Variation of g — with altitude, latitude, depth: Worked examples

Before anything else, here are the tools we will keep re-using. Everything is anchored in the parent note; we only apply them here.

The scenario matrix

Every problem in this topic is one (or a blend) of these cells. If you can do one example per row, you have covered the whole topic.

Cell	What varies	Sign / regime	The trap or edge
A	Altitude, small $h ≪ R$	$g$ decreases, use linear	forgetting the factor 2
B	Altitude, large $h$	$g$ decreases, must use exact	linear formula breaks
C	Depth $d$	$g$ decreases, linear exact	using the " $2$ " by mistake
D	Degenerate: $d = R$ (center) and $h \to \infty$	limiting values	$g \to 0$ both ways, different reasons
E	"Equal drop" compare — same $g$ up vs down	ratio of $h$ to $d$	the $2 \times$ factor decides
F	Latitude, endpoints $λ = 0, 9 0^{\circ}$	equator smallest, pole largest	$cos^{2}$ not $cos$
G	Latitude, general $λ$ (real-world word problem)	intermediate value	radians vs degrees
H	Exam twist: spin-up limit / weight change	when does $g_{eff} = 0$ ?	you can't make $g < 0$

We now hit every cell.

Cell A — Altitude, small height

Worked example A · A drone at 3.2 km

How much does $g$ drop for a drone hovering at $h = 3.2 km$ above the surface? Give the percentage.

Forecast: Guess first — is the drop bigger or smaller than $0.1%$ ? Write your guess down.

Step 1. Check the regime: $h = 3.2 km$ vs $R = 6400 km$ , so $h / R = 0.0005 ≪ 1$ . Why this step? The linear formula is only valid when $h ≪ R$ ; we must confirm it before using it, otherwise we'd need the exact form.

Step 2. Apply $g_{h} \approx g (1 - \frac{2 h}{R})$ : $\frac{2 h}{R} = \frac{2 ( 3.2 )}{6400} = 0.001$ Why this step? The fractional drop is exactly $2 h / R$ — that's what "percentage change" means, so we compute this ratio directly.

Step 3. Convert to percent: $0.001 = 0.1%$ . So $g_{h} \approx 0.999 g$ . Why this step? Multiply a fraction by $100$ to state it as a percent (write $%$ , not the bare symbol).

Verify: Units — $2 h / R$ is (km)/(km) = dimensionless ✅, correct for a ratio. Sanity: at $32 km$ the parent got $1%$ ; we are at one-tenth the height, so one-tenth the drop → $0.1%$ ✅.

Cell B — Altitude, large height (linear formula FORBIDDEN)

Worked example B · Geostationary-ish:

$h = R$ A satellite sits at height $h = R$ (one Earth-radius up). Find $g_{h}$ as a fraction of $g$ . Then show why the linear formula is nonsense here.

Forecast: The naive linear formula gives $g (1 - 2 h / R) = g (1 - 2) = - g$ . Negative gravity?! Guess the true value.

Step 1. Here $h / R = 1$ , which is not $≪ 1$ . So we MUST use the exact form. Why this step? The binomial approximation $(1 + x)^{- 2} \approx 1 - 2 x$ only holds for small $x$ . At $x = 1$ it is wildly wrong — it even goes negative, which is impossible for gravity.

Step 2. Exact form with $r = R + h = 2 R$ : $g_{h} = \frac{GM}{( R + h ) ^{2}} = \frac{GM}{( 2 R ) ^{2}} = \frac{GM}{4 R ^{2}} = \frac{1}{4} g$ Why this step? Double the distance from the center → inverse-square gives one-quarter the pull. Look at the figure: the field lines spread over $2 \times$ radius, so their density (the strength) falls by $2^{2} = 4$ .

Step 3. Compare: exact gives $+ 0.25 g$ ; linear gave $- 1.0 g$ . The linear one is not just off — it has the wrong sign. Why this step? To burn into memory that "small- $h$ " formulas are lies once $h$ is comparable to $R$ .

