Exercises — Escape velocity — derivation

$v_e=\sqrt{\frac{2GM}{R}}\text{and}{v}_e=\sqrt{2gR}.$ Escaping object ki mass $m$ cancel ho jaati hai kyunki dono — jo kinetic energy tumhe supply karni hai ($\frac{1}{2}m{v}_e^2$) aur jo potential energy tumhe pay karni hai ($GMm/R$) — dono $m$ ke proportional hain. Energy equation $\frac{1}{2}m{v}_e^2=GMm/R$ ko $m$ se divide karo aur $m$ chala jaata hai. (Dekho Conservation of mechanical energy.)

$v_e=\sqrt{2gR}$ use karo — yeh isliye choose kiya kyunki hume seedha $g$ aur $R$ diye gaye hain, toh $G$ ya $M$ ki zaroorat nahi. $v_e=\sqrt{2(4.0)(3.0\times 10^6)}=\sqrt{2.4\times 10^7}\approx 4.9\times 10^3\text{m/s}\approx 4.9\text{km/s}.$

Parent note se, $v_e=\sqrt{2}{v}_o$, toh $v_o=\frac{v_e}{\sqrt{2}}=\frac{11.2}{1.41}\approx 7.9\text{km/s}.$ Yahi low-Earth-orbit speed hai (dekho Orbital velocity & circular motion).

$v_e=\sqrt{2gR}$ se shuru karo aur $g$ ke liye solve karo (yeh unknown hai, toh hum ise isolate karte hain): $v_e^2=2gR\Rightarrow g=\frac{v_e^2}{2R}.$ $g=\frac{(2.4\times 10^3{)}^2}{2(1.74\times 10^6)}=\frac{5.76\times 10^6}{3.48\times 10^6}\approx 1.66{\text{m/s}}^2.$ Yeh Moon ki $g\approx 1.62{\text{m/s}}^2$ se match karta hai — thodi si rounding $2.4$ use karne ki wajah se hai instead of $2.37$ km/s.

Yahan hume $v_e$ aur $R$ pata hain lekin $g$ nahi, toh hum $G,M$ form use karte hain aur $M$ ke liye solve karte hain: $v_e=\sqrt{\frac{2GM}{R}}\Rightarrow v_e^2=\frac{2GM}{R}\Rightarrow M=\frac{v_e^{2}R}{2G}.$ $M=\frac{(5.0\times 10^3{)}^2(2.0\times 10^6)}{2(6.674\times 10^{-11})}=\frac{(2.5\times 10^7)(2.0\times 10^6)}{1.3348\times 10^{-10}}.$ $M=\frac{5.0\times 10^{13}}{1.3348\times 10^{-10}}\approx 3.75\times 10^{23}\text{kg}.$

$v_e=\sqrt{2(9.8)(6.4\times 10^6)}=\sqrt{1.2544\times 10^8}\approx 1.12\times 10^4\text{m/s}.$ Energy per kilogram ($\frac{1}{2}m{v}_e^2$ mein $m=1$ set karo): $\frac{K}{m}=\frac{1}{2}{v}_e^2=\frac{1}{2}(1.12\times 10^4{)}^2\approx 6.27\times 10^7\text{J/kg}.$ Matlab ek kilogram ko planet se bahar phenkhne ke liye $\approx 62.7\text{MJ}$ — yeh ek useful feel hai ki rockets itne bade kyun hote hain.

Figure ko algebra ke saath saath padhna. Figure mein, Planet A (blue) ki radius $R$ hai ek chhota yellow escape arrow $v_e$ length ka; Planet B (pink) ki radius $2R$ hai ek yellow escape arrow jo twice as long drawn hai, length $2v_e$. Woh arrow-length ratio exactly wahi hai jo neeche ki algebra produce karti hai — dekho kaise figure mein $R$ ka har factor formula mein ek factor mein convert hota hai.

