Escape velocity — derivation: Worked examples — Hinglish deep dive

1.2.23 · D3 · Physics › Newton's Laws & Dynamics › Escape velocity — derivation

Everything here rests on the parent: Escape velocity — derivation, aur Gravitational potential energy, Newton's law of universal gravitation aur Conservation of mechanical energy par bhi.

The scenario matrix

Solve karne se pehle, escape-velocity problems ke har distinct type ki list banate hain. Baad ke har example pe woh cell tag ki gayi hai jisme woh fit hoti hai.

Cell	Kya cheez alag banati hai	Example
A. Given $g, R$	Surface gravity + radius pata hai, $G, M$ nahi	Ex 1
B. Given $G, M, R$	Mass directly pata hai	Ex 2
C. Scaling / ratio	"double the mass", "half the radius" — koi numbers plug nahi	Ex 3
D. Launch above the surface	$r = R + h$ se shuru, $R$ se nahi	Ex 4
E. Leftover speed (not minimum)	$v_{e}$ se tez launch: infinity par speed?	Ex 5
F. Real-world word problem	Gas molecule vs. planet — kya atmosphere survive karti hai?	Ex 6
G. Limiting / extreme value	$R \to$ chota karte jao jab tak $v_{e} \to c$ (black hole edge)	Ex 7
H. Degenerate input	Kya toot ta hai jab $R \to 0$ ya $M \to 0$ ?	Ex 8
I. Exam twist	Same $ρ$ (density), different size — $v_{e}$ kaise scale hota hai?	Ex 9

Recall Formula ke do forms — kaunsa kab use karein

Agar problem mein == $g$ and $R$ == diya ho, toh $v_{e} = 2 g R$ use karo. Agar == $G$ , $M$ , $R$ == (ya density) diya ho, toh $v_{e} = 2 GM / R$ use karo. Dono same equation hain kyunki $g = GM / R^{2}$ , yani $GM = g R^{2}$ .

Poore page mein use hone waale constants: $G = 6.674 \times 1 0^{- 11} N\cdotm^{2} / kg^{2}, c = 3.0 \times 1 0^{8} m/s$

Neeche wali figure poore page ka map hai: yeh dikhati hai ki $v_{e}$ un two levers, mass $M$ aur radius $R$ , ke saath kaise respond karta hai, jinhe matrix ki har cell kisi na kisi combination mein kheenchti hai.

Isse abhi padho: har curve ek fixed mass hai; daayein kheenchna (bada $R$ ) $v_{e}$ ko ghata deta hai (Cells D, I), aur curve ko upar uthana (bada $M$ ) isse badhata hai (Cells C, G). Steep left edge Cell H ke degenerate $R \to 0$ blow-up ka preview hai.

Cell A — surface gravity aur radius diya ho

Cell B — mass directly diya ho

Cell C — numbers plug kiye bina scaling

Worked example Ex 3 — "Twice the mass, half the radius" (Cell C)

Ek planet ka mass Earth se $2 \times$ aur radius Earth se $\frac{1}{2} \times$ hai. $v_{e}$ mein Earth ke $11.2$ km/s ke relative kitna factor aayega?

Forecast: Zyada mass aur chota radius dono gravity ko strengthen karte hain. Kya factor $1.4$ ke kareeb hoga, $2$ ke kareeb, ya $3$ ke kareeb?

Step 1. $v_{e} \propto M / R$ likho (constants $2, G$ ratio mein cancel ho jaate hain). Yeh step kyun? "Kitna factor" wale question mein sirf new aur old ka ratio matter karta hai, toh sirf variable pieces rakhte hain.

Step 2. Factors $M \to 2 M$ , $R \to \frac{1}{2} R$ substitute karo: $\frac{v _{e}^{'}}{v _{e}} = \frac{2 M}{\frac{1}{2} R} \cdot \frac{R}{M} = \frac{2}{1/2} = 4 = 2.$ Yeh step kyun? New $M / R$ ko old se divide karne par sirf change factors bachte hain.

Step 3. Toh $v_{e}^{'} = 2 \times 11.2 = 22.4 km/s$ .

Verify: Sanity check — sirf mass double karne se $2 \approx 1.41$ milta hai; sirf radius half karne se $2 \approx 1.41$ milta hai; multiply karein: $1.41 \times 1.41 = 2$ . ✓ Dono effects ek hi direction mein push karte hain, jaisa expect tha.

Cell D — surface ke upar se launch karna

Worked example Ex 4 — Height

$h$ se escape (Cell D) Agar tum altitude $h = R$ (yani $r = 2 R$ , Earth ke centre se double radius) se start karo toh kis speed se escape kar sakte ho? $v_{e, surf} = 11.2$ km/s lo.

