9478 / 2027
Gravitational Fields overview

Lesson 1 of 5

Mass, force and field strength

  • I can recall and use Newtonian gravitation, F = Gm1m2/r^2.Syllabus

    Syllabus 9478, 8(a). Recall and use Newtonian gravitation, F = Gm1m2/r^2.

  • I can derive point-mass gravitational field strength from Newtonian gravitation and force per unit mass.Syllabus

    Syllabus 9478, 8(b). Derive point-mass gravitational field strength from Newtonian gravitation and force per unit mass.

  • I can recall and use g = GM/r^2 to solve point-mass gravitational-field problems.Syllabus

    Syllabus 9478, 8(c). Recall and use g = GM/r^2 to solve point-mass gravitational-field problems.

  • I can explain approximately constant near-Earth field strength and its equality to free-fall acceleration.Syllabus

    Syllabus 9478, 8(d). Explain approximately constant near-Earth field strength and its equality to free-fall acceleration.

Make a guess. You are not marked.

A satellite is moved from a distance r to a distance 2r from the centre of the Earth. What happens to the gravitational field strength it experiences?

  1. It halves, because the satellite is now twice as far from Earth's centre.
  2. It falls to one quarter, because the field follows an inverse square law.
  3. It stays the same, because the satellite is in orbit.
Show the answer

It falls to one quarter, because the field follows an inverse square law.

Field strength follows an inverse square law, g = GM / r2. Doubling r multiplies r2 by four, so g is divided by four.

Gravitational force depends on both interacting masses. Field strength describes the force per unit test mass at a position, so it depends on the source and the position rather than the small test mass you choose.

Newtonian attraction

Two point masses M and m attract each other with force magnitude:

F = GMm/r2

G is the universal gravitational constant, in N m2 kg-2. Here M names the source mass and m the small test mass, both in kg. The distance r joins the masses, in metres. The force acts along that joining line, towards the other mass.

The forces on the two bodies are equal and opposite, but act on different bodies. They do not cancel in the equation for either body. Treating a much more massive source as fixed neglects its very small motion, not its interaction with the test mass.

For a point outside a spherically symmetric body, use the body's mass as concentrated at its centre. At altitude h above a surface of radius R, the required separation is r = R + h. This rule applies outside the body; it does not give the field at points inside it.

At fixed masses, doubling r quarters the force. At fixed r, doubling either mass doubles the force on each partner.

Derive the point-mass field

Gravitational field strength is gravitational force per unit test mass at a point. Dividing the force law by m gives:

g = F/m = (GMm/r2)/m
g = GM/r2

Here g is the magnitude, in N/kg. The field points towards the source centre. The small test mass cancels: a heavier test body has more gravitational force, but the same force per unit mass at that position.

If outward radial direction is positive, the signed component is gr = -GM/r2. Keep the positive magnitude g separate from the negative outward-axis component gr. A minus sign in that component means inward, not repulsion.

Estimate the field scale

Using supplied rough scales GM about 4 × 1014 m3/s2 and surface radius R about 6 × 106 m, the ratio GM/R2 is of order 10 N/kg. These rounded scales give a plausibility check before the more precise model calculation.

Worked spherical-body model

Use centre distance, not altitude

For the following examples, use the supplied model product GM = 4.0 × 1014 m3/s2 and body radius R = 6.4 × 106 m. The source is spherical, its centre is treated as fixed, and other bodies' gravity is neglected. GM is a supplied product of G and source mass, not a separate universal constant.

At the surface, r = R:

g(R) = (4.0 × 1014)/(6.4 × 106)2
= 9.765625 N/kg ≈ 9.8 N/kg

At centre distance r = 8.0 × 106 m, the altitude is:

h = r - R = 1.6 × 106 m
g = (4.0 × 1014)/(8.0 × 106)2
= 6.25 N/kg

A 400 kg body there experiences gravitational force F = mg = 2500 N inward. Substituting the altitude 1.6 × 106 m for r would give the wrong field and force.

A radial field can look nearly uniform in a small region

Exterior field of a spherically symmetric source

External gravitational field lines point towards the source centreA spherical source is drawn as a circle centred on O. Eight representative external radial field lines extend to its surface, with all green arrows directed inward. The lines do not cross in the exterior region. The source interior is masked and has no field construction. A marked small surface region near the top is the location for the following local view. This is a schematic field pattern, not a numerical field-strength scale or a drawing of field inside the body.MCentre OSmall regionField direction is inward.

