Escape velocity — derivation — Hinglish notes

1.2.23 · Physics › Newton's Laws & Dynamics

HUM KYA DHUNDH RAHE HAIN?

YEH $m$ par kyun depend nahi karta? Jо kinetic energy tum supply karte ho ( $\propto m$ ) aur jо gravitational energy tumhe pay karni padti hai ( $\propto m$ ), dono $m$ ke saath scale karti hain, isliye $m$ cancel ho jaata hai. Ek kankar aur ek spaceship dono ko same speed chahiye.

HOW to derive it — energy method (first principles)

Step 1 — Gravitational potential energy (WHY this form). $M$ ke centre se $r$ door $m$ par Newton ki gravity force: $F (r) = \frac{GM m}{r ^{2}} (attractive, points inward)$

PE woh work hai jo $m$ ko infinity se $r$ tak gravity ke against laane mein kiya jaata hai. $U (\infty) = 0$ choose karo (natural zero: jab infinitely alag hain toh koi interaction nahi). Tab

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

Yeh step kyun? Andar move karte waqt, gravity positive work karti hai, isliye system PE khota hai → $U$ negative ho jaati hai. Evaluate karo:

$U (r) = - GM m [- \frac{1}{r ^{'}}]_{\infty}^{r} = - \frac{GM m}{r}$

Step 2 — Total mechanical energy conserve karo. Koi air nahi, launch ke baad koi engine nahi ⇒ sirf gravity act karti hai ⇒ mechanical energy conserved hai: $E = K + U = \frac{1}{2} m v^{2} - \frac{GM m}{r} = constant$

Step 3 — Boundary conditions apply karo. Launch par ( $r = R$ , speed $v_{e}$ ): $E_{i} = \frac{1}{2} m v_{e}^{2} - \frac{GM m}{R}$ Just-escaping limit par ( $r \to \infty$ , speed $\to 0$ ): $E_{f} = 0 - 0 = 0$

Final speed = 0 kyun set karte hain? "Minimum" speed ka matlab hai koi energy waste nahi — tum barely pahunchte ho.

Step 4 — Equate karo aur solve karo. $\frac{1}{2} m v_{e}^{2} - \frac{GM m}{R} = 0 \Rightarrow \frac{1}{2} v_{e}^{2} = \frac{GM}{R}$

Worked examples

Common mistake Steel-manned common errors

(1) "Escape velocity rocket ki mass par depend karti hai." Kyun sahi lagta hai: bhaari cheezein "throw karna mushkil" lagti hain. Fix: KE aur PE dono $m$ ke saath scale karte hain, isliye $\frac{1}{2} m v^{2} = GM m / R$ mein $m$ cancel ho jaata hai. Sabhi masses ke liye required speed same hoti hai.

(2) "Infinity par object abhi bhi move kar raha hona chahiye, warna wapas gir jaayega." Kyun sahi lagta hai: zero speed ka matlab "ruk gaya → gir gaya" lagta hai. Fix: gravity $\propto 1/ r^{2} \to 0$ infinity par, isliye $r = \infty$ par $v = 0$ ke saath pahunchna matlab woh kabhi return nahi karega. Yahi minimum case hai.

(3) "Escape velocity ek velocity hai — direction matter karta hai." Kyun sahi lagta hai: word mein "velocity" hai. Fix: yeh actually ek speed hai (energy condition). Direction sirf atmosphere mein path/efficiency change karta hai, energy threshold nahi (air aur rotation ignore karke).

(4) Sign error: $U = + GM m / r$ likhna. Fix: bound systems mein $U (\infty) = 0$ convention ke saath negative PE hoti hai. Agar minus bhool gaye, toh tumhein imaginary $v_{e}$ milegi.

Recall Feynman: ek 12-saal ke bachche ko samjhao

Socho Earth ek giant bowl hai aur tum ek marble ho usme. Agar marble ko halka flick karo toh woh thoda upar chadh ke wapas aa jaata hai. Zyada tez flick karo toh aur upar jaata hai. Ek perfect flick speed hoti hai jis par marble bas bahut upar edge tak (infinitely dur) pahunchta hai aur wahan ruk jaata hai wapas roll karne ki bajaye. Wohi magic speed escape velocity hai — Earth par lagbhag 11 km/s, jo awaaz ki speed se 30 guna zyada tez hai! Ek marble ke liye aur ek school bus ke liye same hai, kyunki bhaari cheezein push dono zaroori hoti hain aur milti bhi hain proportion mein.

Active recall

Escape velocity definition

Surface par minimum speed jo infinity tak pahunchne ke liye chahiye jahan final speed zero ho, kabhi wapas na gire.

Gravitational PE with U(∞)=0

U (r) = - \frac{GM m}{r}

(negative ⇒ bound).

Escape velocity formula (G, M, R)

v_{e} = \frac{2 GM}{R}

Escape velocity in terms of g, R

v_{e} = 2 g R

, using

GM = g R^{2}

Why is

v_{e}

independent of escaping mass

m

KE aur PE dono

\propto m

hain, isliye energy equation mein

m

cancel ho jaata hai.

Earth's escape velocity value

≈ 11.2 km/s.

Moon's escape velocity (and why small)

≈ 2.4 km/s; chhota

g

aur

R

⇒ shallow gravity well.

Relation between escape & orbital speed

v_{e} = 2 v_{o} \approx 1.41 v_{o}

Boundary condition used in the derivation

r \to \infty

par

K = 0

aur

U = 0

, isliye total

E = 0

If planet mass doubles (same R), how does

v_{e}

change?

2

(kyunki

v_{e} \propto M

), ×2 nahi.

Connections

Gravitational potential energy
Newton's law of universal gravitation
Conservation of mechanical energy
Orbital velocity & circular motion
Black holes — Schwarzschild radius ( $v_{e} = c$ set karo)
Why the Moon has no atmosphere

HUM KYA DHUNDH RAHE HAIN?

HOW to derive it — energy method (first principles)

Worked examples

Active recall

Connections

Concept Map