Escape velocity — derivation: Visual walkthrough

1.2.23 · D2 · Physics › Newton's Laws & Dynamics › Escape velocity — derivation

Step 1 — Players se milo: ek mass ek pull mein

KYA HAI. Ek bade gole world ki picture banao jiska mass $M$ hai (maan lo Earth) aur ek choti object jiska mass $m$ hai (ek ball) jo world ke centre se $r$ distance par baithi hai. Do letters ka matlab:

$M$ = bade body ka mass jo pull kar raha hai (kilograms).
$m$ = choti body ka mass jo pull ho rahi hai.
$r$ = $M$ ke centre se ball tak ki straight-line distance.

YEH TEEN KYUN. Gravity ki strength sirf kitna stuff pull karta hai ( $M$ ), kitna stuff pull hota hai ( $m$ ), aur centres ke beech kitni doori hai ( $r$ ) par depend karti hai. Kuch aur nahi. Toh yeh teen hi complete cast hain.

PICTURE. Lal arrow pull hai. Notice karo yeh andar ki taraf point kar raha hai, centre ki taraf — gravity hamesha attract karti hai. Iska length $r$ badhne ke saath ghatta hai: dur hone par pull kamzor hoti hai.

Step 2 — Hum force se energy par kyun switch karte hain

KYA HAI. Hum pull $F$ ke baare mein baat karna band karne wale hain aur energy ke baare mein baat shuru karne wale hain. Do types:

Kinetic energy $K = \frac{1}{2} m v^{2}$ — move karne ki energy. Yahan $v$ ball ki speed hai.
Potential energy $U$ — pull mein position se stored energy, jo hum Step 3 mein banate hain.

TOOL KYU SWITCH KAREIN? Force $F$ har height par badalta hai (woh pareshan karne wala $1/ r^{2}$ ). Ek changing force ko $F = ma$ ke saath infinity tak poore climb mein track karna ek nightmare hai. Energy ise bypass karti hai: isse sirf start aur end se matlab hai, beech ki messy journey se nahi. Yeh "kya yeh kabhi wapas aayega?" wale sawaal ka sahi tool hai — ek start-vs-end sawaal.

PICTURE. Do bars: ek moving ball ka tall $K$ bar hai; climb karte waqt woh $K$ ko $U$ mein trade karta hai, jaise do glasses ke beech paani daalna. Total level rehta hai.

Step 3 — Bahar climb karke potential energy banana

KYA HAI. Potential energy $U (r)$ ko gravity ke against wo work ke roop mein define kiya jaata hai jo tumhe ball ko infinitely far away se andar distance $r$ tak le jaane ke liye karni padti hai. Hum natural zero choose karte hain: $U (\infty) = 0$ , kyunki infinitely apart do objects ka koi interaction nahi hota.

YEH CONVENTION KYUN. Zero ka matlab hona chahiye "koi interaction nahi." Infinity hi akela place hai jahan koi pull nahi, toh wahi $U = 0$ rehta hai. Har closer position uske relative measure hoti hai.

PICTURE — the well. $U$ ko $r$ ke against plot karo. Yeh ek valley hai: surface ke paas deep aur negative, $r \to \infty$ jaane par $0$ ki taraf rise karti hai. Well mein baithne ka matlab hai tum trapped ho — infinity par rim tak climb karne ke liye tumhare paas energy ka hona zaroori hai.

Gravity dwara kiya gaya work jab ball $\infty$ se andar $r$ tak move karti hai: $W_{grav} = \int_{\infty}^{r} (- \frac{GM m}{r ^{' 2}}) d r^{'}$ (pull andar ki taraf point karti hai, yaani decreasing $r^{'}$ ki taraf, isliye minus). Potential energy ko us work ke minus ke roop mein define kiya jaata hai, toh do minus signs combine ho jaate hain:

$U (r) = - W_{grav} = - \int_{\infty}^{r} (- \frac{GM m}{r ^{' 2}}) d r^{'} = \int_{\infty}^{r} \frac{GM m}{r ^{' 2}} d r^{'} = - \frac{GM m}{r}$

Term by term:

Integral sign ka bas matlab hai "har tiny slice of work ko add karo jaise ball move karti hai."
$r^{'}$ — ek moving dummy distance jo $\infty$ se $r$ tak sweep karta hai.
$- \frac{GM m}{r}$ mein final minus — $[- \frac{1}{r ^{'}}]_{\infty}^{r}$ evaluate karne se $- \frac{1}{r}$ milta hai, toh stored energy zero se neeche hai: mass bound hai.

