Gravitational Field Strength

Spec mapping: OCR H556 Module 5.4 — Gravitational fields (the gravitational field as a region in which a mass experiences a force; gravitational field strength $g$g as force per unit mass; representation of fields using field lines and equipotentials; uniform fields near the Earth's surface; radial fields around spherical bodies). Refer to the official OCR H556 specification document for exact wording.

Every body that has mass surrounds itself with a gravitational field — a region of space in which a second mass will experience an attractive force. The field is invisible, but its effects are intimately familiar: dropped books fall, the Moon stays bound to the Earth, the Earth orbits the Sun, and every star in the Milky Way is gravitationally bound to its galactic centre. The first job of A-Level gravitation is to put a number on the strength of that field at any given point. That number is $g$g, the gravitational field strength.

This lesson opens OCR Module 5.4. We define $g$g as a force per unit mass, distinguish the two physically distinct contexts in which $g$g appears (the definition $g = F/m$g=F/m, valid for any field, and the spherical-mass result $g = GM/r^{2}$g=GM/r2, derived next lesson), draw uniform and radial field-line diagrams, evaluate $g$g at familiar locations, and apply the field concept to several specimen calculations. The lesson lands two essential A* discriminators: the unit equivalence $\text{N kg}^{-1} \equiv \text{m s}^{-2}$N kg−1≡m s−2 (and why the first form is conceptually preferable) and the vector superposition of fields from multiple sources.

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The Field Concept — Why It Replaced "Action at a Distance"

Eighteenth-century physics following Newton spoke of gravity as action at a distance: two distant masses simply pulled on one another across empty space, with no intermediate carrier or medium. The mathematics worked, but the physical picture was uncomfortable. How does the Sun "know" that the Earth is $1.5 \times 10^{11}$1.5×1011 m away?

In the nineteenth century, Michael Faraday introduced the field concept while wrestling with the analogous problem in electromagnetism. The picture: each mass (or charge) modifies the space around it, creating a vector quantity defined at every point; a second mass placed at a point feels a force because of the local field there, not because of any direct line to the distant source. The field becomes the carrier of the interaction.

For A-Level gravity the field picture and the action-at-a-distance picture are mathematically equivalent — both give $F = GMm/r^{2}$F=GMm/r2 — but the field picture extends much more naturally to time-varying situations (gravitational waves, general relativity) and to multi-source problems (where the net field is just the vector sum of contributions). Throughout Module 5.4 you should think of any mass as being immersed in a field, not as being pulled by a distant body.

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Definition of Gravitational Field Strength

The gravitational field strength at a point is defined as the gravitational force per unit mass acting on a small test mass placed at that point:

$g = \frac{F}{m}$g=mF​

Symbol: $g$g (with vector form $\vec{g}$g​ when direction matters). SI unit: newtons per kilogram, $\text{N kg}^{-1}$N kg−1, which is dimensionally identical to $\text{m s}^{-2}$m s−2 because $F/m = ma/m = a$F/m=ma/m=a.

Gravitational field strength is a vector: it has the same direction as the force on a positive test mass — which, since all masses are positive and gravity is attractive, means towards the source.

There are two distinct equations called "$g$g" at A-Level:

1. $g = F/m$g=F/m — the definition of the field. This is a measurement instruction (place a test mass, measure the force, divide); it is true in any gravitational field whatever the source.
2. $g = GM/r^{2}$g=GM/r2 — the spherical-mass field result. This is a consequence of Newton's inverse-square law applied to a spherically symmetric body of mass $M$M, evaluated outside the body. It is derived in the next lesson.

Confusing these two — for example, writing $g = GM/r^{2}$g=GM/r2 inside an asteroid where the geometry is not spherical, or applying $g = F/m$g=F/m to "deduce" Newton's law — costs marks routinely. Always know which one you are using and why.

Does the Test Mass Matter?

In principle, yes: a "test" mass should be small enough that it does not perturb the source. In practice for A-Level, $F$F on the test mass is proportional to $m$m (because gravity is always proportional to the mass of the body it acts on), so $F/m$F/m is independent of $m$m — the field is a property of the source, not the probe.

This proportionality is the weak equivalence principle and it is the reason Galileo's leaning-tower argument works: a heavy and a light body fall at the same rate because $a = F/m = g$a=F/m=g depends only on the source field, not on the falling body.

