The Gravitational Field

Each mass reshapes the space around it; other masses feel the shape.

14 minUpdated 21 October 2026gravitational fieldfield and potentialshell theoremfield inside a sphere

Lesson at a glanceShow the key points

Gravitational field E = F/m (11.13), in N/kg; near the Earth it equals g in size and direction.
Field and potential: dV = −E·dr, so Ex = −∂V/∂x etc. (11.17); V(r) = −∫E·dr from the reference point.
Point mass: E = GM/r² toward the mass. Ring on its axis: E = GMr/(a² + r²)^{3/2} (11.20).
Disc on its axis: E = (2GM/a²)(1 − cos θ), where θ is the half-angle the disc subtends (11.21).
Thin shell: GM/r² outside, zero inside. Solid sphere: GM/r² outside, GMr/a³ inside (11.24), maximum at the surface.

How does the Earth pull the Moon across 384 000 km of empty space? The modern answer: it doesn’t reach across. It sets up a gravitational field around itself, and the field acts on the Moon where the Moon is.

A two-step picture

“Action at a distance” has a problem: if the Sun moved, would distant planets feel it instantly? The field picture splits the interaction in two. A mass creates a field throughout the space around it, a physical thing with energy and momentum of its own. Any other mass placed in that field feels a force from the field at its own location.

Definition · Gravitational field intensity

The force per unit mass on a body placed at a point:

\vec E = \vec F/m

(11.13). Unit: N/kg (equivalent to m/s²).

Fields from several sources add as vectors. Near the Earth’s surface the field is just $\vec g$ : the same size and direction as the acceleration due to gravity, though the two are different quantities.

Worked example

Reading a field from a force

A 250 g mass placed at a certain point feels a gravitational force of 1.5 N. What is the gravitational field there?

Field from potential, and back

Moving a mass m by $d\vec r$ , the field does work $m\vec E\cdot d\vec r$ , so the potential energy changes by $-m\vec E\cdot d\vec r$ and the potential by

dV = -\vec E\cdot d\vec r \quad\Rightarrow\quad V(\vec r) = -\int_{\text{ref}}^{\vec r}\vec E\cdot d\vec r \qquad (11.14\text{–}11.16)

Field as the slope of potential(11.17)

E_x = -\frac{\partial V}{\partial x}, \quad E_y = -\frac{\partial V}{\partial y}, \quad E_z = -\frac{\partial V}{\partial z}

the field points downhill on the potential landscape

(∂/∂x is a partial derivative: differentiate with respect to x holding y and z fixed.) Know the field, integrate for the potential; know the potential, differentiate for the field.

Worked example

A field that grows with distance

Inside a uniform solid sphere the field along a line through the centre is E = −Cx (pointing toward the centre). Taking V = V₀ at the centre, find V(x).

Fields of common shapes

Point mass

Point mass(11.18)

E = \frac{GM}{r^2}\ \text{toward the mass}, \qquad \vec E = -\frac{GM}{r^2}\hat e_r

Ring, on its axis

Each bit dm pulls along the line to itself, at distance $z = \sqrt{a^2+r^2}$ . Sideways components cancel round the ring; the components along the axis add, each a fraction $\cos\alpha = r/z$ of the whole:

Ring on its axis(11.20)

E = \frac{GMr}{(a^2 + r^2)^{3/2}}\ \text{toward the centre}

Disc, on its axis

Build the disc from rings of radius x and mass $2Mx\,dx/a^2$ , add their (parallel) fields and substitute $z^2 = r^2 + x^2$ :

E = \frac{2GMr}{a^2}\left[\frac1r - \frac1{\sqrt{r^2+a^2}}\right] = \frac{2GM}{a^2}(1 - \cos\theta) \qquad (11.21)

where θ is the angle a radius of the disc subtends at P.

