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.

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.

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.

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.

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.

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.

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.