Gravitational Field Strength Revision Notes | A Level AQA Physics
Gravitational field strength is the force per unit mass on a small test mass, g = F/m, and in a radial field g = GM/r². Free notes and 12 questions answered.
Updated 12 Sept 2026 · Revise Right Team
Gravitational Field Strength is part of Gravitational fields (A-level only) in A Level AQA Physics.
7 min
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A Level AQA
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What this lesson covers
Gravitational field strength is the force per unit mass on a small test mass, g = F/m, and in a radial field g = GM/r². Free notes and 12 questions answered.
Describe gravitational field lines for uniform and radial fields, including direction conventions
Define gravitational field strength g as force per unit mass using g = F/m
Calculate g in a radial field using Newton's law of gravitation: g = GM/r²
Analyse how g varies with distance from a massive body and interpret this graphically
Key ideas to keep in view
Use this lesson to stay inside Gravitational fields (A-level only) while connecting the detail back to the wider Fields and their consequences (A-level only) topic.
Gravitational field strength (g): Vector quantity defined as the gravitational force per unit mass acting on a small test mass placed at a point in the field. Measured in N kg⁻¹ (equivalent to m s⁻²).
Radial gravitational field: Field in which field lines point radially inward toward a central mass, and field strength decreases with the square of the distance from the centre of that mass.
Uniform gravitational field: Field in which g has the same magnitude and direction at every point, represented by equally spaced parallel field lines. Approximated near the Earth's surface over small regions.
Newton's gravitational constant (G): Universal constant of proportionality in Newton's law of gravitation. G = 6.67 × 10⁻¹¹ N m² kg⁻².
Revision notes
Read the core explanations from Gravitational Field Strength before testing yourself from memory.
1. Why Gravitational Field Strength Matters
Why does an astronaut weigh less on the Moon but still have the same mass? The answer lies in gravitational field strength — and once you understand it, satellite orbits, planetary surfaces, and free-fall all click into place.
At A Level, gravity is treated as a field: a region of space where any mass experiences a force. The strength of that field at any point is what we call $g$. This lesson builds from the field-line picture all the way to the inverse-square law, giving you the tools to calculate $g$ anywhere in the universe.
2. Representing Gravitational Fields with Field Lines
Picture the Earth floating in space. Now imagine drawing arrows that show the direction a small test mass would be pulled at every point around it. Those arrows are gravitational field lines, and their pattern tells you everything about the shape of the field.
For a spherical mass like a planet, the field lines point radially inward toward the centre of the mass — they converge from all directions like the spokes of a wheel pointing toward the hub. This is a radial field. Notice that the lines get further apart as you move away from the planet: that spacing encodes the weakening of the field with distance.
Close to the Earth's surface, over a small region, those converging lines are nearly parallel. We treat this as a uniform field — equally spaced, parallel lines all pointing straight downward. The uniform approximation is valid whenever the region of interest is tiny compared to the Earth's radius.
Field lines always point in the direction of the force on a positive (i.e. ordinary) test mass — for gravity, that is always toward the attracting mass.
Closer spacing of field lines indicates a stronger field; wider spacing indicates a weaker field.
Field lines never cross — each point in space has one unique field direction.
A radial field has lines converging to a central point; a uniform field has lines that are parallel and equally spaced.
Gravitational fields have no 'outward' lines — gravity is always attractive, never repulsive.
3. Gravitational Field Strength: g = F/m
Field lines show direction, but we need a number to describe how strong the field is at any point. That number is $g$, the gravitational field strength, and its definition is precise: it is the gravitational force per unit mass experienced by a small test mass placed at that point.
Rearranging gives $F = mg$, which you will recognise as the weight equation. Weight is not a fixed property of an object — it depends on the local value of $g$. On the Moon, $g \approx 1.6$ N kg⁻¹, so a 70 kg astronaut experiences a gravitational force of only $70 \times 1.6 = 112$ N, compared with $70 \times 9.8 = 686$ N on Earth.
