The big idea: Every mass pulls on every other mass. The bigger the masses and the closer they are, the stronger the pull.
A mass also fills the space around it with a gravitational field — a region where another mass feels a force.
The field strength g is how strong that pull is per kilogram, in N kg⁻¹.
[Diagram: phys-field-lines] - Available in full study mode
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.
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:
- 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 (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)
[Diagram: phys-formula-triangle] - Available in full study mode
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.
Worked example — surface gravity of a planet
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.)
Solution
- Start with the given formula:
- Put in the numbers (M = 6.0 × 10²⁴, r = 6.4 × 10⁶):
- Work it out — keep the unit:
Final answer
g = 9.8 N kg⁻¹ — that is Earth's surface gravity (and the same as 9.8 m s⁻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 when you move to a different distance (e.g. three times farther → one ninth), or 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.
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.
IB-style question — (a) field strength three times farther out
The gravitational field strength at a point 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).
Solution
- Start with the given formula — g depends on 1/r²:
- Tripling the distance divides g by 3² = 9:
- Work it out — keep the unit:
Final answer
gfar = 0.90 N kg⁻¹ — nine times smaller, because g is proportional to 1/r².
IB-style question — (b) which mass accelerates faster?
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.
Solution
- The acceleration of a falling mass is the field strength g there:
- g = GM/r² depends only on the planet's mass M and the distance r — not on the falling mass.
- Both balls are at the same point, so g is the same for each.
Final answer
Their accelerations are equal — g (the free-fall acceleration) does not depend on the falling mass.