In short

The gravitational field \vec{g} at a point is the force per unit mass that a test mass placed there would feel: \vec{g} = \vec{F}/m. For a point mass or a spherical body, |\vec{g}| = GM/r^2 at distance r from its centre. At Earth's surface, g_0 = GM_E/R_E^2 \approx 9.8 m/s². Above the surface at altitude h, g = g_0 R_E^2/(R_E + h)^2; below the surface at depth d, g = g_0(1 - d/R_E). The small additional variation with latitude comes from Earth's rotation and its slightly flattened shape.

Drop a steel bearing from your balcony. It falls. Drop another one from the roof of a tall building in Mumbai — it falls in exactly the same way, picking up 9.8 m/s of speed for every second of fall. Take the same bearing to the top of Mount Everest and drop it — it still falls, but now it picks up speed a tiny bit more slowly, something like 9.77 m/s every second. Carry it down into the deepest coal mine in Jharia and try again — again slightly slower than at the surface, but for a completely different reason.

Gravity is not quite the same everywhere. Most of the time you can treat g as the constant 9.8 m/s² and nothing goes wrong; that approximation gets you through almost every introductory mechanics problem. But a careful measurement shows g varies with where you are — with how high up you are, how deep below the ground, your latitude, and the rocks under your feet. A geologist searching for iron ore can find it because the ore's extra mass makes g measurably larger right above it. A satellite's orbit drifts slightly because g is a hair stronger over the equatorial bulge.

To make sense of all this, you need a better mental model than "g is a number". You need a field — a quantity defined at every point in space — and a set of rules for how a mass reshapes that field. This article builds both.

From force to field

You already know Newton's law of universal gravitation: every mass M attracts every other mass m with a force

F = \frac{GMm}{r^2}

along the line connecting them, where r is the centre-to-centre distance and G = 6.674 \times 10^{-11} N·m²/kg².

This is a force law — it requires two masses before it says anything. But every mechanics problem in which a falling object is treated as "pulled down by mg" is tacitly doing something different. It is saying: the Earth has already set up gravity, independently of whatever test mass we drop into it. The test mass only responds to the gravity that is there. Before the test mass arrives, something is already filling the space.

That something is the gravitational field. It is the pull-per-unit-mass that any test mass would feel if placed there. If you bring a mass m to a point where the field is \vec{g}, it feels force \vec{F} = m\vec{g} — and that is why "weight equals mass times g" is a true equation, not a coincidence.

Gravitational field

The gravitational field at a point is the gravitational force per unit mass experienced by a small test mass placed at that point:

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

The SI unit is N/kg, which is identical to m/s² (the acceleration a test mass would have if the only force on it were gravity).

Reading the definition. The field \vec{g} is a vector — it has a magnitude (how strong the pull is) and a direction (which way the pull points). At every point in space, the field has some value; you can picture it as a sea of tiny arrows, one at each location, telling any mass that visits that location how it will be pushed.

The field is defined without reference to the test mass. You take the limit of small m to avoid the test mass itself noticeably disturbing the distribution of masses you are measuring — a formality, but the principle is that the field is a property of the source masses alone. Place a 1 kg test mass at a point, measure the force, divide by 1; place a 3 kg test mass at the same point, measure the force, divide by 3 — you get the same \vec{g} both times.

The two units, N/kg and m/s², are numerically equal because Newton's second law says F = ma, so F/m = a. The N/kg units remind you that \vec{g} is a field (force per mass). The m/s² units remind you that a freely falling test mass accelerates at exactly \vec{g}. Both readings are correct; each is useful in a different context.

Field of a point mass

Take a single point mass M sitting at the origin. By Newton's law, a test mass m at position \vec{r} from the point mass feels a force directed toward the point mass:

\vec{F} = -\frac{GMm}{r^2}\,\hat{r}

Why: the unit vector \hat{r} points from the source M outward to the test mass. The minus sign then flips the force inward — gravity always attracts.

Divide by m:

\vec{g} = -\frac{GM}{r^2}\,\hat{r}

The magnitude is

|\vec{g}| = \frac{GM}{r^2}

and the direction points radially inward toward M. This is the cleanest statement of the inverse-square law: the gravitational field of a point mass has magnitude GM/r^2, pointing toward the mass.

