behaves locally like a harmonic oscillator with effective spring constant $2GM/a^3$2GM/a3).

Limit $d \gg a$d≫a: $(a^2 + d^2)^{3/2} \approx d^3$(a2+d2)3/2≈d3, so $g_{\text{net}} \approx 2GMd/d^3 = 2GM/d^2$gnet​≈2GMd/d3=2GM/d2. Far from the system, it looks like a single point mass of total mass $2M$2M at distance $d$d — the inverse-square law re-emerges, as expected from the multipole expansion.

Examiner commentary: Full marks (9/9). The vector decomposition is set up cleanly, the symmetry argument identifies which components cancel, and both limiting cases are interpreted physically (not just algebraically). The harmonic-oscillator interpretation in the near limit shows synoptic awareness of mechanics; the multipole interpretation in the far limit shows awareness of asymptotic methods. This is precisely the level of insight Edexcel reserves the top marks for on Paper 3 synoptic questions.

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A-Level-depth misconceptions

The errors that distinguish A from A* on field strength questions:

1. $g$g varies with altitude. Many candidates treat $g = 9.81\ \text{N kg}^{-1}$g=9.81 N kg−1 as a constant. At the orbital altitude of the ISS (~400 km), $g \approx 8.7\ \text{N kg}^{-1}$g≈8.7 N kg−1 — astronauts experience near-free-fall not because gravity is weak but because they are accelerating with the station. "Microgravity" is a misnomer.

2. Mass vs weight. Mass is invariant (kg); weight is the gravitational force on that mass and changes with location (N). On the Moon, a 70 kg astronaut still has mass 70 kg; her weight changes from $\sim 687\ \text{N}$∼687 N to $\sim 113\ \text{N}$∼113 N. Bathroom scales measure weight (force) and divide by an assumed Earth $g$g to display mass — they would mis-read elsewhere.

3. Treating extended bodies as point masses without justification. The shell theorem ($g$g outside a spherically symmetric body equals that of a point mass at its centre) is the justification for using $g = GM/r^2$g=GM/r2 for planets. Using the formula without naming the theorem leaves an unstated assumption — and on derivation questions, the missing justification can lose a mark.

4. Weight at the equator vs poles. Earth's rotation reduces the apparent weight at the equator (some of the gravitational force provides centripetal acceleration). Earth's oblateness also means the equatorial radius is larger, so true $g$g is slightly smaller there ($\sim 9.78\ \text{N kg}^{-1}$∼9.78 N kg−1 at the equator, $\sim 9.83\ \text{N kg}^{-1}$∼9.83 N kg−1 at the poles). Edexcel does not require numerical knowledge of this, but the qualitative reasoning appears.

5. Confusing $g$g as a vector vs scalar. Field strength is a vector $\vec g$g​. When two masses contribute to the field at a point, vector addition is required — not scalar addition of magnitudes. The midpoint between two equal masses has zero net field by symmetry (vectors cancel), not double the field.

6. The negative sign in $g = -GM/r^2$g=−GM/r2. The negative sign indicates direction (radially inward, toward the source). When asked for magnitude, the answer is $GM/r^2$GM/r2 (positive). Carrying the negative sign into a magnitude calculation is a sign error.

7. Field inside a uniform spherical shell. By the shell theorem, the field is zero everywhere inside a hollow shell — the contributions from all parts of the shell cancel exactly. Candidates often guess "still inverse-square" or "constant" — both wrong. The correct answer is identically zero, and it is examinable.

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Common errors and mark-loss patterns

Three patterns repeatedly cost candidates marks on Topic 12 questions. They are about reading and presentation, not technique.

• Using altitude as the orbital radius. When a question says "satellite at altitude $h$h", the radius from Earth's centre is $R_E + h$RE​+h, not $h$h. Substituting $h$h alone into $g = GM/r^2$g=GM/r2 produces an answer that can be three or four orders of magnitude too large. The fix: always sketch the geometry, label distances from the centre.
• Wrong unit for field strength. $g$g has units $\text{N kg}^{-1}$N kg−1 (force per unit mass) or equivalently $\text{m s}^{-2}$m s−2 (acceleration). Candidates who write $\text{N}$N or $\text{kg m s}^{-2}$kg m s−2 have written units of force, not field strength. The A1 mark for the final answer requires the correct unit — Edexcel mark schemes are unforgiving on this.
• Calculator precision loss. Substituting $G \times M$G×M as a single product into a calculator without parentheses, then dividing by $r^2$r2, produces a string of intermediate roundings that can shift the final answer by several percent. The cure: compute the numerator and denominator separately, write each to 4 s.f., then divide. Final answer to 3 s.f.
• Ignoring vector direction. When asked for the field at a point between two masses, candidates compute magnitudes and add. The correct approach: pick a sign convention (e.g. positive toward mass 1), assign signs to each contribution, then add algebraically.