Density link: mass $M=\rho \cdot \frac{4}{3}\pi R^3\propto R^3$. $v_e$ mein substitute karo: $v_e=\sqrt{\frac{2GM}{R}}\propto \sqrt{\frac{R^3}{R}}=\sqrt{R^2}=R.$ Toh $v_e\propto R$: escape arrow ki length planet ki radius ke proportional hai. Figure mein B ki radius A ki radius se $2\times$ drawn hai, toh uska escape arrow $2\times$ as long drawn hai — diagram is result ke liye ek faithful ruler hai. Radius double karne se $v_e$ double hoti hai: $\frac{v_{e,B}}{v_{e,A}}=2.$ $\sqrt{2}$ kyun nahi? Kyunki yahan $M$ fixed nahi hai — same density se bana hua bada planet kaafi heavier hota hai ($M\propto R^3$), aur woh extra mass badi radius par haavi ho jaati hai. (Agar hum circle ki radius ko arrow ko proportionally lambaa kiye bina bada kar dete, toh figure physics ke baare mein jhooth bol rahi hoti.)

$R$ fixed hone par, $v_e=\sqrt{2GM/R}\propto \sqrt{M}$. $M$ ko triple karne se: $\frac{v_e^{\prime }}{v_e}=\sqrt{3}\approx \mathrm{1.73.}$ Toh $v_e^{\prime }\approx 11.2\times 1.73\approx 19.4\text{km/s}$. (3.1 se contrast karo jahan radius ki jagah density fixed rakhi gayi thi.)

Yeh depend karta hai kya fixed rakha gaya hai — yahi poora point hai.

• X sahi hai jab density fixed ho (Problem 3.1): $M\propto R^3$, toh $v_e\propto R$ ⇒ $R$ double karne se $v_e$ double hoti hai (factor $2$).
• Y sahi hai jab mass $M$ fixed ho: tab $v_e=\sqrt{2GM/R}\propto 1/\sqrt{R}$. $R$ double karne se $v_e$ $1/\sqrt{2}\approx 0.71$ se multiply hoti hai — yeh decrease hoti hai, lekin exact factor $1/\sqrt{2}\approx 0.71$ hai, naa ki one-half. Koi bhi student jo kehta hai "$v_e$ half ho jaati hai" direction mein sahi hai lekin number mein galat.

Moral: tum nahi keh sakte $v_e$ $R$ ke saath kaise scale karti hai jab tak yeh nahi batate ki aur kya fixed rehta hai — aur tab bhi, "smaller" $1/\sqrt{2}$ hai, $1/2$ nahi.

Launch ($r=R$, speed $\frac{1}{2}{v}_e$) aur top ($r=R+h$, speed $0$) ke beech energy conservation use karo, is sheet ke top par stated sign convention $U(r)=-GMm/r$ ke saath: $\frac{1}{2}m{\left(\frac{1}{2}{v}_e\right)}^2-\frac{GMm}{R}=0-\frac{GMm}{R+h}.$ $m$ cancel karo; $v_e^2=2GM/R$ use karo toh ${\left(\frac{1}{2}{v}_e\right)}^2=\frac{1}{4}\cdot \frac{2GM}{R}=\frac{GM}{2R}$: $\frac{GM}{4R}-\frac{GM}{R}=-\frac{GM}{R+h}.$ (Left kinetic term hai $\frac{1}{2}\cdot \frac{GM}{2R}=\frac{GM}{4R}$.) $GM$ se divide karo: $\frac{1}{4R}-\frac{1}{R}=-\frac{1}{R+h}\Rightarrow -\frac{3}{4R}=-\frac{1}{R+h}\Rightarrow R+h=\frac{4R}{3}.$ $\boxed{h=\frac{R}{3}.}$ Half escape speed tumhe sirf ek-third Earth-radius upar le jaati hai — "infinity se halfway" se kaafi door, yeh prove karta hai ki $1/r$ potential mein speed linearly distance se map nahi hoti.

Orbital condition (gravity centripetal force provide karti hai, from Orbital velocity & circular motion): $\frac{GMm}{R^2}=\frac{m{v}_o^2}{R}\Rightarrow v_o^2=\frac{GM}{R}.$ Escape result: $v_e^2=\frac{2GM}{R}=2v_o^2$. Square roots lete hain: $v_e=\sqrt{2}{v}_o.■$ Yeh $M$ aur $R$ se independent hai — yeh har body ke liye hold karta hai, isliye ise yaad rakhna worth it hai.