Gravitational field around a point mass A central mass M with radial arrows pointing inward from all around. Arrows closer to the mass are longer (stronger field); arrows farther out are shorter (weaker field). M weaker (farther) stronger (closer)
Gravitational field around a point mass $M$. Every arrow points radially toward $M$. Arrows close to the mass are long (field is strong); arrows far from the mass are short (field is weak). Magnitude drops as $1/r^2$ — at twice the distance, the field is one-quarter as strong.

Field of a spherical body — the shell theorem at work

Planets are not point masses. The Earth is a roughly spherical ball of mass M_E \approx 6.0 \times 10^{24} kg and radius R_E \approx 6.37 \times 10^6 m. Does the GM/r^2 law still work when the source is a ball of that size rather than a mathematical point?

Newton proved — and this is the shell theorem established in Newton's Law of Universal Gravitation — that a uniform spherical shell attracts an outside mass exactly as if all the shell's mass were concentrated at its centre. Stacking shells of different radii gives a solid sphere with the same property: for any point outside the sphere, the gravitational field is identical to that of a point mass at the sphere's centre.

So for a point at distance r from the centre of the Earth, with r > R_E:

|\vec{g}| = \frac{GM_E}{r^2}, \quad \text{pointing toward Earth's centre}

This is the formula to use for anything above the surface — the International Space Station, a communication satellite, the Moon.

At the surface itself (r = R_E), the field magnitude is what you normally call g:

g_0 = \frac{GM_E}{R_E^2}

Plug in numbers:

g_0 = \frac{(6.674 \times 10^{-11})(5.972 \times 10^{24})}{(6.371 \times 10^6)^2} \approx \frac{3.986 \times 10^{14}}{4.059 \times 10^{13}} \approx 9.82 \text{ m/s}^2

Why: this is the famous 9.8 m/s² derived from only two things — Earth's mass and Earth's radius. Every ball you have ever dropped was accelerating at this number, and the number falls out of GM_E/R_E^2.

The small discrepancy between this computed value and the textbook 9.8 m/s² comes from Earth not being perfectly spherical; the accepted standard value g_0 = 9.80665 m/s² includes small corrections.

Inside the Earth: the field of a solid sphere

Now go below the surface. A mine worker in Jharia at depth 500 m, a geophysicist running experiments down a deep borehole — they are at a distance from Earth's centre that is less than R_E. What is the field there?

The shell theorem has a second half. Consider a uniform spherical shell. For any point inside the shell, the shell exerts zero net gravitational force. The pulls from opposite parts of the shell cancel exactly — and this cancellation is so miraculous that it works for shells of any radius, at every point inside.

For a solid sphere of uniform density \rho, think of yourself at depth d below the surface — distance r = R_E - d from the centre. You can slice the planet into two parts:

Gravitational field inside a solid sphere A cross-section of Earth showing a test point at depth d. The outer shell (shaded) above the test point contributes zero field. Only the inner sphere of radius r (R minus d) pulls the test point toward the centre. O test mass g(r) = GM(r)/r² R r d = R − r outer shell: zero contribution (shell theorem)
Inside a solid sphere of radius $R$, a test mass at distance $r$ from the centre is pulled by only the inner sphere of radius $r$ (shaded red); the outer shell of material at radii between $r$ and $R$ exerts zero net force.

The mass of the inner sphere (uniform density \rho, radius r) is:

M(r) = \rho \cdot \tfrac{4}{3}\pi r^3

The full Earth has total mass M_E = \rho \cdot \tfrac{4}{3}\pi R_E^3, so:

\frac{M(r)}{M_E} = \frac{r^3}{R_E^3}, \qquad M(r) = M_E \cdot \frac{r^3}{R_E^3}

Why: volume scales as radius cubed, and density is assumed uniform. The mass enclosed within radius r is a fraction (r/R_E)^3 of the whole planet.

By the shell theorem, the field at radius r (inside) is that of a point mass M(r) at distance r:

g(r) = \frac{GM(r)}{r^2} = \frac{G M_E}{r^2} \cdot \frac{r^3}{R_E^3} = \frac{G M_E}{R_E^3}\,r

Why: the r^2 in the denominator partially cancels with the r^3 in the enclosed mass, leaving one factor of r in the numerator. Inside the sphere, g grows linearly with distance from the centre — the opposite of the inverse-square behaviour outside.