Interactive

On the axis of a ring or a disc

A ring (or disc) of mass M and radius a, seen edge-on. Drag the probe along the axis. At the centre of a ring the pulls from all sides cancel; far away it looks like a point mass. In between the field peaks, at x = a/√2 for a ring. Field in units of GM/a².

x1.20 a

E (GM/a²)0.315

V (GM/a)-0.640

peak field atx = 0.707 a

Source

Thin shell

Repeating the ring-slicing argument for the field gives $GM/r^2$ outside (as if a point) and exactly zero inside. Inside a shell, the near part of the wall pulls harder per kilogram but there is less of it; the far part pulls more weakly but there is more of it. With a 1/r² law the two balance exactly, in every direction.

Solid sphere

Outside: $GM/r^2$ (11.23). Inside at distance r, shells outside r contribute nothing, and the inner ball of mass $Mr^3/a^3$ acts as a point at the centre:

Inside a uniform solid sphere(11.24)

E = \frac{G}{r^2}\cdot\frac{Mr^3}{a^3} = \frac{GM}{a^3}\,r

grows linearly from zero at the centre to GM/a² at the surface

Interactive

Inside and outside a sphere

Drag the probe in or out (distances in units of the radius a). Outside, a shell and a solid sphere both behave like a point mass at the centre. Inside a shell the potential is constant and the field is zero; inside a solid sphere the field grows in proportion to r.

r0.60 a

E (units of GM/a²)0.000

Source

Show

Worked example

Gravity on Mars

Mars has mass 6.42 × 10²³ kg and radius 3.39 × 10⁶ m. Treating it as spherically symmetric, find the gravitational field at its surface.

What to remember

1 $\vec E = \vec F/m$ (N/kg); fields add as vectors.
2 $dV = -\vec E\cdot d\vec r$ ; $E_x = -\partial V/\partial x$ .
3Point and outside any sphere: GM/r². Ring on axis: GMr/(a² + r²)^{3/2}. Disc: (2GM/a²)(1 − cos θ).
4Inside a shell: zero. Inside a solid sphere: GMr/a³.

Check yourself

An astronaut at the very centre of a hollow spherical space station (ignore other bodies) feels a gravitational field from the station of…

Halfway to the centre of a uniform planet, the field is…

Keep exploring

Hand-picked videos, simulations and readings that explain this idea from a different angle.

WatchGravitational forces and fieldsKhan Academy ReadGravitational fieldWikipedia ReadShell theoremWikipedia ReadGravity of a spherical shellHyperPhysics ReadA hole through the EarthHyperPhysics

PreviousGravitational Potential Next How g Changes

The Gravitational Field

Each mass reshapes the space around it; other masses feel the shape.

14 minUpdated 21 October 2026gravitational fieldfield and potentialshell theoremfield inside a sphere

Lesson at a glanceShow the key points

Gravitational field E = F/m (11.13), in N/kg; near the Earth it equals g in size and direction.
Field and potential: dV = −E·dr, so Ex = −∂V/∂x etc. (11.17); V(r) = −∫E·dr from the reference point.
Point mass: E = GM/r² toward the mass. Ring on its axis: E = GMr/(a² + r²)^{3/2} (11.20).
Disc on its axis: E = (2GM/a²)(1 − cos θ), where θ is the half-angle the disc subtends (11.21).
Thin shell: GM/r² outside, zero inside. Solid sphere: GM/r² outside, GMr/a³ inside (11.24), maximum at the surface.

A two-step picture

Definition · Gravitational field intensity

The force per unit mass on a body placed at a point:

\vec E = \vec F/m

(11.13). Unit: N/kg (equivalent to m/s²).

Worked example

Reading a field from a force

A 250 g mass placed at a certain point feels a gravitational force of 1.5 N. What is the gravitational field there?