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Field-Line Representation

A convenient pictorial tool is the gravitational field-line diagram. Two rules:

1. The field $\vec{g}$g​ is tangent to the field line at each point, in the direction of the arrow.
2. The density of field lines (lines per unit area perpendicular to the field) is proportional to $|\vec{g}|$∣g​∣.

Where the lines are tightly packed, the field is strong; where they are sparse, the field is weak. Gravitational field lines always point towards the source (gravity is attractive); they have no "starting" point in space (they begin at infinity) and "end" on the source mass.

Uniform Fields

Near the Earth's surface, on scales much smaller than the Earth's radius ($R_{E} = 6.37 \times 10^{6}$RE​=6.37×106 m), the gravitational field is to excellent approximation uniform: the magnitude is essentially constant at $g \approx 9.81 \text{ N kg}^{-1}$g≈9.81 N kg−1 and the direction is vertically downward.

Field lines are parallel, equally spaced, and pointing straight down — the geometric signature of a uniform field. This is why the GCSE formula $E_{p} = mgh$Ep​=mgh works as long as the height $h$h is small compared with the Earth's radius: in that regime $g$g is essentially constant and the work done lifting mass $m$m through height $h$h is just (force)(distance) $= mgh$=mgh. We will see in Lesson 9 that the exact gravitational potential energy is $E_{p} = -GMm/r$Ep​=−GMm/r, and that $mgh$mgh is its first-order Taylor expansion about $r = R_{E}$r=RE​.

Radial Fields

On the larger scale of an isolated planet or star, the gravitational field is radial: the field lines all point inward, converging on the centre of mass.

Because the field lines converge as they approach the source, the line density rises and so does $|\vec{g}|$∣g​∣. Far from the planet they are widely spaced and the field is weak. The functional form of that fall-off is $g \propto 1/r^{2}$g∝1/r2 (next lesson), an immediate consequence of the geometry: the total "gravitational flux" $4\pi r^{2} g$4πr2g through any spherical shell is constant for a point mass, so $g \propto 1/r^{2}$g∝1/r2.

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$g$g Near the Earth's Surface

The conventional A-Level value is $g = 9.81 \text{ N kg}^{-1}$g=9.81 N kg−1, and OCR examiners are happy with $9.8$9.8 or even $10$10 when the question warrants. What is less obvious is that the local value of $g$g varies slightly across the Earth's surface:

Location	$g$g / N kg⁻¹
Sea level, equator	9.78
Sea level, geographic poles	9.83
Summit of Mount Everest (8.8 km)	9.77
Aircraft cruising altitude (11 km)	9.77
Low-Earth orbit (400 km, ISS)	8.69
Surface of the Moon	1.62
Surface of Mars	3.71
Surface of Jupiter (cloud tops)	24.8

Three causes of variation at the Earth's surface:

1. Oblateness. The Earth is not a perfect sphere — it bulges at the equator by about 21 km in radius, so the equatorial surface is further from the centre than the polar surface. By $g = GM/r^{2}$g=GM/r2, equatorial $g$g is therefore slightly smaller.
2. Rotation. The Earth spins once per sidereal day. At the equator a co-rotating observer requires a small centripetal force ($v^{2}/r \approx 0.034 \text{ m s}^{-2}$v2/r≈0.034 m s−2) to keep them on the rotating surface, supplied by part of gravity. The measured $g$g (gravity minus the centripetal contribution) is therefore a little smaller at the equator.
3. Local density. Mountains of dense rock or large mineral deposits add a small gravitational pull; gravity surveys exploit this to map the subsurface. Variations are at the part-per-thousand level.

For all A-Level calculations you should take $g_{s} = 9.81 \text{ N kg}^{-1}$gs​=9.81 N kg−1 at the Earth's surface unless told otherwise.

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Worked Example 1 — Weight from Field Strength

A 72 kg astronaut steps out of a spacecraft. Find her weight (a) on the Earth's surface, (b) on the surface of the Moon, (c) at the altitude of the International Space Station (ISS), at 400 km.

(a) Earth's surface. $g = 9.81 \text{ N kg}^{-1}$g=9.81 N kg−1.