Forecast: Bahut bahar matlab shallower well. $11.2$ km/s se bada ya chota?

Step 1. Energy balance likho definition box se wahi $U (r) = - GM m / r$ use karke, lekin $r = 2 R$ se start karo: $\frac{1}{2} m v^{2} - \frac{GM m}{2 R} = 0.$ Yeh step kyun? Derivation parent jaisi hi hai; sirf launch radius badla, toh PE term mein $R$ ki jagah $2 R$ likh do.

Step 2. Solve karo: $v = \frac{2 GM}{2 R} = \frac{GM}{R}$ . Yeh step kyun? "Double the radius" wala factor of 2, " $2 GM$ " ke 2 ko cancel kar deta hai.

Step 3. Surface value se relate karo is page par (koi bahari fact nahi chahiye). Surface formula $v_{e, surf} = 2 GM / R$ hai, toh square karne par $\frac{GM}{R} = \frac{1}{2} v_{e, surf}^{2}$ milta hai. Isliye $v = \frac{GM}{R} = \frac{1}{2} v_{e, surf}^{2} = \frac{v _{e, surf}}{2} = \frac{11.2}{1.414} \approx 7.92 km/s .$

Verify: $7.92 \times 2 = 11.2$ ✓. Aur yeh surface value se chota hai — sahi hai, kyunki upar se shuru karne par tum pehle se well ka kuch hissa chadh chuke ho. (Note: $GM / R$ exactly surface orbital speed hai — ek sundar cross-check.)

Cell E — escape se tez launch: leftover speed

Worked example Ex 5 — Infinity par bacha hua speed (Cell E)

Earth ki surface se $v_{0} = 15 km/s$ par ek probe daago (escape se tez, $v_{e} = 11.2$ ). Bahut door jaane par $v_{\infty}$ kitni hogi? Air aur doosre bodies ignore karo.

Forecast: Escape se thodi zyada hai. Kya $v_{\infty}$ 15 ke kareeb hogi, ya kaafi choti?

Step 1. Energy conservation, ab nonzero final KE ke saath (phir $U (r) = - GM m / r$ use karo, jo infinity par zero ho jaata hai): $\frac{1}{2} m v_{0}^{2} - \frac{GM m}{R} = \frac{1}{2} m v_{\infty}^{2} - 0.$ Yeh step kyun? Infinity par $U = 0$ , lekin object abhi bhi move kar raha hai, toh $K_{\infty} \neq = 0$ . Ise rakhte hain.

Step 2. $m$ cancel karo, aur $\frac{2 GM}{R} = v_{e}^{2}$ pehchano: $v_{\infty}^{2} = v_{0}^{2} - \frac{2 GM}{R} = v_{0}^{2} - v_{e}^{2} .$ Yeh step kyun? Saaf relation $v_{\infty}^{2} = v_{0}^{2} - v_{e}^{2}$ kehta hai: "leftover energy = launch energy minus woh toll jo gravity leta hai."

Step 3. Numbers plug karo (km/s mein, squared): $v_{\infty} = 1 5^{2} - 11. 2^{2} = 225 - 125.44 = 99.56 \approx 9.98 km/s .$

Verify: $9.9 8^{2} + 11. 2^{2} = 99.6 + 125.4 = 225 = 1 5^{2}$ ✓ (ek Pythagorean-jaisi energy sum). Dhyan do $v_{\infty} = 9.98 < 15$ : gravity ne fark kha liya. Agar exactly $v_{e}$ par launch karte, toh formula $v_{\infty} = 0$ deta — minimum case, parent se consistent. ✓

Neeche wali picture yeh leftover-speed curve $v_{\infty} = v_{0}^{2} - v_{e}^{2}$ plot karti hai aur $v_{0} = 15$ km/s par kaam kiya hua point mark karta hai; $v_{e}$ se neeche object wapas gir jaata hai (infinity tak pahuncha hi nahi).

Cell F — real-world word problem

Worked example Ex 6 — Kya Moon atmosphere rakh sakta hai? (Cell F)

Kinetic theory mein, molecules sabhi ek hi speed share nahi karte; standard single number hai root-mean-square (RMS) speed, $v_{rms} = \overline{v^{2}}$ — woh speed jiska square squared speeds ka average equal kare. $300 K$ ke paas air ke liye, $v_{rms} \approx 0.5 km/s$ . Moon ki escape velocity $2.4 km/s$ hai (parent note). Ek rough rule: ek body apni gas waqt ke saath khoti hai agar $v_{rms}$ , $v_{e}$ ke $\frac{1}{6}$ se zyada ho (kyunki distribution ki fast tail tab routinely $v_{e}$ cross kar leti hai). Kya Moon atmosphere rakhega?