Enlarged local view near the surface

Nearby field lines are almost parallel towards a distant centreThe surface is a shallow circular arc with its centre far below the displayed patch. Five green field lines are straight radial lines towards that same distant centre; their downward arrows converge slightly rather than being exactly parallel. All arrows remain outside the surface. The surface curvature and small directional changes explain why a small region and a height change much less than the source radius can be treated as approximately uniform. This panel is an independently enlarged schematic local view, not on the same distance scale as the whole source. Arrow lengths are not numerical magnitudes.Almost parallel, almost unchangedSurfaceSource centre is far below this view.

Both direction and magnitude change little over a sufficiently small near-surface region. The exact magnitude ratio is g(R + h)/g(R) = [R/(R + h)]2; the constant-g approximation does not extend to infinity.

The spatial field model uses distance from the source centre and inward field directions. A small surface region can be treated as nearly uniform when changes in distance and direction are small.

Extra digits retained in the examples allow comparisons without repeated rounding. They describe the supplied model, not that precision in measured planetary constants.

Try it step by step

Work it out in steps: Distance from the centre of the Earth r in km, then Field strength g = GM / r2 in N/kg.

  1. Problem 1. Find the gravitational field strength at an altitude of 400 km above the Earth. Take G = 6.67 x 10-11 N m2/kg2, Earth's mass M = 5.97 x 1024 kg and radius R = 6370 km.
  2. Problem 2. Find the gravitational field strength at an altitude of 2000 km. Use the same G, M and R = 6370 km.
  3. Problem 3. Find the gravitational field strength at an altitude of three Earth radii above the surface. Use the same G, M and R = 6370 km.
Show the answer

Problem 1.

Distance from the centre of the Earth r in km: r = 6370 + 400 = 6770 km = 6.77 x 106 m

Field strength g = GM / r2 in N/kg: g = 6.67 x 10-11 x 5.97 x 1024 / (6.77 x 106)2 = 8.69 N/kg (3 sf)

Problem 2.

Distance from the centre of the Earth r in km: r = 6370 + 2000 = 8370 km = 8.37 x 106 m

Field strength g = GM / r2 in N/kg: g = 3.98 x 1014 / (8.37 x 106)2 = 5.68 N/kg (3 sf)

Problem 3.

Distance from the centre of the Earth r in km: r = 4R = 4 x 6370 = 25480 km = 2.548 x 107 m

Field strength g = GM / r2 in N/kg: g = 3.98 x 1014 / (2.548 x 107)2 = 0.613 N/kg (3 sf)

Check your understanding

A spherical body has GM = 4.0 x 1014 m3/s2 and radius R = 6.4 x 106 m. A small object is at altitude h = 1.6 x 106 m. What is the gravitational field-strength magnitude there?

  1. 156.25 N/kg
  2. 9.765625 N/kg at every altitude
  3. 6.25 N/kg
Show the answer

6.25 N/kg

Use r = 8.0 x 106 m: g = GM/r2 = 4.0 x 1014/(8.0 x 106)2 = 6.25 N/kg, directed inward.

Why a near-surface field can be approximately uniform

At height h, the exact ratio in the spherical model is:

g(R + h)/g(R) = [R/(R + h)]2

When the height change is much smaller than R, the radius barely changes. Within a small surface region, the directions towards the distant centre are also nearly parallel. These are the conditions behind a locally uniform downward field and the approximation ΔEp = mgΔh with nearly constant g.

The approximation does not extend to all altitudes. In the worked example, r = 1.25R, so g = g(R)/1.252 = 0.64g(R). The field has changed substantially, and constant surface g would be unsuitable.

Field strength and free-fall acceleration

If gravity is the only force on a body, Newton's second law gives mg = ma, so its acceleration has magnitude g and points inward. The units are equivalent:

1 N/kg = 1 m/s2

Field strength is defined through force per unit mass; acceleration is defined through velocity change per unit time. They are equal for free fall under gravity alone. Support or drag can change the resultant acceleration without removing the gravitational field.

An astronaut and spacecraft can fall together with little or no usual support force between them. That apparent weightlessness is compatible with the substantial 6.25 N/kg field at the quoted orbital radius; it is not evidence of zero gravity.