Because $g$ is defined as force per unit mass, its unit is N kg⁻¹. You can also write this as m s⁻² — both are correct and equivalent, but N kg⁻¹ is the more natural unit when thinking about field strength rather than acceleration.
4. The Radial Field Equation: g = GM/r²
The definition $g = F/m$ tells us what $g$ means, but it does not tell us how $g$ depends on where you are relative to a planet. For that, we combine the definition with Newton's law of gravitation.
Newton's law states that the gravitational force between two masses $M$ and $m$ separated by distance $r$ (measured centre to centre) is $F = \frac{GMm}{r^2}$. Substituting this into $g = F/m$ gives the magnitude of $g$ in a radial field:
The inverse-square relationship is the most important feature here. Double the distance from the centre of a planet and $g$ falls to one quarter of its original value. Triple the distance and $g$ falls to one ninth. This is why gravitational field lines spread out — the same 'amount' of field is distributed over a larger spherical surface area as $r$ increases.
Notice that $g$ depends only on $M$ (the mass creating the field) and $r$ (your distance from its centre). The mass of the object placed in the field does not appear — $g$ is a property of the field at a point, not of the object experiencing it.
5. Worked Calculations Using Both Equations
Theory is only useful when you can apply it under exam conditions. The two equations $g = F/m$ and $g = GM/r^2$ are often used together in multi-step problems. Here is a structured walkthrough of each type.
Worked example: Calculating g at the Earth's Surface from First Principles
The mass of the Earth is $5.97 \times 10^{24}$ kg and its mean radius is $6.37 \times 10^{6}$ m. Calculate the gravitational field strength at the Earth's surface. Give your answer to 3 significant figures. ($G = 6.67 \times 10^{-11}$ N m² kg⁻²)
$g = \frac{GM}{r^2}$ — [M1] This is the equation for g in a radial field. We are at the surface, so r equals the Earth's radius. Always state the formula first for the method mark.
$G = 6.67 \times 10^{-11}$ N m² kg⁻²
$M = 5.97 \times 10^{24}$ kg
$r = 6.37 \times 10^{6}$ m — Writing out known values separately prevents substitution errors and earns the method mark even if arithmetic slips later.
$r^2 = (6.37 \times 10^{6})^2 = 6.37 \times 6.37 \times 10^{12}$
$6.37 \times 6.37 = 40.5769$
$r^2 = 4.058 \times 10^{13}$ m² — Expand the square explicitly: 6.37 × 6.37 = 40.5769, then apply the power of ten: (10⁶)² = 10¹². Never skip this step — squaring errors are the most common source of wrong answers here.
$GM = 6.67 \times 10^{-11} \times 5.97 \times 10^{24}$
$= 6.67 \times 5.97 \times 10^{13}$
$6.67 \times 5.97 = 39.82$
$GM = 3.982 \times 10^{14}$ N m² kg⁻¹ — [M1] Multiply the coefficients: 6.67 × 5.97 = 39.82, then combine the powers of ten: 10⁻¹¹ × 10²⁴ = 10¹³.
$g = \frac{3.982 \times 10^{14}}{4.058 \times 10^{13}} = \frac{3.982}{4.058} \times 10^{14-13} = 0.9813 \times 10^{1}$
$g = 9.81$ N kg⁻¹ — [A1] Divide coefficients: 3.982 ÷ 4.058 ≈ 0.9813. Combine powers: 10¹⁴ ÷ 10¹³ = 10¹. So g ≈ 9.81 N kg⁻¹, consistent with the accepted surface value.
$g = 9.81$ N kg⁻¹ (3 s.f.) — Always include units in the final answer. N kg⁻¹ is the correct unit for gravitational field strength. This matches the standard value of g at the Earth's surface, which is a useful sanity check.
Answer: $g = 9.81$ N kg⁻¹
Worked example: Finding the Force on a Satellite Using g = F/m
At a certain altitude above the Earth, the gravitational field strength is $g = 6.40$ N kg⁻¹. A satellite of mass $850$ kg orbits at this altitude. Calculate the gravitational force acting on the satellite.
