Visual walkthrough — Variation of g — with altitude, latitude, depth

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Step 1 — What "$g$" even means: a ball pulling a ball

WHAT. Draw the Earth as a big circle and put a small dot (you, mass $m$) on its surface. There is an arrow pulling that dot straight toward the centre. The length of that arrow, per kilogram, is what we call $g$.

WHY start here. Before we can say how $g$ changes, we must pin down what $g$ is: it is the strength of the pull at a point, and by Newton's gravitation law that strength drops off with distance. Everything on this page is one idea — change the distance or the effective pulling mass, and the arrow changes length.

PICTURE.

Here $R$ is Earth's radius (≈ $6400$ km), because the shell theorem lets us treat all of Earth's mass as if squeezed into a point at the centre — so a surface dot sits a distance $R$ from that point.

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Step 2 — Going UP: the distance in the bottom grows

WHAT. Lift the dot to a height $h$ above the surface. Its distance from the centre is no longer $R$ — it is $R+h$. Put that new, longer distance into the same law.

WHY. Nothing about Earth changed; only your distance changed. Since distance lives in the denominator (squared), a bigger distance makes a smaller pull. This is the only mechanism at work for altitude.

PICTURE.

Why divide instead of plugging in numbers? Because we don't care about the value of $g$ here, only the fraction it becomes. The ratio $h/R$ tells the whole story.

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Step 3 — The small-height shortcut (and where the factor 2 is born)

WHAT. For a mountain or a balloon, $h$ is tiny next to $R$ ($32$ km vs $6400$ km). When a small number $x=h/R$ is raised to $(1+x{)}^{-2}$, it behaves almost exactly like $1-2x$. That $-2$ is not a guess — it is the slope of the inverse-square curve near $x=0$.

WHY this tool (the binomial approximation). We use $(1+x{)}^{-2}\approx 1-2x$ because computing $(1.005{)}^{-2}$ by hand is ugly, but the straight-line approximation is instant and, for $x$ this small, wrong by less than a hundredth of a percent. The tool answers the question "how fast does $g$ start to drop the moment I leave the ground?" — and the answer is twice as fast as the height fraction.

PICTURE.

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Step 4 — Going DOWN: the outer shell stops pulling

WHAT. Now sink the dot to a depth $d$ below the surface. Split Earth into two pieces: the inner ball of radius $R-d$ that you are sitting on top of, and the outer shell that is now above your head. Draw the outer shell shaded.

WHY. By the shell theorem, a uniform spherical shell pulls nothing on anything inside it — the tugs from all directions cancel exactly. So the shell above you contributes zero. Only the smaller inner ball pulls you. Less pulling mass → weaker $g$. This is a completely different mechanism from altitude (there the mass stayed the same, the distance grew; here the distance shrank but so did the mass).

PICTURE.

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Step 5 — The depth formula is a straight line (no approximation)

WHAT. At depth $d$ you sit on a ball of radius $R-d$. Reuse the same $g\propto (\text{radius})$ rule, but with radius $R-d$ instead of $R$.

WHY it is exact. Because $g$ vs radius is a genuine straight line (Step 4), there is no curve to approximate — the answer is honest, no $\ll$ needed. Contrast this with altitude, where the inverse-square curve forced us to approximate.

PICTURE.

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Step 6 — Up vs Down on one graph

WHAT. Plot $g$ against "how far from the surface", going right for altitude and left for depth. Two straight lines meet at the surface; the altitude line is twice as steep.

WHY. Seeing both slopes side by side makes the factor-2 unforgettable and shows the ranges: altitude keeps curving toward zero forever (never quite reaching it), depth reaches exactly zero at the centre.

PICTURE.

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Step 7 — Sideways: LATITUDE and the spinning merry-go-round

WHAT. Earth spins. A point at latitude $\lambda$ (angle up from the equator) is carried in a circle. That circle's radius is not $R$ — it is the distance from the spin axis, which is $r=R\cos \lambda$. Draw the axis, the point, and the little horizontal circle it rides.

WHY circular motion enters. Anything going in a circle needs a real inward force to keep it curving — the centripetal force. On spinning Earth, part of gravity's pull is "spent" supplying that inward force, so the leftover pull you actually feel (your apparent weight, see Weight vs Mass) is smaller.

PICTURE.

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Step 8 — Every latitude case, including the extremes

WHAT. Walk the angle from equator to pole and check both ends, plus the "what if Earth spun faster" limit.

WHY. The contract: no reader should meet an untested case. We cover $\lambda =0$, $\lambda =90^{\circ }$, and the runaway limit.

PICTURE.

• Equator, $\lambda =0$: ${\cos }^{2}0=1$, so $g_{\text{eq}}=g-{\omega }^{2}R$ — the biggest subtraction, $g$ smallest. Combine with the equatorial bulge (you're also farther from the centre there) and $g$ is smallest of all here.
• Pole, $\lambda =90^{\circ }$: ${\cos }^{2}90^{\circ }=0$, subtraction vanishes, $g_{\text{pole}}=g$ — largest.
• Spin-faster limit: if $\omega$ grew until ${\omega }^{2}R=g$ at the equator, $g_{\lambda }\to 0$ and loose objects would lift off (relevant to orbital motion — that's literally the orbit condition at ground level). Solving ${\omega }^{2}R=g$ gives a day of about $1.4$ hours; you cannot make $g$ negative by spinning.

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The one-picture summary

Recall Feynman retelling — the walkthrough in plain words

Earth is a giant ball that pulls everything toward its middle, and how hard it pulls is $g$. Climb up a mountain and you're farther from the middle; because the pull weakens with the square of distance, $g$ drops fast — for every step up, twice as fast as the height fraction. Dig down instead and something sneaky happens: all the Earth now above your head pulls you in every direction at once and cancels itself out, so only the smaller ball beneath you pulls — $g$ drops, but only half as fast as climbing, and it reaches exactly zero when you reach the centre. Finally, Earth spins like a merry-go-round. Near the equator you're whipping around the biggest circle, and gravity has to spend some of its strength just keeping you on that curve, so you feel lighter there; at the poles you barely move in a circle at all, so you feel gravity's full pull. Heaviest at the poles, lighter down a mine, lighter up a mountain — three different reasons, one simple picture.

Recall Quick self-test

The factor in front of $h/R$ for altitude ::: $2$ (from the inverse-square law) The factor in front of $d/R$ for depth ::: $1$ (from $g\propto R$, exact and linear) Why two cosines at latitude $\lambda$ ::: one from circle radius $R\cos \lambda$, one from projecting onto local vertical Value of $g$ at Earth's centre ::: $0$ — everything is shell above you Where is spin's subtraction largest ::: equator, where ${\cos }^2\lambda =1$

Connections

• Newton's Law of Universal Gravitation — the $1/r^2$ law of Step 1.
• Shell Theorem — kills the outer shell in the depth derivation (Steps 4–5).
• Centripetal Force & Circular Motion — the inward force behind the latitude term (Step 7).
• Weight vs Mass — what "apparent" $g$ measures.
• Oblate Spheroid Earth — the shape effect that adds to the equator's smaller $g$.
• Escape Velocity & Orbital Mechanics — the spin-faster limit is the ground-level orbit condition.