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Going further — university and academic signposting

Gravitational field strength is the gateway to several Year 1 and Year 2 university topics:

• General Relativity (Year 3/4 physics; Year 4 maths). Newton's $g = GM/r^2$g=GM/r2 is the weak-field limit of Einstein's field equations. In the full theory, gravity is not a force but the curvature of spacetime caused by mass-energy. Predictions diverge from Newton's only in extreme conditions: near black holes, around neutron stars, or for high-precision tests like GPS satellite timing (which corrects for ~38 microseconds per day of relativistic effects).
• Gravitational time dilation. Clocks run slower in stronger gravitational fields. A clock at sea level ticks slightly slower than a clock on a mountain top — measurable with modern atomic clocks. The effect scales as $\Delta t / t \approx gh/c^2$Δt/t≈gh/c2 for height differences $h$h in a uniform field $g$g. This is the empirical evidence that gravity is geometry, not force.
• Equivalence principle. Einstein's "happiest thought": a person in free fall feels no gravity. This equivalence between gravitational and inertial mass — the experimental observation that all bodies fall with the same acceleration in a given field — was the seed from which General Relativity grew. The Eötvös experiment (1908) verified this to one part in $10^9$109; modern satellite tests (MICROSCOPE, 2022) verify it to one part in $10^{15}$1015.
• Gravitational wave astronomy (Year 3 onwards). LIGO detected the first gravitational waves in 2015 — ripples in spacetime caused by accelerating massive objects. The analysis requires understanding how field strength varies near merging black holes, a regime where Newton's formula fails completely.

Oxbridge interview prompt: "An astronaut on the ISS is in 'free fall' yet experiences a gravitational field strength of about $8.7\ \text{N kg}^{-1}$8.7 N kg−1. Reconcile these statements. Then describe what would happen if the ISS suddenly stopped orbiting and hovered in place."

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Required practical reference

Core Practical 1 (CP1) — Determination of $g$g by free-fall (~150 words).

A small ball-bearing is held by an electromagnet and released from rest. The release simultaneously starts an electronic timer; the ball passes through a light gate at distance $h$h below the release point, stopping the timer. From $h = \tfrac{1}{2} g t^2$h=21​gt2, plotting $h$h on the $y$y-axis against $t^2$t2 on the $x$x-axis yields a straight line through the origin with gradient $g/2$g/2. The expected value is $g \approx 9.81\ \text{N kg}^{-1}$g≈9.81 N kg−1.

Key uncertainties:

• Reaction-time error is eliminated by the electronic timing, but the electromagnet's residual magnetism may delay release by a few milliseconds — the dominant systematic error.
• Air resistance is negligible for a dense ball over short distances (~50 cm) but should be acknowledged.
• Repeats at each height (typically 5) reduce random uncertainty in $t$t. A range of heights (10 cm to 100 cm) produces enough data for a reliable gradient.

The graphical approach (linearisation) is preferred over single-shot calculation because it averages out timing fluctuations and exposes systematic offsets as a non-zero $y$y-intercept.

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Edexcel A-Level alignment footer

This content is aligned with the Pearson Edexcel GCE A Level Physics (9PH0) specification, Paper 2 — Advanced Physics II, Topic 12: Gravitational Fields. For the most accurate and up-to-date information, please refer to the official Pearson Edexcel specification document.

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Visual summary

graph TD
    A["Two masses<br/>interact gravitationally"] --> B["Newton's law:<br/>F = Gm₁m₂/r²"]
    B --> C["Define field strength:<br/>g = F/m (force per unit mass)"]
    C --> D{"Source geometry?"}
    D -->|"Point mass<br/>or sphere (outside)"| E["Radial field:<br/>g = GM/r²<br/>inverse-square"]
    D -->|"Inside spherical shell"| F["Shell theorem:<br/>g = 0"]
    D -->|"Near Earth's surface<br/>(small Δh)"| G["Uniform field:<br/>g ≈ 9.81 N kg⁻¹"]
    E --> H{"Multiple sources?"}
    F --> H
    G --> H
    H -->|"Yes"| I["Vector superposition:<br/>add g⃗ contributions"]
    H -->|"No"| J["Single radial field"]
    I --> K["Net g⃗ at point"]
    J --> K
    K --> L["Use g for:<br/>weight (W = mg)<br/>orbital motion (g = v²/r)<br/>potential gradient (g = -dV/dr)"]

    style E fill:#27ae60,color:#fff
    style F fill:#e67e22,color:#fff
    style L fill:#3498db,color:#fff