Verify: $g$ can never be negative (gravity always pulls in), so $- g$ is instantly wrong; $\frac{1}{4} g$ is positive and less than $g$ (we went up) ✅. At $h = R$ , $(1 + h / R)^{- 2} = (2)^{- 2} = 1/4$ ✅.

Cell C — Depth (exact linear, no factor 2)

Worked example C · The world's deepest mine

A mine reaches $d = 4 km$ . By what fraction does $g$ change at the bottom (uniform-Earth model)?

Forecast: Bigger or smaller change than the same distance up? Guess.

Step 1. Depth uses the exact linear formula $g_{d} = g (1 - \frac{d}{R})$ . Why this step? Inside a uniform sphere $g \propto r$ (parent note), so there is NO factor 2 and NO approximation — it is exact.

Step 2. Plug in: $\frac{d}{R} = \frac{4}{6400} = 6.25 \times 1 0^{- 4} \approx 0.0625%$ So $g_{d} \approx 0.999375 g$ . Why this step? The fractional change is just $d / R$ .

Step 3. Compare to going up $4 km$ : that would be $2 d / R = 0.125%$ . Why this step? Same distance, up vs down — up drops twice as fast. This is the whole point of cell C.

Verify: Depth change $0.0625%$ is exactly half the altitude change $0.125%$ for equal $4 km$ ✅ (factor-2 rule confirmed). $g_{d} < g$ (we removed the outer shell's would-be pull) ✅.

Cell D — Degenerate limits (center of Earth, infinite height)

Worked example D · Both roads lead to zero — for opposite reasons

Find $g$ (a) at Earth's center $d = R$ , and (b) infinitely far up $h \to \infty$ . Both are $0$ — explain why the reasons differ.

Forecast: Which one reaches zero "smoothly and linearly", which "asymptotically"?

Step 1. Center: $g_{d} = g (1 - \frac{d}{R})$ with $d = R$ : $g_{R} = g (1 - 1) = 0$ Why this step? At the exact center there is no inner sphere left — ALL Earth's mass is in shells around you, and by the shell theorem each shell pulls zero. Look at the figure: symmetric pulls cancel.

Step 2. Infinity: $g_{h} = \frac{GM}{( R + h ) ^{2}}$ ; as $h \to \infty$ the denominator $\to \infty$ , so $lim_{h \to \infty} g_{h} = 0$ Why this step? Distance grows without bound, and $1/ r^{2} \to 0$ . But it never equals zero at finite height — it only approaches it.

Step 3. Contrast the two curves:

Depth: a straight line from $g$ at surface down to $0$ at the center.
Altitude: a curve that dives fast then flattens, touching $0$ only at infinity.

Verify: Depth zero is reached at finite $d = R$ (center) ✅. Altitude zero is a limit, never finite ✅. Both $\geq 0$ everywhere — gravity never reverses ✅.

Cell E — Equal-drop comparison

Cell F — Latitude endpoints (equator vs pole)

Worked example F · Heaviest and lightest spots

Using only the rotation effect, find the difference $g_{pole} - g_{eq}$ in $m/s^{2}$ , and the fractional weight change for a $70 kg$ person.

Forecast: A few grams, or a few kilograms of apparent weight difference? Guess.

Step 1. Endpoints of $g_{λ} = g - ω^{2} R cos^{2} λ$ :

Pole $λ = 9 0^{\circ}$ : $cos^{2} 9 0^{\circ} = 0 \Rightarrow g_{pole} = g$ .
Equator $λ = 0^{\circ}$ : $cos^{2} 0^{\circ} = 1 \Rightarrow g_{eq} = g - ω^{2} R$ .

Why this step? $cos^{2}$ (not $cos$ ) because one factor comes from the spin-circle radius $r = R cos λ$ and one from projecting the outward push onto local vertical (parent note).

Step 2. The difference is the whole rotation term: $g_{pole} - g_{eq} = ω^{2} R = (7.27 \times 1 0^{- 5})^{2} (6.4 \times 1 0^{6}) \approx 0.0338 m/s^{2}$ Why this step? Subtracting kills $g$ , leaving the pure spin contribution.

