Gravitational Force Near Earth's Surface

Reading tiny variations in $g$ has become a way of seeing underground. Modern quantum gravimeters drop clouds of ultracold atoms and time their fall with lasers, measuring $g$ to a part in a billion; carried across a city, they reveal buried tunnels, voids, and pipes by their missing mass — a 2022 Birmingham experiment did exactly this through metres of urban ground. Geophysicists watch volcanoes inflate as magma arriving reads as $g$ creeping upward, and the GRACE-FO satellites weigh droughts and ice sheets from orbit by the same principle at continental scale. Every one of these instruments is Derivation B run in reverse: the corrections we neglected in the classroom, promoted into the signal. The apparent-weight ledger is working infrastructure too. Astronaut-training aircraft fly repeated parabolic arcs — engine-assisted free fall — buying 20-second windows of genuine $N = 0$; onboard the ISS, where scales are useless, crew mass is tracked with a spring-oscillator device that extracts $m$ from a measured period, inertia stepping in where weight has vanished. And every phone in every pocket knows $g$ personally: the accelerometer that flips your screen is measuring the direction of local gravity dozens of times a second, while fighter pilots' g-suits and crash-test standards are engineered directly from $N = m(g+a)$ — apparent weight, not weight, is what stresses a human body.

Watch a dropped stone with a stopwatch in hand. After one second it moves at 9.8 m/s; after two, 19.6; after three, 29.4. Gravity near the surface is a metronome of speed: plus 9.8 every second, unconditionally. That steady beat — not the force, the beat — is what $g$ names. Now the astonishing part. Swap the stone for one ten times heavier. The Earth pulls it ten times harder — that's Newton's law, no exemption. But the heavier stone is also ten times more sluggish to accelerate — that's inertia, $a = F/m$. Ten times the pull, delivered to ten times the reluctance: the ratio is untouched, and the beat stays 9.8. Gravity is the only force in nature with this self-adjusting property; a stronger magnet pulls a bigger nail with more acceleration, but gravity meters out force in exact proportion to what each object can absorb. Every object falls to the same drumbeat — which is why the beat deserves its own symbol.

Why "uniform," when the true field is inverse-square and radial? Imagine the field lines fanning out from Earth's centre. Zoom into a tiny patch at the surface — a lab, a stadium — and those lines look parallel and equally dense; the fan is invisible at that magnification. "Uniform field" is just the statement that you are looking at a small enough patch, and the local direction of the lines defines down: a plumb line hangs along $\vec g$, which is why "vertical" and "horizontal" are gravitational, not geometric, concepts. And because the beat is set by the planet alone — its mass pulling, its radius-squared diluting — every world compresses into a single number of its own:

One more picture completes the set: what travels with you and what stays behind. Your mass — the amount of matter — is identical on every world; your weight is the local field's verdict on that mass; and in free fall no support force develops at all:

The starting point is Newton's universal law, evaluated where you live. A body of mass $m$ at the surface sits one planetary radius from the centre, so the pull on it is

where: $G = 6.674\times10^{-11}\ \mathrm{N\,m^2/kg^2}$ — the gravitational constant; $M_\oplus \approx 5.97\times10^{24}\ \mathrm{kg}$ and $R_\oplus \approx 6.37\times10^{6}\ \mathrm{m}$ — Earth's mass and mean radius; the force points toward the centre, i.e. down. Everything in this article is equation (1) plus bookkeeping: the coefficient multiplying $m$ is about to become $g$, and the question of how fast it changes as you move around is about to become the "uniform field" approximation.

With $g$ built, the force law of the whole near-surface world is one line. The weight of a body — the gravitational force on it — is its mass times the local field:

where: $\vec F$ — the weight, in newtons, pointing along the local vertical; $m$ — the mass, in kilograms. The crucial reading: $m$ belongs to you and never changes; $g$ belongs to the place. A 72 kg astronaut is 72 kg on Earth, on Mars, and adrift in deep space — but she weighs 706 N here, 268 N on Mars, and it is the newtons, not the kilograms, that belong on a free-body diagram. How local is "local"? Climb to altitude $h$ and your distance from the centre grows from $R$ to $R+h$, so the inverse-square fade must shave something off:

where: $g_0$ — the surface value; the second form is a pure discount factor, no $G$ or $M$ required. Because $R_\oplus$ is 6371 km, the fraction stays stubbornly close to 1 for any human altitude: Everest ($h = 8.8$ km) reads $9.82 \times (6371/6380)^2 = 9.79\ \mathrm{m/s^2}$ — a 0.3% discount for the world's highest climb. At space-station altitude ($h = 400$ km), $g = 9.82 \times (6371/6771)^2 = 8.69\ \mathrm{m/s^2}$ — fully 89% of surface strength, the number that dismantles the "no gravity in space" myth. Only at geostationary distance ($r \approx 6.6\,R_\oplus$) does the field fade to $9.82/6.6^2 = 0.22\ \mathrm{m/s^2}$ — a whisper, but exactly the whisper a 24-hour orbit needs.