$F = mg$ — [M1] Rearranging g = F/m gives F = mg. We know both m and g, so this is a direct substitution.
$F = 850 \times 6.40$ — [M1] m = 850 kg, g = 6.40 N kg⁻¹. Both values are already in SI units, so no conversion is needed.
$F = 850 \times 6.40 = 5440$ N — [A1] 850 × 6 = 5100, 850 × 0.40 = 340, total = 5440 N.
$F = 5440$ N (or $5.44 \times 10^3$ N) — The unit of force is the newton (N). This force is the satellite's weight at that altitude — it provides the centripetal force for the orbit.
Answer: $F = 5440$ N
6. How g Varies with Distance: Graphical Interpretation
The equation $g = GM/r^2$ predicts a specific shape for the graph of $g$ against $r$. Understanding this shape is essential for data-analysis questions in the exam.
A graph of $g$ against $r$ (for $r$ measured from the centre of the planet, outside the surface) produces a curve that falls steeply at first and then flattens — an inverse-square curve. It never reaches zero, but approaches it asymptotically as $r \to \infty$. The curve is not a straight line and not a simple $1/r$ curve.
To confirm the inverse-square relationship from data, plot $g$ against $1/r^2$. If the relationship is truly $g \propto 1/r^2$, this graph will be a straight line through the origin with gradient equal to $GM$. This is a standard linearisation technique that examiners test directly.
At the Earth's surface ($r = R_E$), $g \approx 9.81$ N kg⁻¹. At $r = 2R_E$, $g = 9.81/4 \approx 2.45$ N kg⁻¹. At $r = 3R_E$, $g = 9.81/9 \approx 1.09$ N kg⁻¹. Each value follows directly from the inverse-square law — check these against the chart above.
7. Exam Technique and Common Mistakes
The calculations in this topic are straightforward in structure, but several specific traps account for the majority of lost marks. Addressing them directly is more useful than a general reminder to 'check your work'.
For extended-response questions asking you to 'explain how g varies with distance from a planet', a full-mark answer requires three elements: the mathematical relationship ($g \propto 1/r^2$), the direction of change (g decreases as r increases), and a quantitative statement (doubling r reduces g by a factor of 4).
Worked example: Exam-Style: Finding g at Altitude Above a Planet
A planet has mass $M = 3.20 \times 10^{23}$ kg and radius $R = 3.40 \times 10^{6}$ m. A spacecraft orbits at an altitude of $h = 2.00 \times 10^{5}$ m above the surface. Calculate the gravitational field strength at the spacecraft's orbital altitude. ($G = 6.67 \times 10^{-11}$ N m² kg⁻²) [3 marks]
$r = R + h = 3.40 \times 10^{6} + 2.00 \times 10^{5}$
$= 3.40 \times 10^{6} + 0.200 \times 10^{6}$
$= 3.60 \times 10^{6}$ m — [M1] r must be measured from the planet's centre, not from the surface. Convert h to the same power of ten as R before adding.
$g = \frac{GM}{r^2} = \frac{2.134 \times 10^{13}}{1.296 \times 10^{13}}$
$= \frac{2.134}{1.296} \times 10^{0}$
$= 1.646...$
$g \approx 1.65$ N kg⁻¹ — [A1] Divide coefficients: 2.134 ÷ 1.296 ≈ 1.646. Powers: 10¹³ ÷ 10¹³ = 10⁰ = 1. Round to 3 significant figures to match the data given.
$g = 1.65$ N kg⁻¹ (3 s.f.) — Units are N kg⁻¹. This is significantly less than the surface value, which makes physical sense — the spacecraft is further from the planet's centre.
Answer: $g = 1.65$ N kg⁻¹
Both spec equations — $g = F/m$ and $g = GM/r^2$ — are now fully covered. The first defines what $g$ means physically; the second tells you how to calculate it at any distance from a spherical mass. Together, they underpin every gravitational field calculation at A Level.
