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

KYA karte hain: altitude identify karo, depth nahi. Upar jaane par inverse-square law use hoti hai. $g_h\approx g\left(1-\frac{2h}{R}\right)⟹\frac{g-g_h}{g}=\frac{2h}{R}.$ 2 ka factor KYU aata hai: $1/r^2$ ka shape, chhote $h$ ke liye expand karne par, exponent $2$ neeche le aata hai. Answer: fractional drop $\frac{2h}{R}$ hai — yaani $h/R$ ki value se do guna.

Altitude drop $=\frac{2x}{R}$, depth drop $=\frac{x}{R}$. $\frac{\text{altitude drop}}{\text{depth drop}}=\frac{2x/R}{x/R}=2.$ Answer: upar jaana $g$ ko do guna zyada change karta hai utne hi distance neeche jaane se. Yeh "UP gets a 2, DOWN gets a 1" rule hai.

KYU exact use karein, approximate nahi: yahan $h/R=0.5$ hai, jo $\ll 1$ NAHI hai, isliye linear approximation bahut galat hogi. Poora inverse-square form use karo. $g_h=g{\left(1+\frac{h}{R}\right)}^{-2}=9.8{\left(1+0.5\right)}^{-2}=\frac{9.8}{(1.5{)}^2}=\frac{9.8}{2.25}.$ $g_h=4.356{\text{m/s}}^2\approx 4.36{\text{m/s}}^2.$ Approximation se check karo (galti dikhane ke liye): linear formula deta $9.8(1-1.0)=0$ — bilkul galat. Yahi reason hai exact form use karne ka.

KYU exact directly kaam karta hai: depth linear hai, kisi bhi depth par koi approximation ki zaroorat nahi. $g_d=g\left(1-\frac{d}{R}\right)=9.8\left(1-0.5\right)=9.8\times 0.5=4.9{\text{m/s}}^2.$ Answer: $g_d=4.9{\text{m/s}}^2$. Note karo ki half-radius depth par, exactly aadhi gravity bachti hai.

$\Delta g={\omega }^{2}R{\cos }^{2}0={\omega }^{2}R=(7.27\times 10^{-5}{)}^2(6.4\times 10^6).$ $=(5.285\times 10^{-9})(6.4\times 10^6)=0.0338{\text{m/s}}^2\approx 0.034{\text{m/s}}^2.$ Answer: rotation equator par $g$ ko lagbhag $0.034{\text{m/s}}^2$ kam kar deta hai.

KYA invert karte hain: hum chahte hain $g_h/g=1/2$, aur $h/R$ bada hai, isliye exact form use karo. ${\left(1+\frac{h}{R}\right)}^{-2}=\frac{1}{2}⟹{\left(1+\frac{h}{R}\right)}^2=2⟹1+\frac{h}{R}=\sqrt{2}.$ $h=R(\sqrt{2}-1)=6.4\times 10^6(0.4142)=2.65\times 10^6\text{m}.$ Answer: $h\approx 2.65\times 10^6\text{m}\approx 2650\text{km}$ surface se upar.

$g\left(1-\frac{d}{R}\right)=\frac{g}{2}⟹1-\frac{d}{R}=\frac{1}{2}⟹d=\frac{R}{2}=3.2\times 10^6\text{m}.$ Comparison: $g$ ko aadha karne ke liye $2650$ km upar chadhna padta hai lekin sirf $3200$ km neeche khodna padta hai. Surface ke paas altitude faster hai (factor 2 ki wajah se), lekin half-$g$ tak pahunchte-pahunchte dono cross kar jaate hain kyunki altitude inverse-square curve use karti hai jo dheemi padti jaati hai jabki depth linear rehti hai. Answer: $d=3200\text{km}$, $h=2650\text{km}$ se bada.

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

Step 1 — KYA: altitude $g$ reduce karta hai. $g_h=g\left(1-\frac{2h}{R}\right)=9.8\left(1-\frac{2(5000)}{6.4\times 10^6}\right)=9.8(1-0.0015625)=9.7847{\text{m/s}}^2.$ Step 2 — KYA: equator par rotation ${\omega }^2{R}^{\prime }$ subtract karta hai. Strictly spin radius $R+h$ hai, lekin ${\omega }^{2}h$ lagbhag $3\times 10^{-8}$ hai, bilkul negligible, isliye ${\omega }^{2}R=0.0338$ use karo: $g_{\text{eff}}=g_h-{\omega }^{2}R=9.7847-0.0338=9.7509{\text{m/s}}^2.$ 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 $686\text{N}$ sea level par bina spin ke). Dekho Weight vs Mass — mass poore time $70$ kg rehta hai.