Rewrite the prefactor in terms of g_0 = GM_E/R_E^2:

g(r) = \frac{GM_E}{R_E^2} \cdot \frac{r}{R_E} = g_0 \cdot \frac{r}{R_E}, \qquad r \le R_E

Substitute r = R_E - d (where d is depth below the surface):

\boxed{g_{\text{depth}}(d) = g_0\left(1 - \frac{d}{R_E}\right)}

At d = 0 (surface): g = g_0. ✓

At d = R_E (Earth's centre): g = 0. ✓ — every direction around you is now "up", and the pulls cancel.

At an intermediate depth, say d = R_E/2 (halfway to the centre): g = g_0/2 \approx 4.9 m/s².

Outside the Earth: altitude dependence

Back above the surface. A point at altitude h is at distance r = R_E + h from the centre, so:

g_{\text{altitude}}(h) = \frac{GM_E}{(R_E + h)^2}

Divide by g_0:

\frac{g(h)}{g_0} = \frac{GM_E/(R_E + h)^2}{GM_E/R_E^2} = \frac{R_E^2}{(R_E + h)^2} = \left(\frac{R_E}{R_E + h}\right)^2
\boxed{g_{\text{altitude}}(h) = g_0 \left(\frac{R_E}{R_E + h}\right)^2}

For small altitudes h \ll R_E, expand:

g_{\text{altitude}}(h) \approx g_0\left(1 - \frac{2h}{R_E}\right)

Why: for |x| \ll 1, (1+x)^{-2} \approx 1 - 2x (the leading-order binomial expansion). Substituting x = h/R_E gives the approximation. It is accurate to about 0.01% for h under 100 km and within 1% for h below 1000 km.

Notice the factor-of-2 difference between altitude and depth:

So a 10-km ascent into the atmosphere decreases g more than a 10-km descent into the crust would — you lose about 0.3% of g going up 10 km, versus about 0.16% going down 10 km. Both are tiny, but the altitude effect dominates.

Variation of g with distance from Earth's centre A graph showing g as a function of distance r from Earth's centre. Inside the sphere, g rises linearly from 0 at the centre to g-naught at radius R. Outside the sphere, g decreases as 1 over r squared. The peak is at the surface. distance from centre, r gravitational field g(r) R (surface) 2R 3R g₀ 0 inside: g ∝ r outside: g ∝ 1/r²
The gravitational field of a uniform solid sphere (Earth-like) as a function of distance from the centre. Inside ($r < R$), $g$ grows linearly from zero at the centre to the peak value $g_0$ at the surface. Outside ($r > R$), $g$ decreases as $1/r^2$, falling to one-quarter at $r = 2R$ and one-ninth at $r = 3R$. The peak is exactly at the surface.

The graph has a kink at the surface — the derivative is discontinuous there (inside, dg/dr = g_0/R_E; outside, dg/dr = -2g_0/R_E), but g itself is continuous. This kink is the signature of the shell theorem: the transition from "growing with r" to "shrinking with r^2" happens precisely at the boundary.

Explore the altitude dependence

The interactive figure below lets you drag the altitude h from 0 (surface) to 6000 km (roughly the altitude of Chandrayaan's transfer orbit). Watch how g drops, and how the exact formula g_0 R_E^2/(R_E + h)^2 compares with the approximate g_0(1 - 2h/R_E).

Interactive: g versus altitude A graph of gravitational field g as a function of altitude h above Earth's surface. A draggable point lets the reader move along the curve and read live values of g. 1000 2000 3000 4000 5000 6000 altitude h (km) 0 2 4 6 8 10 g (m/s²) exact: g₀(R/(R+h))² linear approx drag the red point to change altitude
Drag the red point to change the altitude. The red curve is the exact formula $g = g_0(R_E/(R_E + h))^2$; the dashed grey line is the linear approximation $g \approx g_0(1 - 2h/R_E)$, which fits well for altitudes below a few hundred kilometres but drifts noticeably past 1000 km. At the altitude of the ISS ($h \approx 400$ km), $g$ is still about 8.7 m/s² — about 89% of surface gravity. Weightlessness in orbit is free fall, not the absence of gravity.

