Exercises — Gravitational potential energy — U = −GMm - r (not mgh)

WHAT hum karte hain: $-GMm/r$ ko do radii par compare karo — koi numbers nahi chahiye, bas $1/r$ ki shape dekho. WHY: bade $r$ se divide karne par $GMm/r$ size mein chhota ho jaata hai, aur aage minus hone ki wajah se smaller-size number zero ke kareeb hota hai, matlab number line par bada hota hai.

• $r_1=8000$ km par: chhota $r$ → bada $GMm/r$ → bahut negative $U$.
• $r_2=20000$ km par: bada $r$ → chhota $GMm/r$ → kam negative $U$.

Toh bahar jaane par, $U$ increase hoti hai (zero ki taraf upar jaati hai). $r_1$ (8000 km) wali value zyada negative hai. Picture: number line par, $r_1$ zero se zyada neeche baitha hai; $r_2$ zero ke kareeb hai. Tumne satellite ko upar uthaya, toh uski energy badi. ✔

Tay karne wala sawaal hamesha yeh hota hai: kya motion ke dauran $g$ effectively constant rehta hai? $mgh$ maanta hai ki field uniform hai; yeh $-GMm/r$ ka small-$h$ approximation hi hai.

• (a) $h=2\text{ m}\ll R_{\oplus }$ → $g$ constant → $mgh$ theek hai.
• (b) height $R_{\oplus }$ ke comparable aur bahut aage tak → $g$ bahut zyada gir jaata hai → $-GMm/r$ use karna zaroori hai.
• (c) $h=10\text{ m}\ll R_{\oplus }$ → $mgh$ theek hai.

Rule of thumb: agar $h$ kuch kilometres ya kam hai, $mgh$ use karo; agar yeh planet radius ka bada fraction ya zyada hai, $-GMm/r$ use karo. Dekho Acceleration due to gravity g.

WHY yeh formula: escape ka matlab hai "barely infinity tak zero speed se pahunchna", matlab total energy $=0$: $\frac{1}{2}m{v}^2-GMm/R=0$. $m$ cancel ho jaata hai, bacha $v_{\text{esc}}=\sqrt{2GM/R}$ (dekho Escape Velocity). $v_{\text{esc}}=\sqrt{\frac{2(6.67\times 10^{-11})(7.35\times 10^{22})}{1.74\times 10^6}}.$ Andar numerator: $2\cdot 6.67\times 10^{-11}\cdot 7.35\times 10^{22}=9.805\times 10^{12}$. $1.74\times 10^6$ se divide karo: $5.635\times 10^6$. Square root: $\approx 2.37\times 10^3\text{m/s}$. $\boxed{v_{\text{esc}}\approx 2.37\text{km/s}}$ Yeh Earth ke $11.2$ km/s ka almost paanchwan hissa hai — Moon ka shallower gravity well climb karna aasaan hai.

WHAT: $U=-GMm/r$ use karo jahan $r=R_{\oplus }+h$ hai. $r=6.37\times 10^6+0.41\times 10^6=6.78\times 10^6\text{m}.$ $U=-\frac{(6.67\times 10^{-11})(5.97\times 10^{24})(4.2\times 10^5)}{6.78\times 10^6}.$ Numerator: $6.67\times 10^{-11}\cdot 5.97\times 10^{24}=3.982\times 10^{14}$; times $4.2\times 10^5=1.673\times 10^{20}$. $6.78\times 10^6$ se divide karo: $2.467\times 10^{13}$. $\boxed{U\approx -2.47\times 10^{13}\text{J}}$ Negative, jaisa ki har bound object ka hona chahiye.

WHY energy conservation: gravity conservative hai, toh $K_i+U_i=K_f+U_f$.

• Start: rest par ($K_i=0$) $r\to \infty$ par ($U_i=0$). Toh total energy $E=0$.
• End: $r=R_{\oplus }$ par, $U_f=-GMm/R$, $K_f=\frac{1}{2}m{v}^2$.

$E=0$ set karo: $\frac{1}{2}m{v}^2-\frac{GMm}{R}=0\Rightarrow v=\sqrt{\frac{2GM}{R}}.$ Yeh bilkul escape speed hi hai — infinity se girna waqt ka ulta hai barely escape karne ka. Earth ke liye: $v=\sqrt{\frac{2(6.67\times 10^{-11})(5.97\times 10^{24})}{6.37\times 10^6}}\approx 1.12\times 10^4\text{m/s}=\boxed{11.2\text{km/s}}.$

Exact: surface ($r=R$) aur top ($r=R+h$, rest par) ke beech energy conservation: $\frac{1}{2}m{v}^2-\frac{GMm}{R}=0-\frac{GMm}{R+h}\Rightarrow v=\sqrt{2GM\left(\frac{1}{R}-\frac{1}{R+h}\right)}.$ $R=6.37\times 10^6$, $R+h=6.67\times 10^6$ ke saath: $\frac{1}{R}-\frac{1}{R+h}=1.5699\times 10^{-7}-1.4993\times 10^{-7}=7.06\times 10^{-9}.$ $v=\sqrt{2(3.982\times 10^{14})(7.06\times 10^{-9})}=\sqrt{5.62\times 10^6}\approx 2.37\times 10^3\text{m/s}.$ Approximate: $v=\sqrt{2gh}=\sqrt{2(9.8)(3\times 10^5)}=\sqrt{5.88\times 10^6}\approx 2.42\times 10^3$ m/s. $\boxed{v_{\text{exact}}\approx 2.37\text{km/s},v_{mgh}\approx 2.42\text{km/s}}$ $mgh$ ka answer ~$2\%$ zyada hai kyunki yeh maanta hai ki $g$ poore upar tak $9.8$ rehta hai, lekin $g$ actually height ke saath kamzor hota hai, isliye actually kam speed chahiye.

