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.

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:

$g = \frac{F}{m}$

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.

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):

$g_h \approx g_0 \left(1 - \frac{2h}{R}\right)$

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

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

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

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:

$g = \frac{GM}{r^2}$

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.

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:

$\frac{GM_E}{x^2} = \frac{GM_M}{(d - x)^2}$

$\frac{M_E}{x^2} = \frac{M_M}{(d - x)^2}$

$\frac{d - x}{x} = \sqrt{\frac{M_M}{M_E}} = \sqrt{\frac{7.35 \times 10^{22}}{5.97 \times 10^{24}}} = 0.1109$

$d - x = 0.1109x$

$d = 1.1109x$

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

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

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.

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:

$g = \frac{GM}{R^2} = \frac{6.674 \times 10^{-11} \times 5.97 \times 10^{24}}{(6.37 \times 10^6)^2} = \frac{3.984 \times 10^{14}}{4.058 \times 10^{13}} = 9.81 \text{ N kg}^{-1}$

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.

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