Step 3. Weight change for $m = 70 kg$ : $Δ W = m Δ g = 70 \times 0.0338 \approx 2.37 N$ , i.e. about $0.24 kg$ of apparent weight (since $2.37/9.8 \approx 0.24$ ). Why this step? Weight $= m g_{eff}$ ; the drop in $g$ is what a scale reads as lost weight.

Verify: $0.0338 m/s^{2}$ matches the parent's $\approx 0.034$ ✅. $70 \times 0.0338/9.8 \approx 0.24 kg$ — a couple hundred grams, exactly as the parent's big-picture claimed ✅.

Cell G — Latitude, general angle (word problem)

Cell H — Exam twist: the spin-up limit

Worked example H · How fast a day makes you weightless

If Earth spun fast enough that equatorial $g_{eff} = 0$ , how long would one day be? Show $g$ cannot go negative.

Forecast: Guess the day length — minutes? hours? days?

Step 1. Set effective equatorial gravity to zero: $g - ω^{2} R = 0 ⟹ ω^{2} R = g ⟹ ω = \frac{g}{R}$ Why this step? At the equator $cos^{2} λ = 1$ , so the full spin term $ω^{2} R$ is subtracted. Weightlessness means the outward requirement equals the whole pull.

Step 2. Compute $ω$ : $ω = \frac{9.8}{6.4 \times 1 0 ^{6}} \approx 1.24 \times 1 0^{- 3} rad/s$ Why this step? This is the critical spin rate where objects at the equator go into orbit at ground level.

Step 3. Convert to a day $T = \frac{2 π}{ω}$ : $T = \frac{2 π}{1.24 \times 1 0 ^{- 3}} \approx 5063 s \approx 1.41 hours$ Why this step? Period is $2 π / ω$ ; convert seconds to hours by $/3600$ .

Step 4. Beyond this speed, would $g$ go negative? No — for $ω^{2} R > g$ , objects at the equator simply leave the ground (they'd need MORE inward force than gravity provides, so they fly off). Gravity itself, $GM / R^{2}$ , is unchanged; only the apparent $g$ hits zero and stops. Why this step? To kill the misconception that spinning can invert gravity.

Verify: $T \approx 1.41 h$ matches the parent's "about 1.4 hours" ✅. $ω^{2} R = (1.24 \times 1 0^{- 3})^{2} (6.4 \times 1 0^{6}) \approx 9.8 = g$ ✅ (weightless condition satisfied).

Recall Which cell is each formula for? (self-test)

Small altitude, want percent drop — which formula? ::: linear $g (1 - 2 h / R)$ (Cell A) Height comparable to $R$ — which formula? ::: exact $GM / (R + h)^{2}$ , never the linear one (Cell B) Depth $d$ — is there a factor 2? ::: No — depth is $g (1 - d / R)$ , exact and linear (Cell C) At Earth's center, $g = ?$ and why ::: $0$ , because all mass is in shells above you (Cell D) Same $g$ reduction: depth vs height ::: depth must be twice the height, $d = 2 h$ (Cell E) Pole minus equator $g$ (rotation only) ::: $ω^{2} R \approx 0.034 m/s^{2}$ (Cell F) Latitude $4 0^{\circ}$ : radians or degrees in $cos$ ? ::: convert to radians first (Cell G) Can spinning make $g$ negative? ::: No — apparent $g$ bottoms out at $0$ ; objects then fly off (Cell H)

Connections

Parent topic — the four formulas these examples apply.
Newton's Law of Universal Gravitation — the $1/ r^{2}$ behind cells A, B, D.
Shell Theorem — why $g = 0$ at the center (cell D) and the depth formula (cell C).
Centripetal Force & Circular Motion — the spin term in cells F, G, H.
Weight vs Mass — cell F's apparent-weight change.
Oblate Spheroid Earth — the shape effect that adds to rotation at the equator.
Escape Velocity & Orbital Mechanics — cell H is ground-level orbit.