$v_e=\sqrt{2GM/R}$ mein $v_e=c$ set karo aur radius ke liye solve karo, jo ab $R_s$ kehlata hai: $c=\sqrt{\frac{2GM}{R_s}}\Rightarrow c^2=\frac{2GM}{R_s}\Rightarrow R_s=\frac{2GM}{c^2}.$ Sun ke liye: $R_s=\frac{2(6.674\times 10^{-11})(1.99\times 10^{30})}{(3.0\times 10^8{)}^2}=\frac{2.656\times 10^{20}}{9.0\times 10^{16}}\approx 2.95\times 10^3\text{m}.$ Lagbhag 3 km. (Yeh Newtonian shortcut sahi General-Relativity answer par land karta hai — ek khush-naseeb coincidence. Dekho Black holes — Schwarzschild radius.)

Derivation ki logic unchanged hai — sirf starting radius different hai. Is sheet ke top par stated sign convention $U(r)=-GMm/r$ use karte hue, $r_0$ (speed $v_e^{\prime }$) se infinity (speed $0$, jahan $U=0$) tak energy conservation apply karo: $\frac{1}{2}m{v}_e^{\prime 2}-\frac{GMm}{r_0}=0\Rightarrow v_e^{\prime }(r_0)=\sqrt{\frac{2GM}{r_0}}=\sqrt{\frac{2GM}{R+H}}.$ Degenerate check ($H=0\Rightarrow r_0=R$): $v_e^{\prime }(R)=\sqrt{\frac{2GM}{R}}=v_e,$ surface value — jaisi honi chahiye. Reasoning: upar, tum pehle se kuch part well se bahar climb kar chuke ho, isliye kam speed chahiye — $v_e^{\prime }$ badhte $H$ ke saath shrink hoti hai, aur $\to 0$ jab $H\to \infty$.

Altitude result use karte hue, hum chahte hain $v_e^{\prime }(r_0)=\frac{1}{2}{v}_e$: $\sqrt{\frac{2GM}{r_0}}=\frac{1}{2}\sqrt{\frac{2GM}{R}}.$ Dono sides square karo (roots ek saath undo karo): $\frac{2GM}{r_0}=\frac{1}{4}\cdot \frac{2GM}{R}\Rightarrow \frac{1}{r_0}=\frac{1}{4R}\Rightarrow r_0=4R.$ Kyunki $r_0=R+H$: $H=4R-R=3R.$ Escape speed ko half hone ke liye tumhe surface se teen Earth-radii upar hona chahiye. Kyunki $v_e\propto 1/\sqrt{r}$, $v_e$ ko half karne ke liye $r$ ko quadruple karna padta hai — $1/\sqrt{r}$ scaling ka ek achha consistency check.

Step 1 — molecular speed compute karo. Diye gaye numbers $v_{rms}=\sqrt{3kT/m_{mol}}$ mein plug karo: $v_{rms}=\sqrt{\frac{3(1.38\times 10^{-23})(400)}{4.65\times 10^{-26}}}=\sqrt{\frac{1.656\times 10^{-20}}{4.65\times 10^{-26}}}=\sqrt{3.56\times 10^5}\approx 597\text{m/s}.$ Step 2 — escape speed se compare karo. Ratio $v_{rms}/v_e$ banao: $\frac{v_{rms}}{v_e}=\frac{597}{2370}\approx \mathrm{0.25.}$ Toh ek typical nitrogen molecule pehle se Moon ki escape speed ki ek-quarter par chal raha hai.

Step 3 — atmospheric-loss argument. Temperature koi ek speed nahi deti; yeh speeds ka poora spread deti hai, ek lambi high-speed tail ke saath jahan bahut saari molecules average $v_{rms}$ se kaafi tez chalti hain. Jab $v_{rms}$ $v_e$ ka $\frac{1}{4}$ jitna bada ho, woh fast tail regularly escape threshold $2.37\text{km/s}$ se upar pooch deti hai, toh molecules continuously leak away hoti rehti hain. Billions of years mein atmosphere poori tarah drain ho jaata hai — isliye Moon ka essentially koi atmosphere nahi hai. Contrast ke liye, Earth ka $v_e=11.2\text{km/s}$ is molecular speed se lagbhag $19\times$ hai ($11200/597\approx 18.8$), escape threshold ko negligible tail mein kaafi door rakhta hai — isliye Earth apni air rakhti hai. (Dekho Why the Moon has no atmosphere.)