Why g varies with latitude

Strictly speaking, "g_0 = 9.8 m/s²" is an average. The actual measured value at Earth's surface varies with latitude:

That is a difference of about 0.5%, small but measurable. Two effects combine to produce it.

First: Earth's rotation. A test mass sitting at the equator is carried along in a circle of radius R_E once every 24 hours. For it to move in that circle, there must be a centripetal force pointing toward the Earth's axis, equal to m\omega_E^2 R_E, where \omega_E = 2\pi/(86400 \text{ s}) \approx 7.27 \times 10^{-5} rad/s. Part of the gravitational pull goes into supplying this centripetal force; only the remainder shows up as the weight you measure on a bathroom scale.

The centripetal correction at the equator:

\Delta g_{\text{rot}} = \omega_E^2 R_E = (7.27 \times 10^{-5})^2 \times 6.371 \times 10^6 \approx 0.0337 \text{ m/s}^2

Why: the "effective" gravity felt at the equator is the true gravitational pull minus the centripetal acceleration, because the test mass is going around in a circle with the Earth. At the poles, the distance from the rotation axis is zero, so there is no centripetal correction.

At a general latitude \lambda, the distance from the rotation axis is R_E \cos\lambda, so the rotation correction scales as \cos^2\lambda: maximum at the equator, zero at the poles.

Second: Earth's oblate shape. The Earth is not a perfect sphere — centripetal flinging has stretched the equator outward, so the equatorial radius (6378 km) is about 21 km larger than the polar radius (6357 km). A point at the equator is therefore farther from Earth's centre, making the gravitational pull smaller. A point at the pole is closer to the centre, so the pull is larger.

Both effects add in the same direction — both make g smaller at the equator and larger at the poles. Together they explain the observed 0.5% difference.

Worked examples

Example 1: Weight at the top of Mount Everest

A climber of mass 70 kg stands at the summit of Mount Everest, 8848 m above sea level. Taking R_E = 6.371 \times 10^6 m and g_0 = 9.81 m/s² at sea level, find (a) the value of g at the summit and (b) the climber's weight there. How much lighter is the climber compared with sea level?

Earth's surface with Mount Everest and a climber A curved Earth surface with a triangular mountain rising 8.85 km. A stick figure is at the summit, with an arrow labelled g = 9.78 m/s² pointing downward. g h = 8848 m sea level
A climber at the summit of Mount Everest stands 8848 m above sea level. The gravitational field is slightly weaker there than at sea level.

Step 1. Use the small-altitude approximation.

g(h) \approx g_0\left(1 - \frac{2h}{R_E}\right)
g(8848) \approx 9.81 \times \left(1 - \frac{2 \times 8848}{6.371 \times 10^6}\right) = 9.81 \times (1 - 0.002778)
g(8848) \approx 9.81 - 0.02725 \approx 9.783 \text{ m/s}^2

Why: Everest is only about 0.14% of Earth's radius, so the linear approximation is extremely accurate here. You lose about 0.28% of g by ascending 8.85 km.

Step 2. Verify with the exact formula.

g(h) = g_0\left(\frac{R_E}{R_E + h}\right)^2 = 9.81 \times \left(\frac{6.371 \times 10^6}{6.371 \times 10^6 + 8848}\right)^2
= 9.81 \times \left(\frac{6371000}{6379848}\right)^2 = 9.81 \times (0.99861)^2 = 9.81 \times 0.99722 \approx 9.783 \text{ m/s}^2

Why: the exact formula agrees with the linear approximation to four significant figures at this altitude. The approximation is vindicated.

Step 3. Compute the weight.

At sea level: W_0 = mg_0 = 70 \times 9.81 = 686.7 N.

At the summit: W_{\text{Everest}} = mg(h) = 70 \times 9.783 = 684.8 N.

Step 4. The difference.

W_0 - W_{\text{Everest}} = 686.7 - 684.8 = 1.9 \text{ N}

Why: the climber weighs about 1.9 N less at the summit than at sea level — about the weight of a half-full cup of water. Not enough to notice by hand, but easily measurable on a laboratory spring balance.