Sources and content standards
Published by Revise Right. Use the official specification reference to check the requirements for your course. Content updated: 2026-09-12.
What is the formula for gravitational field strength in a radial field?
In a radial field, g = GM/r², where M is the mass creating the field and r is the distance from its centre. For Earth, M = 5.97 × 10²⁴ kg and r = 6.37 × 10⁶ m give g = 3.982 × 10¹⁴ ÷ 4.058 × 10¹³ = 9.81 N kg⁻¹. The mass of the object placed in the field does not appear.
What is gravitational field strength?
Gravitational field strength g is the gravitational force per unit mass on a small test mass at a point: g = F/m, measured in N kg⁻¹ (equivalent to m s⁻²). Rearranging gives weight, F = mg, so weight depends on the local g: a 70 kg astronaut weighs 112 N on the Moon, where g ≈ 1.6 N kg⁻¹, but 686 N on Earth.
How does g change with distance from a planet?
g decreases with distance following an inverse-square law, g ∝ 1/r²: doubling r reduces g to one quarter and tripling it to one ninth. For Earth, g ≈ 9.81 N kg⁻¹ at the surface, about 2.45 N kg⁻¹ at 2R and 1.09 N kg⁻¹ at 3R. A graph of g against 1/r² is a straight line through the origin with gradient GM.
What is the difference between a radial and a uniform gravitational field?
In a radial field, such as around a planet, field lines point inward towards the centre and spread out with distance, showing the field weakening. In a uniform field the lines are parallel and equally spaced; close to Earth's surface, over a small region, the field is treated as uniform with lines pointing straight downward.
How do you find g at a height above a planet?
Add the altitude to the planet's radius to get r from the centre, then use g = GM/r². For a planet of mass 3.20 × 10²³ kg and radius 3.40 × 10⁶ m, a spacecraft at an altitude of 2.00 × 10⁵ m is at r = 3.60 × 10⁶ m, so g = 2.134 × 10¹³ ÷ 1.296 × 10¹³ ≈ 1.65 N kg⁻¹.
What are the units of gravitational field strength?
Gravitational field strength is measured in N kg⁻¹, because it is force per unit mass, g = F/m. N kg⁻¹ is equivalent to m s⁻², but N kg⁻¹ is the more natural unit when thinking about field strength rather than acceleration.
How do you calculate weight on the Moon?
Weight on the Moon is F = mg with g ≈ 1.6 N kg⁻¹, so a 70 kg astronaut experiences a gravitational force of 70 × 1.6 = 112 N, compared with 70 × 9.8 = 686 N on Earth. The astronaut's mass stays the same; only the local value of g changes.
How do you calculate g at Earth's surface?
g at Earth's surface comes from g = GM/r² with M = 5.97 × 10²⁴ kg and r = 6.37 × 10⁶ m. Here r² = 4.058 × 10¹³ m² and GM = 3.982 × 10¹⁴ N m² kg⁻¹, so g = 3.982 × 10¹⁴ ÷ 4.058 × 10¹³ ≈ 9.81 N kg⁻¹.
Why plot g against 1/r²?
A graph of g against 1/r² tests the inverse-square relationship: if g ∝ 1/r², the graph is a straight line through the origin with gradient equal to GM. A plain graph of g against r is a curve that falls steeply at first and then flattens.
Why do field lines never cross?
Field lines never cross because each point in space has one unique field direction. The tangent to a gravitational field line gives the direction of the force on a small test mass, so two lines crossing at a point would wrongly give two directions there.
Does g depend on the mass of the object in the field?
Gravitational field strength does not depend on the mass of the object placed in the field: g = GM/r² depends only on the mass M creating the field and the distance r from its centre. g is a property of the field at a point, not of the object experiencing it.
How do you find the gravitational force on a satellite from g?
The gravitational force on a satellite is F = mg, using the field strength at its altitude. An 850 kg satellite where g = 6.40 N kg⁻¹ experiences F = 850 × 6.40 = 5440 N, which is its weight at that altitude and provides the centripetal force for the orbit.