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

WHAT we do: identify altitude, not depth. Going up uses the inverse-square law. $g_h\approx g\left(1-\frac{2h}{R}\right)⟹\frac{g-g_h}{g}=\frac{2h}{R}.$ WHY the factor 2: the $1/r^2$ shape, expanded for small $h$, brings down the exponent $2$. Answer: the fractional drop is $\frac{2h}{R}$ — i.e. twice the value $h/R$.

Altitude drop $=\frac{2x}{R}$, depth drop $=\frac{x}{R}$. $\frac{\text{altitude drop}}{\text{depth drop}}=\frac{2x/R}{x/R}=2.$ Answer: rising changes $g$ twice as much as descending the same distance. This is the "UP gets a 2, DOWN gets a 1" rule.

WHY exact, not approximate: here $h/R=0.5$, which is NOT $\ll 1$, so the linear approximation would be badly wrong. Use the full inverse-square form. $g_h=g{\left(1+\frac{h}{R}\right)}^{-2}=9.8{\left(1+0.5\right)}^{-2}=\frac{9.8}{(1.5{)}^2}=\frac{9.8}{2.25}.$ $g_h=4.356{\text{m/s}}^2\approx 4.36{\text{m/s}}^2.$ Check with approximation (to expose the error): the linear formula would give $9.8(1-1.0)=0$ — completely wrong. That's the whole point of using the exact form.

WHY exact works directly: depth is linear, no approximation needed at any depth. $g_d=g\left(1-\frac{d}{R}\right)=9.8\left(1-0.5\right)=9.8\times 0.5=4.9{\text{m/s}}^2.$ Answer: $g_d=4.9{\text{m/s}}^2$. Note that at half-radius depth, exactly half the gravity remains.

$\Delta g={\omega }^{2}R{\cos }^{2}0={\omega }^{2}R=(7.27\times 10^{-5}{)}^2(6.4\times 10^6).$ $=(5.285\times 10^{-9})(6.4\times 10^6)=0.0338{\text{m/s}}^2\approx 0.034{\text{m/s}}^2.$ Answer: rotation lowers $g$ by about $0.034{\text{m/s}}^2$ at the equator.

WHAT we invert: we want $g_h/g=1/2$, and $h/R$ is large, so use the exact form. ${\left(1+\frac{h}{R}\right)}^{-2}=\frac{1}{2}⟹{\left(1+\frac{h}{R}\right)}^2=2⟹1+\frac{h}{R}=\sqrt{2}.$ $h=R(\sqrt{2}-1)=6.4\times 10^6(0.4142)=2.65\times 10^6\text{m}.$ Answer: $h\approx 2.65\times 10^6\text{m}\approx 2650\text{km}$ above the surface.

$g\left(1-\frac{d}{R}\right)=\frac{g}{2}⟹1-\frac{d}{R}=\frac{1}{2}⟹d=\frac{R}{2}=3.2\times 10^6\text{m}.$ Comparison: to halve $g$ you must climb $2650$ km up but only dig $3200$ km down. Near the surface altitude is faster (the factor 2), but by half-$g$ the two have crossed because altitude uses the slowing inverse-square curve while depth stays linear. Answer: $d=3200\text{km}$, larger than $h=2650\text{km}$.

$1-\frac{2h}{R}=0.98⟹\frac{2h}{R}=0.02⟹h=0.01R.$ $h=0.01\times 6.4\times 10^6=6.4\times 10^4\text{m}=64\text{km}.$ Justified? $h/R=0.01\ll 1$, so the linear approximation is excellent (next term $\sim (h/R{)}^2=10^{-4}$, negligible). Answer: $h=64\text{km}$, approximation valid.

Step 1 — WHAT: altitude reduces $g$. $g_h=g\left(1-\frac{2h}{R}\right)=9.8\left(1-\frac{2(5000)}{6.4\times 10^6}\right)=9.8(1-0.0015625)=9.7847{\text{m/s}}^2.$ Step 2 — WHAT: rotation at the equator subtracts ${\omega }^2{R}^{\prime }$. Strictly the spin radius is $R+h$, but ${\omega }^{2}h$ is $\sim 3\times 10^{-8}$, utterly negligible, so use ${\omega }^{2}R=0.0338$: $g_{\text{eff}}=g_h-{\omega }^{2}R=9.7847-0.0338=9.7509{\text{m/s}}^2.$ Step 3 — apparent weight: $W=m{g}_{\text{eff}}=70\times 9.7509=682.6\text{N}.$ Answer: $g_{\text{eff}}\approx 9.751{\text{m/s}}^2$, weight $\approx 682.6\text{N}$ (vs $686\text{N}$ at sea level with no spin). See Weight vs Mass — mass stays $70$ kg throughout.