Field from potential, and back

Moving a mass m by $d\vec r$ , the field does work $m\vec E\cdot d\vec r$ , so the potential energy changes by $-m\vec E\cdot d\vec r$ and the potential by

dV = -\vec E\cdot d\vec r \quad\Rightarrow\quad V(\vec r) = -\int_{\text{ref}}^{\vec r}\vec E\cdot d\vec r \qquad (11.14\text{–}11.16)

Field as the slope of potential(11.17)

E_x = -\frac{\partial V}{\partial x}, \quad E_y = -\frac{\partial V}{\partial y}, \quad E_z = -\frac{\partial V}{\partial z}

the field points downhill on the potential landscape

(∂/∂x is a partial derivative: differentiate with respect to x holding y and z fixed.) Know the field, integrate for the potential; know the potential, differentiate for the field.

Worked example

A field that grows with distance

Inside a uniform solid sphere the field along a line through the centre is E = −Cx (pointing toward the centre). Taking V = V₀ at the centre, find V(x).

Fields of common shapes

Point mass

Point mass(11.18)

E = \frac{GM}{r^2}\ \text{toward the mass}, \qquad \vec E = -\frac{GM}{r^2}\hat e_r

Ring, on its axis

Ring on its axis(11.20)

E = \frac{GMr}{(a^2 + r^2)^{3/2}}\ \text{toward the centre}

Disc, on its axis

Build the disc from rings of radius x and mass $2Mx\,dx/a^2$ , add their (parallel) fields and substitute $z^2 = r^2 + x^2$ :

E = \frac{2GMr}{a^2}\left[\frac1r - \frac1{\sqrt{r^2+a^2}}\right] = \frac{2GM}{a^2}(1 - \cos\theta) \qquad (11.21)

where θ is the angle a radius of the disc subtends at P.

Interactive

On the axis of a ring or a disc

x1.20 a

E (GM/a²)0.315

V (GM/a)-0.640

peak field atx = 0.707 a

Source

Thin shell

Solid sphere

Outside: $GM/r^2$ (11.23). Inside at distance r, shells outside r contribute nothing, and the inner ball of mass $Mr^3/a^3$ acts as a point at the centre:

Inside a uniform solid sphere(11.24)

E = \frac{G}{r^2}\cdot\frac{Mr^3}{a^3} = \frac{GM}{a^3}\,r

grows linearly from zero at the centre to GM/a² at the surface

Interactive

Inside and outside a sphere

r0.60 a

E (units of GM/a²)0.000

Source

Show

Worked example

Gravity on Mars

Mars has mass 6.42 × 10²³ kg and radius 3.39 × 10⁶ m. Treating it as spherically symmetric, find the gravitational field at its surface.

What to remember

1 $\vec E = \vec F/m$ (N/kg); fields add as vectors.
2 $dV = -\vec E\cdot d\vec r$ ; $E_x = -\partial V/\partial x$ .
3Point and outside any sphere: GM/r². Ring on axis: GMr/(a² + r²)^{3/2}. Disc: (2GM/a²)(1 − cos θ).
4Inside a shell: zero. Inside a solid sphere: GMr/a³.

Check yourself

An astronaut at the very centre of a hollow spherical space station (ignore other bodies) feels a gravitational field from the station of…

Halfway to the centre of a uniform planet, the field is…

Keep exploring

Hand-picked videos, simulations and readings that explain this idea from a different angle.

WatchGravitational forces and fieldsKhan Academy ReadGravitational fieldWikipedia ReadShell theoremWikipedia ReadGravity of a spherical shellHyperPhysics ReadA hole through the EarthHyperPhysics

PreviousGravitational Potential Next How g Changes

A two-step picture#

Field from potential, and back#

Fields of common shapes#

Point mass

Ring, on its axis

Disc, on its axis

Thin shell

Solid sphere

Check yourself

Keep exploring

A two-step picture#

Field from potential, and back#

Fields of common shapes#

Point mass

Ring, on its axis

Disc, on its axis

Thin shell

Solid sphere

Check yourself

Keep exploring

A two-step picture

Field from potential, and back

Fields of common shapes

A two-step picture

Field from potential, and back

Fields of common shapes