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

Before you can read a single line of the parent note, you need to earn every symbol it throws at you. Below, each idea is built from nothing, drawn as a picture, and justified by the question it answers. Read top to bottom — every item leans on the one above it.

1. Mass $m$ — "how much stuff"

Why the topic needs it: the whole question "does a 70 kg person weigh less at the equator?" only makes sense if we agree the 70 kg (the stuff) never changes, and only the pull on it does. Mass is the unchanging anchor. See Weight vs Mass for the full split.

2. Force $F$ and weight — "the pull"

In the figure, the same brick (mass $m$, unchanged) hangs from a spring at two places. The brick is the same; the red arrow — the pull, its weight — is longer in one place than the other. That difference is the entire subject of this topic.

Why the topic needs it: the parent note writes $F=mg$. That sentence says "the pull $F$ equals mass $m$ times the pull-per-kilogram $g$." To read it, you must already know $F$ is the pull and $m$ is the stuff.

3. Gravitational acceleration $g$ — "pull per kilogram"

Why the topic needs it: the entire topic is titled "Variation of $g$." $g$ is the star. Notice $g$ is a ratio — the mass $m$ has been divided out — which is exactly why $g$ describes the place, not the object. A feather and a boulder at the same spot feel the same $g$.

4. The inverse-square distance rule $1/r^2$

This is the tool the topic leans on hardest, so we build it with care.

Look at the figure: the red patch of fixed size catches lots of arrows up close, but far away the same-size patch catches far fewer, because the arrows have fanned out over a bigger sphere.

Why the topic needs it: the altitude section (go up by $h$) is nothing but "$r$ got bigger, so $1/r^2$ got smaller." The factor of 2 in $g_h\approx g(1-2h/R)$ is a direct fingerprint of the square in $r^2$. If you don't feel why it's a square, that mysterious 2 will always look arbitrary.

5. The constants $G$, $M$, $R$, and the height/depth symbols $h$, $d$

Now we specialise item 4's two generic masses: one of them is Earth, so from here on we rename it $M$. The other is your object $m$ from items 1–2.

The figure shows Earth as a circle with centre $O$. The red line is $r$ — your actual distance from the centre. Notice how it stretches to $R+h$ when you climb a height $h$, and shrinks to $R-d$ when you descend a depth $d$.

Why the topic needs the split $r=R\pm x$: every section of the parent is just this formula with $r$ nudged. Altitude nudges $r$ up by $h$, depth nudges the effective $r$ down by $d$, and both are compared back to the surface value $GM/R^2$. Recognising which of $h$, $d$, $R$ is which prevents every algebra slip in the chapter.

6. The Shell Theorem — why the roof above you doesn't pull

Before density, we must earn the fact that item 5 leaned on: underground, the Earth above your head pulls with zero net force. This is the Shell Theorem, and it deserves its own picture.

In the figure you stand at the red dot, off-centre. The near patch (short arrow, small patch) and the far patch (long arrow — but the arrows are equal length because the bigger far patch exactly makes up for its distance) pull in opposite directions and cancel.

Why the topic needs it: when you dig to depth $d$, split Earth into (a) the inner ball of radius $R-d$ below you and (b) all the shells above you. The shells above are exactly the "you're inside a shell" case — they contribute zero. So only the inner ball pulls. This single fact is what makes the depth formula (item 7) possible.

7. Density $\rho$, mass = density × volume, and the depth formula

Now we can build the depth result promised by the parent, using items 5, 6 and this one together. Assume Earth has uniform density $\rho$ (same everywhere).

Why the topic needs it: without density you cannot write the inner-ball mass, and without the Shell Theorem you cannot drop the outer shells — the two combine to give this exact, approximation-free depth law. Contrast it with altitude's $1-2h/R$: depth is linear (no factor 2) precisely because $g\propto R$ inside a uniform sphere.

8. Angular speed $\omega$ and the circle a spinning point traces

In the figure the vertical line is the spin axis. A point at ==latitude $\lambda$== (the angle up from the equator) sits on a circle whose red radius is $r=R\cos \lambda$. At the equator ($\lambda =0$) that radius is the full $R$; at the pole ($\lambda =90^{\circ }$) it shrinks to zero — a pole point just spins in place.

Why the topic needs it: the latitude section subtracts a term ${\omega }^{2}R{\cos }^2\lambda$ from $g$. You cannot understand that term until you see (a) $\omega$ is the spin rate, (b) the spin-circle radius is $R\cos \lambda$, giving one cosine, and (c) projecting that horizontal effect onto "down" gives the second cosine — hence ${\cos }^2\lambda$. This connects to Centripetal Force & Circular Motion.

9. Centripetal acceleration ${\omega }^{2}r$ — why spinning "steals" gravity

Now the geometry that produces the second cosine. The centripetal acceleration points toward the spin axis (horizontally, along the little circle's radius), but $g$ points toward Earth's centre (along the local "down"). At latitude $\lambda$ these two directions are not the same — the angle between them is exactly $\lambda$.

In the figure, the red arrow ${\omega }^{2}R\cos \lambda$ points horizontally toward the axis. To see how much of it lies along "down" (the line to Earth's centre), drop its shadow onto that line — that projection costs a factor $\cos \lambda$. So the amount subtracted from vertical $g$ is $({\omega }^{2}R\cos \lambda )\times \cos \lambda ={\omega }^{2}R{\cos }^2\lambda .$ The first $\cos \lambda$ came from the circle's radius $R\cos \lambda$ (item 8); the second $\cos \lambda$ comes from this projection onto the vertical. Two different cosines, two different reasons — that is the whole story behind ${\cos }^2\lambda$.

10. The binomial shortcut $(1+x{)}^n\approx 1+nx$

Why the topic needs it: the exact altitude ratio is ${\left(1+\frac{h}{R}\right)}^{-2}$. Since a mountain's height $h$ is tiny beside Earth's radius $R$, we set $x=h/R$ and get the clean $g_h\approx g(1-2h/R)$. That surviving 2 (from the exponent $-2$) is why altitude weakens $g$ twice as fast as depth.

11. Ratios — the trick used everywhere

Example the topic uses: $\frac{g_h}{g}=\frac{R^2}{(R+h{)}^2}$ — the $GM$ vanished, and only shapes remain.

Prerequisite map

(Alt text / reading order: Mass → Force/weight → g as a ratio. Separately, the inverse-square rule combines with g to give Newton's law and surface gravity. Height h and depth d feed the altitude and depth cases; the Shell Theorem plus density build the depth case; angular speed ω and centripetal acceleration build the latitude case; the binomial shortcut builds the altitude case. All three cases feed the topic "Variation of g".)

Equipment checklist

Cover the right side and see if you can state each from memory.

Connections

• Newton's Law of Universal Gravitation — the $1/r^2$ law all of this rests on.
• Shell Theorem — why outer shells give zero pull (needed for depth).
• Centripetal Force & Circular Motion — where the ${\omega }^{2}r$ term comes from.
• Weight vs Mass — the mass-vs-weight split built in items 1–2.
• Oblate Spheroid Earth — the shape effect that adds to rotation in the latitude case.
• Escape Velocity & Orbital Mechanics — same $GM/r^2$ machinery, applied to leaving Earth.
• Hinglish version →