Altitude drop from Step 1: $\Delta g=9.8-9.7847=0.01531{\text{m/s}}^2$. Depth drop: $\Delta g=g\cdot \frac{d}{R}$. Set equal: $\frac{d}{R}=\frac{0.01531}{9.8}=0.0015625⟹d=0.0015625\times 6.4\times 10^6=10,000\text{m}.$ Interpretation: a $5000\text{m}$ climb equals a $10,000\text{m}$ dig — the depth is twice the height, the "UP gets a 2" rule in action ($2h=d$). Answer: $d=10\text{km}$.

WHY this matters: it is the hard physical limit of the rotation effect — you cannot make $g_{\lambda }$ negative. Set $g_{\text{eq}}=g-{\omega }_c^{2}R=0$: ${\omega }_c=\sqrt{\frac{g}{R}}=\sqrt{\frac{9.8}{6.4\times 10^6}}=\sqrt{1.531\times 10^{-6}}=1.2374\times 10^{-3}\text{rad/s}.$ $T_c=\frac{2\pi }{{\omega }_c}=\frac{6.2832}{1.2374\times 10^{-3}}=5078\text{s}\approx 1.41\text{hours}.$ Answer: ${\omega }_c\approx 1.24\times 10^{-3}\text{rad/s}$, day $\approx 1.41$ h. At this spin the required centripetal acceleration ${\omega }_c^{2}R$ exactly consumes all of true gravity — this is the equatorial orbit condition, closely tied to Escape Velocity & Orbital Mechanics and Centripetal Force & Circular Motion.

WHAT — build $g$ inside. By the Shell Theorem, at radius $r=R-d$ only the inner sphere pulls. Its mass with uniform density $\rho$: $M^{\prime }(r)=\rho \cdot \frac{4}{3}\pi r^3.$ WHY linear: the gravity there is $g(r)=\frac{G{M}^{\prime }(r)}{r^2}=\frac{G\rho \frac{4}{3}\pi r^3}{r^2}=\frac{4}{3}\pi G\rho r.$ The $r^3$ (mass) beats the $r^2$ (distance), leaving one clean power of $r$: $g\propto r$, a straight line, no approximation. Ratio to surface ($r=R$ gives surface $g$): $\frac{g(r)}{g}=\frac{r}{R}=\frac{R-d}{R}=1-\frac{d}{R}.■$ Boundary checks:

• $d=0$ (surface): $g_d=g(1-0)=g$. ✓
• $d=R$ (center): $g_d=g(1-1)=0$. ✓ — all mass is in shells above you, net pull zero.

Step 1 — undo rotation. At the equator the true (radial) gravity is $g_{\text{true}}=g_{\text{eq}}+{\omega }^{2}R=9.780+0.0338=9.8138{\text{m/s}}^2.$ Step 2 — pole has no rotation term (${\cos }^{2}90^{\circ }=0$), so if Earth were a sphere, $g_{\text{pole}}=g_{\text{true}}=9.814{\text{m/s}}^2$. Step 3 — why reality is bigger. Earth is an Oblate Spheroid Earth: the pole is closer to the center than the equator, so by $1/r^2$ the pole feels extra pull beyond the rotation restoration. That shape effect adds the remaining $\approx 9.832-9.814=0.018{\text{m/s}}^2$. Answer: rotation alone predicts $\approx 9.814{\text{m/s}}^2$; oblateness accounts for the further rise to $9.832$.

Halving $g$ by going up needs $h=?$ ::: $R(\sqrt{2}-1)\approx 2650$ km (use exact form, $h/R$ large) Halving $g$ by digging needs $d=?$ ::: $R/2=3200$ km (linear, exact) Equatorial rotation drop in $g$ ::: ${\omega }^{2}R\approx 0.034$ m/s${}^2$ Spin day that makes equatorial $g=0$ ::: $T_c=2\pi \sqrt{R/g}\approx 1.41$ h A 5 km climb equals how deep a dig for equal $\Delta g$? ::: 10 km (because $2h=d$)