What a uniform field buys you

The payoff for "constant $\vec g$" is that the hardest force problem in the room becomes the easiest differential equation. First, energy: a constant force integrates to a linear potential. With $\hat y$ up, $F_y = -mg$, and demanding $F_y = -\mathrm{d}U/\mathrm{d}y$ gives

where: $U$ — gravitational potential energy relative to the level you declare zero; $y$ — height above it. Every "raise it, then let it fall" energy audit near the surface — pile drivers, pendulums, roller coasters — is equation (5) traded against $\tfrac12 mv^2$. (It is the small-$h$ limit of the exact $-GM_\oplus m/r$; that is why it only works near the surface.) Second, motion: with $a = -g$ constant, one integration gives the velocity and a second gives the position,

where: $y_0, v_0$ — initial height and upward velocity; the second (time-free) form comes from eliminating $t$ and answers "how fast at this height?" directly. These two lines are one-dimensional kinematics; gravity is where they earn their keep. Third, the showpiece: launch at angle $\theta_0$ and speed $v_0$, and because $\vec g$ has no horizontal component the motions decouple — $\ddot x = 0$, $\ddot y = -g$. Integrating each and eliminating time bends the trajectory into a parabola,

where: the range on level ground follows by setting $y = 0$: $R = v_0^2\sin 2\theta_0/g$, maximised at $\theta_0 = 45°$. Projectile motion has its own article; the point here is that the entire parabola is a consequence of field uniformity — curve the field lines and the parabola bends into an orbit's ellipse.

Weight, apparent weight, and what the scale reads

Hold a brick: you feel its heaviness in your straining muscles. But notice what is actually happening — gravity pulls the brick down, and your hand pushes it up. What your nerves report is the push you must supply, not the pull the Earth exerts. Gravity acts on every atom of the brick, and of you, in exact proportion to mass, accelerating all of them identically — a force that moves every part of you together leaves nothing to strain, nothing to feel. The sensation of weight is entirely a story about support.

One last honesty clause. The $g$ you actually measure at ground level is not the pure Newtonian $GM_\oplus/R_\oplus^2$: standing on a spinning planet, part of gravity's pull is diverted into keeping you on your daily circle, and the spin has flattened the Earth so the equator sits farther from the centre. In the rotating frame the effective field is

where: $\omega = 7.29\times10^{-5}\ \mathrm{rad/s}$ — Earth's rotation rate; $\lambda$ — latitude; the centrifugal term peaks at the equator at $\omega^2 R_\oplus \approx 0.034\ \mathrm{m/s^2}$. Rotation plus oblateness make the measured $g$ run from about $9.78\ \mathrm{m/s^2}$ at the equator to $9.83$ at the poles — the reason pendulum clocks shipped across latitudes in the 1600s refused to keep time, and the first hint (Derivation A's story, told by Newton himself) that 9.8 is a readable property of the planet.

Which formula answers which question

1 · The dropped stone (confidence builder)

A stone is released from rest at the top of a 20 m building. How long does it fall, and how fast does it land? (Ignore air resistance; take $g = 9.81\ \mathrm{m/s^2}$.)

Step 1. Fall time from equation (6) with $v_0 = 0$: $h = \tfrac12 g t^2$, so $t = \sqrt{2h/g} = \sqrt{40/9.81} = \sqrt{4.08} = \mathbf{2.0\ s}$.

Step 2. Landing speed, time-free form: $v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 20} = \sqrt{392} = \mathbf{19.8\ m/s}$ — about 71 km/h from just six storeys, which is why falls are dangerous.

Step 3. Notice what was never asked: the stone's mass. Both answers came from $g$ alone — the universality of Derivation A doing its quiet work.

2 · Gravity on Mars, from two numbers (intermediate)

Mars has mass $M = 6.42\times10^{23}\ \mathrm{kg}$ and radius $R = 3.39\times10^6\ \mathrm{m}$. (a) Find $g$ on the Martian surface. (b) A 72 kg astronaut walks there — find her weight on Mars and on Earth.