$W_{\text{Earth}} = mg = 72 \times 9.81 = 706 \text{ N}$WEarth​=mg=72×9.81=706 N

(b) Moon's surface. From the table, $g_{\text{Moon}} = 1.62 \text{ N kg}^{-1}$gMoon​=1.62 N kg−1.

$W_{\text{Moon}} = 72 \times 1.62 = 117 \text{ N}$WMoon​=72×1.62=117 N

Her mass is still 72 kg on the Moon; only the weight has changed.

(c) ISS altitude. $g_{\text{ISS}} = 8.69 \text{ N kg}^{-1}$gISS​=8.69 N kg−1 from the table.

$W_{\text{ISS}} = 72 \times 8.69 = 626 \text{ N}$WISS​=72×8.69=626 N

Notice that the ISS is not in zero gravity — the field there is only 11% smaller than at the surface. Astronauts on the ISS feel weightless because they are in free fall together with the station, not because gravity is absent (Lesson 11 returns to this).

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Worked Example 2 — $g$g from a Free-Fall Experiment

A small steel ball is dropped from rest in a vacuum tube and falls 1.225 m, breaking a light gate at the bottom 0.500 s after release. Use this data to estimate $g$g at the laboratory.

Using $s = ut + \tfrac{1}{2}at^{2}$s=ut+21​at2 with $u = 0$u=0:

$g = \frac{2s}{t^{2}} = \frac{2 \times 1.225}{(0.500)^{2}} = \frac{2.450}{0.250} = 9.80 \text{ m s}^{-2} = 9.80 \text{ N kg}^{-1}$g=t22s​=(0.500)22×1.225​=0.2502.450​=9.80 m s−2=9.80 N kg−1

The two units are dimensionally identical: $\text{N kg}^{-1} = \text{kg m s}^{-2} / \text{kg} = \text{m s}^{-2}$N kg−1=kg m s−2/kg=m s−2. In gravitational-field contexts the $\text{N kg}^{-1}$N kg−1 form is preferred because it emphasises that $g$g is a force per unit mass; in kinematics contexts the $\text{m s}^{-2}$m s−2 form emphasises that $g$g is also the acceleration of free fall. They are the same number with two different labels.

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Worked Example 3 — A Lander on an Asteroid

A 250 kg lander hovers 100 m above the surface of a small asteroid where the gravitational field strength is $g = 0.025 \text{ N kg}^{-1}$g=0.025 N kg−1, taken as approximately uniform over the lander's working range. What constant upward thrust must its rocket engine supply?

The downward weight is

$W = mg = 250 \times 0.025 = 6.25 \text{ N}$W=mg=250×0.025=6.25 N

For hover (Newton's first law, zero acceleration) the thrust must exactly balance the weight: $F_{\text{thrust}} = 6.25$Fthrust​=6.25 N upward.

The small numerical answer is physically reasonable — a lander on a small asteroid genuinely needs only a fraction of the engine output it would need on the Moon, let alone Earth. This is why asteroid-hopping missions (Hayabusa, OSIRIS-REx) can manoeuvre with tiny cold-gas thrusters.

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Worked Example 4 — Vector Superposition of Two Fields

Two point masses sit 0.30 m apart: $M_{1} = 2.0 \times 10^{6}$M1​=2.0×106 kg on the left, $M_{2} = 5.0 \times 10^{5}$M2​=5.0×105 kg on the right. Find the magnitude and direction of the gravitational field at a point P that is 0.10 m to the right of $M_{1}$M1​ (so 0.20 m to the left of $M_{2}$M2​). Use $G = 6.67 \times 10^{-11} \text{ N m}^{2} \text{ kg}^{-2}$G=6.67×10−11 N m2 kg−2.