WHY $E=-GMm/2r$: circular orbit mein, gravity centripetal force supply karti hai, jo force karti hai $K=\frac{1}{2}|U|$, toh total energy $E=K+U=-GMm/2r$ hai (parent note aur Kepler's Laws & Orbital Energy). Energy chahiye $=E_2-E_1$: $\Delta E=-\frac{GMm}{2r_2}-\left(-\frac{GMm}{2r_1}\right)=\frac{GMm}{2}\left(\frac{1}{r_1}-\frac{1}{r_2}\right).$ $GMm=(6.67\times 10^{-11})(5.97\times 10^{24})(1200)=4.778\times 10^{17}$. $\frac{1}{r_1}-\frac{1}{r_2}=1.4286\times 10^{-7}-1.0\times 10^{-7}=4.286\times 10^{-8}.$ $\Delta E=\frac{1}{2}(4.778\times 10^{17})(4.286\times 10^{-8})\approx 1.02\times 10^{10}\text{J}.$ $\boxed{\Delta E\approx 1.02\times 10^{10}\text{J}>0}$ Positive: higher orbit par jaane ke liye energy daalni padti hai — parent ke Forecast-then-Verify se match karta hai. ✔

WHAT: radial fall, energy conserved $r_1=R+500$ km aur $r_2=R+100$ km ke beech, rest se shuru. $0-\frac{GMm}{r_1}=\frac{1}{2}m{v}^2-\frac{GMm}{r_2}\Rightarrow v=\sqrt{2GM\left(\frac{1}{r_2}-\frac{1}{r_1}\right)}.$ $r_1=6.87\times 10^6$, $r_2=6.47\times 10^6$. $\frac{1}{r_2}-\frac{1}{r_1}=1.5456\times 10^{-7}-1.4556\times 10^{-7}=9.00\times 10^{-9}.$ $v=\sqrt{2(3.982\times 10^{14})(9.00\times 10^{-9})}=\sqrt{7.168\times 10^6}\approx 2.68\times 10^3\text{m/s}.$ $\boxed{v\approx 2.68\text{km/s}}$ (Ek quick $mgh$ estimate $g\approx 9.8$ se $400$ km par deta hai $\sqrt{2\cdot 9.8\cdot 4\times 10^5}\approx 2.80$ km/s — kuch percent zyada kyunki wahan $g$ chhota hota hai.)

Step 1 — exact form factor karo: $\Delta U=\frac{GMm}{R^2}\cdot \frac{h}{1+h/R}=mgh\cdot \frac{1}{1+h/R},\text{since }g=\frac{GM}{R^2}.$ WHY yeh shape: yeh pure $mgh$ aur ek correction factor $\frac{1}{1+h/R}$ ko alag karta hai. Step 2 — limit lo: jaise $h/R\to 0$, factor $\to 1$ ho jaata hai, deta hai $\Delta U\to mgh$. Yahi hai schoolbook formula, recover ho gayi. Step 3 — error ka size: true value $mgh$ se factor $\frac{1}{1+h/R}$ se chhoti hai. Fractional error $\approx \frac{h}{R}$ (from $\frac{1}{1+x}\approx 1-x$). $h=100$ km par: $\frac{h}{R}=\frac{1.0\times 10^5}{6.37\times 10^6}\approx 0.0157=\boxed{1.6\%}.$ Toh $mgh$ 100 km par lagbhag $1.6\%$ overestimate karta hai — L3 ke comparisons se consistent hai. Dekho Conservative Forces and Potential Energy aur Work-Energy Theorem.

Step 1 — force balance. Gravity centripetal force hai: $\frac{GMm}{r^2}=\frac{m{v}^2}{r}\Rightarrow m{v}^2=\frac{GMm}{r}.$ Step 2 — kinetic energy. $K=\frac{1}{2}m{v}^2=\frac{GMm}{2r}$. Step 3 — $U$ se compare karo. Kyunki $U=-\frac{GMm}{r}$, hume milta hai $K=-\frac{1}{2}U$. ✔ Step 4 — total energy. $E=K+U=\frac{GMm}{2r}-\frac{GMm}{r}=-\frac{GMm}{2r}=\frac{1}{2}U=-K.$ Interpretation: kinetic energy exactly well ki depth ki aadhi hai. Yeh inverse-square force ke liye virial theorem hai. Physically: ek bound orbit energy ke hisaab se bottom se adha "upar" baithti hai, isliye $E<0$ (bound) hone ke bawajood object move karta hai.

Bade $r$ par $U$ ka sign? ::: Zero ki taraf increase hota hai (kam negative). $U=-GMm/r$ mein $r$ kahan se measure hota hai? ::: Planet ke centre se, toh $r=R+h$. Infinity se girne ki impact speed barabar hai…? ::: Escape speed ke, $\sqrt{2GM/R}$. Circular orbit ki total energy? ::: $E=-GMm/2r=\frac{1}{2}U=-K$. $mgh$ ka height $h$ par fractional error? ::: Lagbhag $h/R$ (yeh over-estimate karta hai).