Result: g \approx 9.78 m/s² at the summit of Everest; the 70 kg climber weighs 684.8 N there, about 1.9 N less than at sea level.

What this shows: ascending even an 8 km mountain measurably reduces your weight. The ISS, at an altitude of 400 km, has g \approx 8.7 m/s² — astronauts still have 89% of surface gravity, but they feel weightless because they are falling alongside the station.

Example 2: g in a deep mine

A physicist takes a precision pendulum 3.0 km down a mineshaft to measure g. Assuming Earth has uniform density, what value of g does she measure? What is the fractional change from the surface value, and is it larger or smaller than the change at the summit of Everest?

A mineshaft going 3 km below Earth's surface A vertical cross-section showing the ground, a vertical shaft descending 3 km, and a physicist at the bottom with a pendulum. surface d = 3.0 km g(d)
A physicist at the bottom of a 3 km mineshaft. Only the mass *below* her contributes to gravity — the rock above forms a spherical shell whose net pull is zero.

Step 1. Use the depth formula for a uniform sphere.

g_{\text{depth}}(d) = g_0\left(1 - \frac{d}{R_E}\right)

Step 2. Plug in d = 3000 m and R_E = 6.371 \times 10^6 m.

g_{\text{depth}}(3000) = 9.81 \times \left(1 - \frac{3000}{6.371 \times 10^6}\right) = 9.81 \times (1 - 4.71 \times 10^{-4})
g_{\text{depth}}(3000) = 9.81 \times 0.999529 \approx 9.8054 \text{ m/s}^2

Why: at 3 km depth you are 3000/6.371 \times 10^6 \approx 0.047\% closer to Earth's centre, so g decreases by 0.047% from the surface value.

Step 3. Fractional change from the surface.

\frac{\Delta g}{g_0} = \frac{9.81 - 9.8054}{9.81} \approx 4.7 \times 10^{-4} = 0.047\%

Step 4. Compare with Everest.

For Everest at h = 8848 m: the fractional change was 2h/R_E = 2 \times 8848/6.371 \times 10^6 \approx 2.78 \times 10^{-3} = 0.28\%.

Ratio: the Everest change is 0.28/0.047 \approx 6 times larger than the 3 km mine change.

Why: going up by h loses a fraction 2h/R_E of g (inverse-square, linearised). Going down by d loses a fraction d/R_E of g (linear-inside result). At similar distances, altitude matters roughly twice as much per kilometre — and Everest is three times taller than the mine is deep, so the combined ratio is about 6.

Result: g = 9.805 m/s² at 3 km depth, a reduction of 0.047%. The Everest reduction (0.28%) is about six times larger.

What this shows: going down in a mine reduces g much less, per kilometre, than going up in altitude. Both effects are small by everyday standards, but precision gravimetry can measure them — and real mines show a bit more complexity, because the Earth's density is not uniform (the core is denser than the mantle), so the uniform-density formula is a first-order estimate. Still, it captures the central idea: the shell above you pulls in every direction and cancels out.

Common confusions

You now have the field equations and know how to compute g anywhere above or below Earth's surface. What follows is for readers who want to see the field computed for two non-spherical sources — a ring and a straight rod — and a careful derivation of the "effective gravity" felt in a rotating frame.

Field on the axis of a uniform ring

Consider a uniform ring of mass M, radius a, with its centre at the origin and its plane perpendicular to the z-axis. Find the gravitational field on the axis at distance z from the centre.

Slice the ring into tiny mass elements dM. Each element sits at distance \sqrt{a^2 + z^2} from the axial point. The field contribution from dM points from the axial point toward the element — it has a component along the axis (pointing toward the ring's plane) and a component perpendicular to the axis.

By symmetry, when you add contributions from elements on opposite sides of the ring, the perpendicular components cancel. Only the axial component survives. The axial component from dM is:

dg_z = \frac{G\,dM}{(a^2 + z^2)} \cdot \frac{z}{\sqrt{a^2 + z^2}} = \frac{G\,z\,dM}{(a^2 + z^2)^{3/2}}

Why: the magnitude of the field from dM is G\,dM/\text{distance}^2. The fraction z/\sqrt{a^2+z^2} is the cosine of the angle between the axis and the line from the axial point to dM — the geometric factor that projects the contribution onto the axis.