Field from $M_{1}$M1​ at P, directed leftward (towards $M_{1}$M1​):

$g_{1} = \frac{GM_{1}}{r_{1}^{2}} = \frac{6.67 \times 10^{-11} \times 2.0 \times 10^{6}}{(0.10)^{2}} = \frac{1.334 \times 10^{-4}}{0.010} = 1.334 \times 10^{-2} \text{ N kg}^{-1}$g1​=r12​GM1​​=(0.10)26.67×10−11×2.0×106​=0.0101.334×10−4​=1.334×10−2 N kg−1

Field from $M_{2}$M2​ at P, directed rightward (towards $M_{2}$M2​):

$g_{2} = \frac{GM_{2}}{r_{2}^{2}} = \frac{6.67 \times 10^{-11} \times 5.0 \times 10^{5}}{(0.20)^{2}} = \frac{3.335 \times 10^{-5}}{0.040} = 8.34 \times 10^{-4} \text{ N kg}^{-1}$g2​=r22​GM2​​=(0.20)26.67×10−11×5.0×105​=0.0403.335×10−5​=8.34×10−4 N kg−1

Taking rightward as positive:

$g_{\text{net}} = -g_{1} + g_{2} = -1.334 \times 10^{-2} + 8.34 \times 10^{-4} = -1.25 \times 10^{-2} \text{ N kg}^{-1}$gnet​=−g1​+g2​=−1.334×10−2+8.34×10−4=−1.25×10−2 N kg−1

The negative sign tells us $\vec{g}_{\text{net}}$g​net​ points to the left, towards $M_{1}$M1​. Magnitude $1.25 \times 10^{-2} \text{ N kg}^{-1}$1.25×10−2 N kg−1.

Three lessons from this example: (i) gravitational fields from multiple sources add as vectors, with direction tracked carefully; (ii) the closer and more massive source dominates; (iii) at no finite point between two attractive masses does the net field point along the line of the smaller mass — that would require negative mass, which does not exist in classical physics.

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Worked Example 5 — Decision Tree for Choosing the Right Formula

Which form of $g$g do you reach for in a given problem? The decision tree below summarises the workflow used throughout Module 5.4.

graph TD
    A["Need g at a point"] --> B{"Spherical body<br/>outside surface?"}
    B -- "Yes" --> C["Use g = GM/r²<br/>(spherical-mass result)"]
    B -- "No" --> D{"Near Earth surface,<br/>h is small vs R_E?"}
    D -- "Yes" --> E["Use g ≈ 9.81 N/kg<br/>(uniform field)"]
    D -- "No" --> F{"Multiple sources?"}
    F -- "Yes" --> G["Vector-sum each contribution<br/>g_i = GM_i/r_i²"]
    F -- "No, exotic geometry" --> H["Use g = F/m definition<br/>with measured F"]
    C --> I["Check r is from centre<br/>of mass, not altitude"]
    G --> I

The most common slip is using $g = GM/r^{2}$g=GM/r2 with $r$r as the altitude above the surface rather than the distance from the centre. Always write $r = R + h$r=R+h explicitly when working with planetary altitudes.

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Scalar vs Vector Addition of Fields

The example above generalises: at any point P where multiple sources contribute, the net gravitational field is the vector sum

$\vec{g}_{\text{net}} = \sum_{i} \vec{g}_{i} = \sum_{i} \frac{GM_{i}}{r_{i}^{2}} \, \hat{r}_{i}$g​net​=∑i​g​i​=∑i​ri2​GMi​​r^i​

where $\hat{r}_{i}$r^i​ is the unit vector from P to the i-th source (so $\vec{g}_{i}$g​i​ points towards source $i$i). Two consequences:

• Adding magnitudes is wrong unless all sources are collinear and pull in the same direction. Two equal masses on opposite sides of P give zero net field, not double field.
• Symmetry is your friend. If two equal masses sit symmetrically about a point P, the field along the symmetry axis cancels by pairs and only the axial component remains. This trick is heavily used in continuous-distribution problems (rings, discs, shells) at undergraduate level.

The Lagrange point L1 between the Earth and the Sun (about 1.5 million km from Earth) is the locus where the combined gravitational field of Sun and Earth, plus the centrifugal field of the rotating frame, all sum to zero — making it the location of choice for solar-observation spacecraft (SOHO, DSCOVR; JWST sits at L2).

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Conceptual Question — Two Symbols, One Letter

The single symbol $g$g does double duty in mechanics:

• In kinematics ("acceleration of free fall"), $g$g has units $\text{m s}^{-2}$m s−2 and represents the acceleration of a body in free fall.
• In field theory ("gravitational field strength"), $g$g has units $\text{N kg}^{-1}$N kg−1 and represents the gravitational force per unit mass.