Step 1. Apply equation (2). Numerator: $GM = (6.674\times10^{-11})(6.42\times10^{23}) = 4.28\times10^{13}$.

Step 2. Denominator: $R^2 = (3.39\times10^6)^2 = 1.15\times10^{13}\ \mathrm{m^2}$.

Step 3. Divide: $g_{\text{Mars}} = \dfrac{4.28\times10^{13}}{1.15\times10^{13}} = \mathbf{3.73\ m/s^2}$ — about 38% of Earth's. Every rover descent and every future boot-print budgets around this number.

Step 4. Weight on Mars: $W = mg = 72 \times 3.73 = \mathbf{268\ N}$. On Earth: $72 \times 9.81 = \mathbf{706\ N}$. Her mass is 72 kg on both worlds — the scale reading changed, not her.

3 · How high must you go? (exam level)

(a) Find $g$ at the altitude of the International Space Station, $h = 400\ \mathrm{km}$, and the true weight of a 70 kg astronaut there. (b) At what altitude does $g$ drop to exactly half its surface value?

Step 1. Part (a) is a ratio via equation (4): $g = 9.82 \times \left(\dfrac{6371}{6771}\right)^2 = 9.82 \times (0.9409)^2 = 9.82 \times 0.885 = \mathbf{8.69\ m/s^2}$.

Step 2. True weight up there: $W = 70 \times 8.69 = \mathbf{608\ N}$ — not remotely zero. Floating is a statement about falling together with your spacecraft (Derivation C, Step 4), not about the field being gone.

Step 3. Part (b): set the discount factor to one half, $\left(\dfrac{R_\oplus}{R_\oplus+h}\right)^2 = \dfrac12$.

Step 4. Square-root both sides: $R_\oplus + h = \sqrt{2}\,R_\oplus$, so $h = (\sqrt{2}-1)R_\oplus = 0.414 \times 6371\ \mathrm{km} = \mathbf{2640\ km}$.

Step 5. Sanity check: that is nearly seven times the ISS altitude — almost half an Earth radius of climbing for only half the gravity. The inverse-square fade is genuinely slow on the scale of a planet; "space" is nowhere near "no gravity."

Texts. Halliday, Resnick & Walker, Fundamentals of Physics, Ch. 5 and 13 · Young & Freedman, University Physics, Ch. 13 · Morin, Introduction to Classical Mechanics, Ch. 3 · Taylor, Classical Mechanics, Ch. 1–2, and Ch. 9 for the rotating-Earth corrections done properly · Goldstein, Classical Mechanics, Ch. 1 · Landau & Lifshitz, Mechanics (Vol. 1), §1–§4.

Provenance. A verifier should recompute $g=GM_\oplus/R_\oplus^2$ from $G=6.674\times10^{-11}\ \mathrm{N\,m^2/kg^2}$, $M_\oplus=5.97\times10^{24}\ \mathrm{kg}$, $R_\oplus=6.37\times10^{6}\ \mathrm{m}$ (yielding $\approx 9.8\ \mathrm{m/s^2}$); check the binomial expansion $(1+h/R)^{-2}\approx 1-2h/R$; confirm $\omega=2\pi/86164\ \mathrm{s}\approx 7.29\times10^{-5}\ \mathrm{rad/s}$ and the equatorial centrifugal term $\omega^2 R_\oplus\approx 0.034\ \mathrm{m/s^2}$; verify free-fall numbers $t=\sqrt{2h/g}$, $v=\sqrt{2gh}$ and range $R=v_0^2\sin2\theta_0/g$. Equations cross-checked against Morin Ch. 3, Taylor Ch. 1–2, Goldstein Ch. 1, and Landau–Lifshitz Vol. 1 §1–4. Planetary data are standard NASA fact-sheet values ($M_{\text{Mars}} = 6.42\times10^{23}$ kg, $R_{\text{Mars}} = 3.39\times10^6$ m); worked-example numbers on this page ($t = 2.0$ s and $v = 19.8$ m/s for the 20 m drop; $g_{\text{Mars}} = 3.73\ \mathrm{m/s^2}$ and $W = 268$ N; $g_{\text{ISS}} = 8.69\ \mathrm{m/s^2}$ and $h_{1/2} = 2640$ km) are illustrative arithmetic recomputed from the boxed formulas.

Builds on: Newton's Second Law · Vectors in Kinematics and Coordinate Systems · Velocity and Acceleration. Leads to: Projectile Motion · One-Dimensional Kinematics · Normal Force · Friction Forces.