Step 1 se altitude drop: $\Delta g=9.8-9.7847=0.01531{\text{m/s}}^2$. Depth drop: $\Delta g=g\cdot \frac{d}{R}$. Equal set karo: $\frac{d}{R}=\frac{0.01531}{9.8}=0.0015625⟹d=0.0015625\times 6.4\times 10^6=10,000\text{m}.$ Interpretation: $5000\text{m}$ ka climb $10,000\text{m}$ ki khudaai ke barabar hai — depth do guni height hai, "UP gets a 2" rule action mein ($2h=d$). Answer: $d=10\text{km}$.

KYU yeh matter karta hai: yeh rotation effect ki hard physical limit hai — $g_{\lambda }$ negative nahi ho sakta. Set karo $g_{\text{eq}}=g-{\omega }_c^{2}R=0$: ${\omega }_c=\sqrt{\frac{g}{R}}=\sqrt{\frac{9.8}{6.4\times 10^6}}=\sqrt{1.531\times 10^{-6}}=1.2374\times 10^{-3}\text{rad/s}.$ $T_c=\frac{2\pi }{{\omega }_c}=\frac{6.2832}{1.2374\times 10^{-3}}=5078\text{s}\approx 1.41\text{hours}.$ Answer: ${\omega }_c\approx 1.24\times 10^{-3}\text{rad/s}$, day $\approx 1.41$ h. Is spin par required centripetal acceleration ${\omega }_c^{2}R$ poori true gravity consume kar leta hai — yeh equatorial orbit condition hai, closely tied to Escape Velocity & Orbital Mechanics aur Centripetal Force & Circular Motion.

KYA — andar $g$ build karo. Shell Theorem se, radius $r=R-d$ par sirf inner sphere pull karta hai. Uniform density $\rho$ ke saath uski mass: $M^{\prime }(r)=\rho \cdot \frac{4}{3}\pi r^3.$ KYU linear: wahan gravity hai $g(r)=\frac{G{M}^{\prime }(r)}{r^2}=\frac{G\rho \frac{4}{3}\pi r^3}{r^2}=\frac{4}{3}\pi G\rho r.$ $r^3$ (mass) $r^2$ (distance) ko beat karta hai, ek clean power of $r$ bachti hai: $g\propto r$, ek straight line, koi approximation nahi. Surface se ratio ($r=R$ surface $g$ deta hai): $\frac{g(r)}{g}=\frac{r}{R}=\frac{R-d}{R}=1-\frac{d}{R}.■$ Boundary checks:

• $d=0$ (surface): $g_d=g(1-0)=g$. ✓
• $d=R$ (center): $g_d=g(1-1)=0$. ✓ — saari mass tumhare upar shells mein hai, net pull zero.

Step 1 — rotation undo karo. Equator par true (radial) gravity hai $g_{\text{true}}=g_{\text{eq}}+{\omega }^{2}R=9.780+0.0338=9.8138{\text{m/s}}^2.$ Step 2 — pole par koi rotation term nahi (${\cos }^{2}90^{\circ }=0$), toh agar Earth sphere hoti, $g_{\text{pole}}=g_{\text{true}}=9.814{\text{m/s}}^2$. Step 3 — reality zyada kyun hai. Earth ek Oblate Spheroid Earth hai: pole center ke zyada paas hai equator se, isliye $1/r^2$ se pole rotation restoration ke upar extra pull feel karta hai. Woh shape effect baaki $\approx 9.832-9.814=0.018{\text{m/s}}^2$ add karta hai. Answer: sirf rotation $\approx 9.814{\text{m/s}}^2$ predict karta hai; oblateness $9.832$ tak ka aagla rise account karta hai.

Upar jaake $g$ aadha karne ke liye $h=?$ chahiye ::: $R(\sqrt{2}-1)\approx 2650$ km (exact form use karo, $h/R$ bada hai) Khood ke $g$ aadha karne ke liye $d=?$ chahiye ::: $R/2=3200$ km (linear, exact) Equatorial rotation mein $g$ ki drop ::: ${\omega }^{2}R\approx 0.034$ m/s${}^2$ Woh spin day jo equatorial $g=0$ kare ::: $T_c=2\pi \sqrt{R/g}\approx 1.41$ h 5 km ki climb equal $\Delta g$ ke liye kitni deep khudaai ke barabar hai? ::: 10 km (kyunki $2h=d$)