Step 4 — Woh energy jo kabhi nahi badlati

KYA HAI. Launch ke baad, engine band karo. Sirf force gravity hai (koi air nahi, koi thrust nahi). Do energies ko ek single total $E$ mein add karo:

$E = K + U = \frac{1}{2} m v^{2} - \frac{GM m}{r}$

YEH CONSTANT KYUN HAI. Jab sirf force gravity ho, toh mechanical energy conserved hoti hai (Conservation of mechanical energy) — Step 2 se glasses-ke-beech-paani wali picture. Jaise ball rise karti hai, $K$ $U$ mein pour hota hai; jaise fall karti hai, $U$ wapas $K$ mein pour hota hai. Sum $E$ ek horizontal line hai.

PICTURE. Ek hi flight ke do moments. Surface par $K$ bar tall hai aur $U$ bar ek deep negative pit hai. Upar $K$ bar shrink ho gaya hai aur $U$ pit shallower hai. Stacked total — dashed line — dono mein identical hai.

Step 5 — "Bas barely" condition

KYA HAI. Minimum launch speed ka matlab kya hai? Matlab ball infinity tak exactly kuch bhi bacha nahi kar pahunche — final speed $0$ . Koi bhi spare speed matlab hum zaroori se zyada zyada se launch kiye.

FINAL SPEED = 0 KYUN, "ABHI BHI MOVE KAR RAHA" NAHI. Infinity par pull fad kar zero ho jaati hai (woh $1/ r^{2} \to 0$ phir se). Toh ek ball jo infinity par pahunchti hai — chahe zero speed par crawl kar rahi ho — free hai; gravity ab ise wapas nahi khich sakti. $v = 0$ par pahunchna escape karne aur wapas girne ke beech ki razor's edge hai.

PICTURE. Ek hi surface se teen flights:

Slow (magenta): climb karti hai, finite height par rukti hai, wapas girती hai — ek bound arc.
Just-right (violet): escape speed; forever coast karti hai, iska speed picture ke edge par zero ho jaata hai.
Fast (orange): spare speed ke saath escape karti hai.

Toh infinity par, just-right flight ke liye: $K_{f} = 0, U_{f} = - \frac{GM m}{\infty} = 0 \Rightarrow E_{f} = 0.$

Step 6 — Start aur end ko equate karo, dekho $m$ gayab ho jaata hai

KYA HAI. Energy constant hai, toh launch par total infinity par total ke barabar hai:

$launch KE \frac{1}{2} m v_{e}^{2} - surface par launch PE r = R \frac{GM m}{R} = E_{f} 0$

Yahan $R$ = world ka radius (ball surface par start karti hai, toh $r = R$ ), aur $v_{e}$ = escape speed jo hum dhundh rahe hain.

YEH BEAUTIFUL KYUN HAI. Har single term mein $m$ ka factor hai. Puri line ko $m$ se divide karo aur yeh gayab ho jaata hai:

$\frac{1}{2} v_{e}^{2} = \frac{GM}{R}$

PICTURE. Ek pebble aur ek spaceship, side by side, dono ko identical launch speed chahiye. Unke $K$ aur $U$ bars alag heights ke hain (spaceship ke huge hain) lekin speed fix karne wala ratio same hai, toh required $v_{e}$ identical hai.

$v_{e}$ ke liye $2$ se multiply aur square root leke solve karo:

Step 7 — Degenerate & edge cases (koi gap mat chhodna)

KYA HAI. Formula ko extreme par push karo aur check karo ki picture abhi bhi sense banati hai.

Case A — Choti duniya (small $M$ ya huge $R$ ): shallow well. $v_{e} = 2 GM / R$ shrink hoti hai. Moon ka well itna shallow hai ki gas molecules routinely $2.4$ km/s beat karti hain aur leak ho jaati hain — yahi hai Why the Moon has no atmosphere.

Case B — Seedha upar launch vs. sideways. Energy sirf speed dekhti hai, direction nahi. Threshold identical hai — escape velocity actually ek speed hai, true velocity nahi. Direction path change karta hai, kabhi bhi needed energy nahi.

