Variation of g — with altitude, latitude, depth

0. The starting point — surface gravity

WHY this form? All of Earth's mass acts as if concentrated at its center (shell theorem), so a surface object at distance $R$ feels $GM/R^2$ . Everything below is a correction to this.

1. Variation with ALTITUDE (height $h$ above surface)

Why the step "divide by $g$ "? It cancels $GM$ , leaving a clean ratio that only depends on geometry $h/R$ — easier and avoids needing $G$ or $M$ numerically.

2. Variation with DEPTH (depth $d$ below surface)

Formula Derivation from scratch

Assume uniform density $\rho$ . Surface: $g = \frac{GM}{R^2} = \frac{G}{R^2}\cdot \rho\cdot\tfrac{4}{3}\pi R^3 = \tfrac{4}{3}\pi G\rho R$ So $g \propto R$ for a uniform sphere — surface gravity is proportional to radius. At depth $d$ , effective radius is $R-d$ , and only the inner mass $M' = \rho\cdot\tfrac43\pi(R-d)^3$ counts: $g_d = \frac{GM'}{(R-d)^2} = \tfrac{4}{3}\pi G\rho (R-d)$ Take the ratio (the $\tfrac43\pi G\rho$ cancels): $\frac{g_d}{g} = \frac{R-d}{R} = 1-\frac{d}{R} \implies \boxed{g_d = g\left(1-\frac{d}{R}\right)}$

Why this is EXACT (not approximate) for depth but altitude needed a binomial: because $g\propto R$ inside a uniform sphere is linear, so the depth formula has no approximation. At the center ( $d=R$ ): $g=0$ .

Figure — Variation of g — with altitude, latitude, depth

3. Variation with LATITUDE (angle $\lambda$ from equator)

Formula Rotation effect — derivation

A point at latitude $\lambda$ moves in a circle of radius $r = R\cos\lambda$ (distance from spin axis). The true gravity $g$ points to the center. The required centripetal acceleration is $\omega^2 r = \omega^2 R\cos\lambda$ , directed toward the axis. Resolving along the radial (vertical) direction, the component of centripetal need is $\omega^2 R\cos\lambda \cdot \cos\lambda$ . Effective gravity: $\boxed{g_\lambda = g - \omega^2 R\cos^2\lambda}$

Equator $\lambda=0$ : $\cos^2 0 =1 \Rightarrow g_{eq} = g-\omega^2 R$ (smallest).
Poles $\lambda=90°$ : $\cos^2 90°=0 \Rightarrow g_{pole}=g$ (largest, no rotation effect).

Why $\cos^2\lambda$ and not $\cos\lambda$ ? One $\cos\lambda$ comes from the circle's radius $r=R\cos\lambda$ ; the second comes from projecting the horizontal centripetal vector onto the vertical (local "down") direction.

4. Forecast-then-Verify table

Recall Predict the direction of change BEFORE reading

Action	$g$ increases or decreases?	Formula	Rate near surface
Go up by $h$	decrease	$g(1-2h/R)$	$-2/R$ per metre
Go down by $d$	decrease	$g(1-d/R)$	$-1/R$ per metre
Move equator→pole	increase	$+\omega^2R\cos^2\lambda$ restored	up to $+0.034$
Key cross-check: altitude falls twice as fast as depth. ✅

Flashcards

Surface gravity in terms of

G,M,R

g=\dfrac{GM}{R^2}

Formula for

g

at altitude

h

(exact)

g_h=\dfrac{GM}{(R+h)^2}=g\left(1+\frac hR\right)^{-2}

Approximate

g

at small altitude

g_h\approx g\left(1-\dfrac{2h}{R}\right)

Formula for

g

at depth

d

(uniform Earth)

g_d=g\left(1-\dfrac dR\right)

Why does

g=0

at Earth's center?

All mass is now in shells above you → shell theorem gives zero net force.

Why does altitude reduce

g

twice as fast as depth near surface?

Altitude uses inverse-square (

-2h/R

); depth uses linear

g\propto R

(

-d/R

Effective

g

at latitude

\lambda

g_\lambda=g-\omega^2R\cos^2\lambda

Where is

g

largest/smallest due to rotation?

Largest at poles (

\cos^2 90°=0

), smallest at equator (

\cos^2 0°=1

Why two cosines in latitude formula?

One from circle radius

r=R\cos\lambda

, one from projecting onto local vertical.

Inside a uniform sphere,

g \propto

g\propto r

(distance from center).

Recall Feynman: explain to a 12-year-old

Imagine Earth is a giant ball that pulls you toward its middle. If you climb a tall mountain, you're a bit farther from the middle, so the pull is a little weaker — you'd weigh slightly less. If you dig a deep hole and stand in it, the part of Earth above your head now pulls you up a little and cancels itself out, so only the smaller ball beneath you pulls — again weaker. And because Earth spins like a merry-go-round, standing near the equator (the fattest, fastest-spinning part) flings you outward a tiny bit, making you feel lighter than at the icy poles. So you're heaviest at the poles, in a deep mine you're lighter, and on a mountain you're lighter too!

Connections

Newton's Law of Universal Gravitation — the parent $1/r^2$ law.
Shell Theorem — justifies why outer shells contribute zero (depth case).
Centripetal Force & Circular Motion — source of the rotation (latitude) term.
Weight vs Mass — weight $=mg$ changes with location, mass does not.
Oblate Spheroid Earth — geometric part of latitude variation.
Escape Velocity & Orbital Mechanics — uses the same $GM/r^2$ field.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, $g$ koi fixed number nahi hai — yeh depend karta hai ki aap Earth pe kahan khade ho. Basic formula hai $g = GM/R^2$ , matlab jitna door center se, utna kam pull. Isliye agar aap upar jao (altitude $h$ ), toh distance $R+h$ ho jaata hai, aur $g$ kam ho jaata hai: $g_h = g(1 - 2h/R)$ . Yahan 2 ka factor inverse-square law se aata hai — yaad rakhna.

Neeche jao (depth $d$ ) toh ek interesting baat hoti hai: aapke upar wali Earth ki shell ka net gravity zero ho jaata hai (shell theorem), sirf andar wali chhoti ball pulls. Isliye $g_d = g(1 - d/R)$ . Yahan factor 1 hai, 2 nahi. Toh seedha rule: upar jaane pe $g$ , neeche jaane se DOUBLE speed se girta hai (near surface). Earth ke center pe $g = 0$ ho jaata hai.

Latitude ka effect Earth ke rotation se hai. Earth ghoom rahi hai, toh kuch gravity centripetal force dene mein "use" ho jaati hai. Equator pe yeh effect maximum ( $g$ sabse kam), poles pe zero ( $g$ sabse zyada). Formula: $g_\lambda = g - \omega^2 R\cos^2\lambda$ . Do cosine isliye — ek circle ke radius $R\cos\lambda$ se, doosra vertical direction pe project karne se.

Kyun important? Pendulum clocks, accurate weighing, satellites, rockets — sabko exact $g$ chahiye. Aur exam mein yeh formulas direct aate hain, especially "altitude 2x faster than depth" wala trap. Bas mnemonic yaad rakho: UP-2, DOWN-1, SPIN steals at equator.

Variation of g — with altitude, latitude, depth

0. The starting point — surface gravity

1. Variation with ALTITUDE (height $h$ above surface)

2. Variation with DEPTH (depth $d$ below surface)

3. Variation with LATITUDE (angle $\lambda$ from equator)

4. Forecast-then-Verify table

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Newton's Laws & Dynamics

Connections