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c059741
NotesPhysics HLTopic 4.1Newton's law of gravitation and field strength
Back to Physics HL Topics
4.1.12 min read

Newton's law of gravitation and field strength (Physics HL)

IB Physics • Unit 4

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Contents

  • Gravity as a field
  • The two given equations
  • Exam-style question
The big idea: Let go of your phone and it drops straight down — the Earth is pulling on it, as it pulls on every mass nearby. That invisible pull filling the space around a mass is a gravitational field.

The field strength g measures how strong the pull is per kilogram (N kg⁻¹) — stronger for bigger masses and closer in.

A gravitational field points inward, towards the mass M, because gravity always pulls. The lines get further apart as you move out — the field is weaker further away.

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Spot it: Gravity is always attractive — the field lines point inward, towards the mass.

The lines spread out as you move away, so the field gets weaker the further out you go.

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Newton's law of gravitation gives the pull between any two masses. The force grows with the masses and shrinks with the square of the distance between them:

Newton's law of gravitation (in the data booklet). Double the distance and the force drops to a quarter.
gravitational force between the masses (N)
gravitational constant, 6.67 × 10⁻¹¹ N m² kg⁻²
the two masses (kg)
distance between their centres (m)

The gravitational field strength g is the force per kilogram on a small mass placed in the field. Dividing Newton's law by that small mass m gives a neat form that only needs the big mass M and the distance:

Gravitational field strength (in the data booklet). It equals the free-fall acceleration, so its unit N kg⁻¹ is the same as m s⁻².
gravitational field strength (N kg⁻¹), also the free-fall acceleration
gravitational force on the small mass (N)
the small mass placed in the field (kg)
mass of the planet or star making the field (kg)
distance from the centre of M (m)

F = m g. Cover the one you want: m and g side by side → multiply (F = m g); F above g → divide (m = F ÷ g); F above m → divide (g = F ÷ m).

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g is also the acceleration of free fall: Because F = mg and F = ma, the field strength g equals the acceleration a falling mass would have.

That is why two different masses dropped at the same place fall with the same acceleration — g does not depend on the falling mass m.
IB-style questionCalculate[2 marks]

A planet has mass 6.0 × 10²⁴ kg and radius 6.4 × 10⁶ m. Find the gravitational field strength at its surface. (G = 6.67 × 10⁻¹¹ N m² kg⁻².)

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How this is tested — gravitation and field strength turn up across the Fields theme:

Paper 1A

  • Quick inverse-square reasoning — how g changes at a different distance (three times farther → one ninth).
  • Comparing the accelerations of masses falling from different heights.

Paper 2

  • Calculate g = GM/r² at a planet's surface.
  • Or a star's field at an orbital distance.
The classic trap: Forgetting the square — moving three times farther divides g by 3² = 9, not by 3.
The inverse-square shortcut: Because g is proportional to 1/r², you don't always need G and M.

If the distance is multiplied by a number n, the field strength is divided by n². So r ×2 → g ÷4, and r ×3 → g ÷9.

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IB-style questionDetermine[2 marks]

The gravitational field strength a distance r from the centre of a planet is 8.1 N kg⁻¹. Find the field strength at a point three times as far from the centre (a distance 3r).

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IB-style questionCompare[2 marks]

Two small balls, one of mass 2.0 kg and one of mass 5.0 kg, are released from rest at the same point above the planet's surface. Compare their initial accelerations.

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gravitational field strength at a point, and its SI unit. [2 marks]

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4.1.2Kepler's laws and orbital motion
4.1.3Circular orbits and satellites
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4.1.5Gravitational potential energy and potential (HL)
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