Case C — World ko compress karo (fixed $M$ par $R$ shrink karo). $v_{e}$ bina limit ke climb karta hai. $R$ itna squeeze karo ki $v_{e}$ light ki speed $c$ tak pahunch jaaye, aur light bhi escape nahi kar sakti: tumne black hole bana liya, aur woh $R$ uska Black holes — Schwarzschild radius hai (set $v_{e} = c$ ).

Case D — Circular orbit se compare karo. Surface par orbital speed $v_{o} = GM / R$ hai (Orbital velocity & circular motion). Tab $\frac{v _{e}}{v _{o}} = \frac{2 GM / R}{GM / R} = 2 .$ Escape ke liye sirf $2 \approx 1.41$ guna orbit speed chahiye.

PICTURE. Ek "well depth" axis par teen worlds: Moon (shallow, low $v_{e}$ ), Earth (medium), compressed star (near-vertical wall, $v_{e} \to c$ ). Rim height hi escape energy hai.

Ek-picture summary

Upar sab kuch ek single diagram mein collapse ho jaata hai: well $U (r) = - GM m / r$ , ek launch KE bar jo exactly well ko rim tak fill karta hai, aur coasting curve jo infinity par zero ho jaati hai.

Recall Feynman retelling — poora walkthrough plain words mein

Earth ek bowl hai aur tum neeche ek marble ho. Pehle humne pull draw ki (Step 1): ek rope tumhe centre ki taraf khichti hai, jitna bahar utni kamzor — aur hum ne agree kiya ki bowl ko hold karenge jabki marble move kare, jo theek hai kyunki marble itni light hai ki bowl barely react karta hai. Changing rope ko chase karna mushkil hai, toh hum energy par switch kiye — moving-energy aur stored-energy jo ek doosre mein pour hoti hain jaise do glasses ke beech paani (Step 2). Humne stored-energy ko ek valley ki tarah measure kiya, surface par sab se deep aur dur ke rim par zero tak rise karti hai (Step 3). Kyunki sirf gravity act karti hai, total energy ek flat line hai: climb karo aur tumhari moving-energy well mein empty ho jaati hai, giro aur wapas fill ho jaati hai (Step 4). Sab se sasta escape woh flick hai jo marble ko bilkul rim ko zero speed se kiss karne de — final speed zero hai (Step 5). Start-total = rim-total set karne par, marble ka apna weight dono sides se cancel ho jaata hai — toh ek pebble aur ek bus ko same flick chahiye (Step 6). Nikalta hai $v_{e} = 2 GM / R = 2 g R$ : Earth ke liye lagbhag 11 km/s, shallow Moon ke liye 2.4, aur — bowl ko itna chota squeeze karo — rim light ki speed tak rise karti hai aur tumne black hole bana liya (Step 7).

Active recall

Recall Pictures test karo

Humne force ko energy ke liye kyun chhodha? ::: Force $r$ ke saath badlti hai ( $1/ r^{2}$ ); energy ko sirf start aur end chahiye, "kya yeh wapas aayega?" wale sawaal ke liye perfect. Step 3 mein, surface ke paas $U$ negative kyun hai? ::: Well $U (\infty) = 0$ rim se neeche dip karti hai, toh trapped rehna matlab negative stored energy hona. Step 5 mein, subscript $f$ ka matlab kya hai, aur final speed zero kyun hai? ::: $f$ = "final", $r \to \infty$ par; wahan pull zero hai, toh $v = 0$ par bhi pahunchna matlab gravity kabhi wapas nahi khich sakti — minimum case. Step 6 mein, $m$ kyun cancel hota hai? ::: Har term mein $m$ ka factor hai; divide karne se yeh hat jaata hai, toh sabhi masses ko same $v_{e}$ chahiye. $+ m$ (reduced mass) correction kab chahiye? ::: Jab do masses comparable hon; tab centre-of-mass frame mein $v_{e} = 2 G (M + m) / R$ . $m ≪ M$ ke liye yeh negligible hai. $R$ compress karna black hole kaise banata hai (Step 7)? ::: $R$ shrink karna $v_{e} = 2 GM / R$ raise karta hai; jab $v_{e} = c$ , light bhi escape nahi kar sakti.

Visual walkthrough — Escape velocity — derivation