All the dM contributions to dg_z have the same geometric factor (because every element is equidistant from the axial point), so you can pull it out of the sum:

g_z = \frac{G\,z}{(a^2 + z^2)^{3/2}} \int dM = \frac{GMz}{(a^2 + z^2)^{3/2}}

Sanity checks:

  • At z = 0 (centre of the ring): g_z = 0. ✓ By symmetry, no axial pull exists when you are in the plane of the ring at its centre — every direction has an opposite direction pulling back equally.
  • For z \gg a: g_z \to GM/z^2. ✓ From very far away, the ring looks like a point mass M, and the field reduces to the familiar inverse-square law.
  • The field is maximum at z = a/\sqrt{2} (found by dg_z/dz = 0), with value g_{z,\max} = (2/3\sqrt{3})\,GM/a^2 \approx 0.385\,GM/a^2.

This ring result is a stepping-stone to the field of a disc (integrate over radii from 0 to a) and then a sphere (integrate over shells).

Effective gravity in a rotating frame

A plumb bob hanging from a string at latitude \lambda does not point exactly at Earth's centre. It points along the direction of the effective gravity, which is the true gravitational pull minus the centrifugal correction you feel because you are rotating with the Earth.

Setup. Choose coordinates at latitude \lambda with \hat{r} pointing radially outward from Earth's centre and \hat{\lambda} pointing northward along the surface tangent. The true gravitational pull is \vec{g}_{\text{true}} = -g_0 \hat{r}, pointing radially inward with magnitude g_0.

The centripetal acceleration required to keep a test mass rotating with the Earth is directed toward the axis of rotation, with magnitude \omega_E^2 \rho, where \rho = R_E \cos\lambda is the distance from the axis. In the rotating frame, this appears as an outward centrifugal correction, +\omega_E^2 R_E \cos\lambda pointing perpendicular to the rotation axis (away from it).

Decomposing the centrifugal vector. The outward-from-axis direction, at latitude \lambda, has:

  • A radial component (outward from Earth's centre) of \omega_E^2 R_E \cos\lambda \cdot \cos\lambda = \omega_E^2 R_E \cos^2\lambda.
  • A tangential component (pointing southward, since at northern latitudes "outward from axis" has a southward component) of \omega_E^2 R_E \cos\lambda \cdot \sin\lambda.

Effective gravity. Adding the true gravity and the centrifugal correction:

\vec{g}_{\text{eff}} = (-g_0 + \omega_E^2 R_E \cos^2\lambda)\hat{r} - \omega_E^2 R_E \cos\lambda \sin\lambda\,\hat{\lambda}

The magnitude of \vec{g}_{\text{eff}} at the surface:

|\vec{g}_{\text{eff}}| \approx g_0 - \omega_E^2 R_E \cos^2\lambda

to first order in \omega_E^2 R_E / g_0 \approx 0.0035. At the equator (\lambda = 0): |\vec{g}_{\text{eff}}| = g_0 - \omega_E^2 R_E \approx 9.78 m/s². At the poles (\lambda = \pm 90°): no centrifugal correction, so |\vec{g}_{\text{eff}}| = g_0 \approx 9.81 m/s² (plus the ~0.02 m/s² further increase from the polar radius being smaller, which this calculation neglects but is included in the standard geodetic formula).

The tangential component of \vec{g}_{\text{eff}} is why the plumb bob deflects slightly toward the equator — the "true vertical" that a spirit level defines at latitude \lambda is not quite parallel to the line from you to Earth's centre. The angle is small: \tan\theta = \omega_E^2 R_E \cos\lambda \sin\lambda / g_0, maximum at \lambda = 45° where \theta \approx 0.1°.

A practical note — the standard "g_0 = 9.80665 m/s²"

The International System of Units fixes g_0 = 9.80665 m/s² as the standard acceleration due to gravity — a defined constant, used for calibrating weight scales. This value is a compromise: it is the measured g at latitude ~45° at sea level, chosen as a convenient universal reference so that mass-weight conversions are unambiguous across latitudes. Actual local g varies from this standard by as much as 0.5% depending on where you are. Precision experiments in fundamental physics — Kibble balances, measurements of G — must account for the local g at the specific laboratory, not the conventional value.

Where this leads next