Forecast: Compute karne se pehle "haan" ya "naa" andaza lagao.

Step 1. Retention ratio compute karo $v_{e} / v_{rms} = 2.4/0.5 = 4.8$ . Yeh step kyun? Atmospheric escape depend karta hai iss par ki $v_{e}$ , ek typical (RMS) molecule se kitni baar tez hai — woh ratio slow leak control karta hai.

Step 2. Rule se compare karo. Rule chahta hai $v_{e} / v_{rms} ≳ 6$ gas safely rakhne ke liye. Yahaan $4.8 < 6$ hai. Yeh step kyun? Threshold hi word problem ka poora point hai — ek akela number tab tak kuch nahi kehta jab tak criterion se compare na ho.

Step 3. Conclusion: $4.8 < 6$ , toh molecules ki fast tail $2.4$ km/s se itni baar zyada jaati hai ki escape ho sake → Moon apni atmosphere khota hai.

Verify: Earth ke liye, $v_{e} / v_{rms} = 11.2/0.5 = 22.4 ≫ 6$ → Earth hawa rakhta hai. ✓ Yahi Why the Moon has no atmosphere ki story hai, ab quantitative ban gayi.

Cell G — limiting value: black-hole edge

Worked example Ex 7 — Sun ko tab tak compress karo jab tak

$v_{e} = c$ na ho (Cell G) Sun ki mass $M = 1.99 \times 1 0^{30} kg$ lo. Ise kitne radius tak compress karna padega taaki escape velocity light ki speed $c$ ke equal ho jaaye? (Yeh Newtonian estimate sach-much true Black holes — Schwarzschild radius se match karta hai.)

Forecast: Kuch metres, kuch km, ya kuch thousand km?

Step 1. $v_{e} = 2 GM / R$ mein $v_{e} = c$ set karo aur $R$ ke liye solve karo: $c = \frac{2 GM}{R} \Rightarrow R = \frac{2 GM}{c ^{2}} .$ Yeh step kyun? "Escape ke liye speed of light chahiye" exactly condition $v_{e} = c$ hai; formula invert karte hain kyunki ab $R$ unknown hai.

Step 2. Numerator $2 GM = 2 (6.674 \times 1 0^{- 11}) (1.99 \times 1 0^{30}) = 2.656 \times 1 0^{20}$ . Denominator $c^{2} = 9.0 \times 1 0^{16}$ . Yeh step kyun? Top aur bottom alag karna exponents ko sahi rakhta hai.

Step 3. Divide karo: $R = \frac{2.656 \times 1 0 ^{20}}{9.0 \times 1 0 ^{16}} \approx 2.95 \times 1 0^{3} m \approx 2.95 km .$

Verify: Sun ka jaana-maana Schwarzschild radius $\approx 2.95$ km hai. ✓ Newtonian escape argument, apni limit tak push kiya, exactly relativistic answer par aata hai — form ka ek sundar coincidence.

Cell H — degenerate inputs

Worked example Ex 8 —

$R \to 0$ ya $M \to 0$ par kya toot ta hai? (Cell H) $v_{e} = 2 GM / R$ ke do edge cases examine karo: (a) fixed $M$ par $R \to 0$ , aur (b) fixed $R$ par $M \to 0$ .

Forecast: Kis case mein $v_{e}$ blow up hota hai, aur kis mein zero ho jaata hai?

Step 1 — case (a), $R \to 0$ . $v_{e} = 2 GM / R \to 2 GM / 0^{+} \to \infty$ . Yeh step kyun? Ek shrinking positive number se divide karna ratio ko infinity bhejna hai; infinity ka root bhi infinity hai. Physically: ek point mass ki bottomless well hoti hai — koi finite speed escape nahi kar sakti, jo phir se black-hole intuition hai.

Step 2 — case (b), $M \to 0$ . $v_{e} = 2 G \cdot 0/ R = 0 = 0$ . Yeh step kyun? Koi mass nahi matlab koi gravity nahi matlab koi well nahi — tum "kuch nahi" se zero speed par escape karte ho.

Step 3 — the wrong sign trap. Definition box yaad karo: $U (r) = - GM m / r$ , minus ke saath. Agar kisi ne galti se $U = + GM m / r$ likha, toh energy balance $\frac{1}{2} m v_{e}^{2} + \frac{GM m}{R} = 0$ deta $v_{e}^{2} = - 2 GM / R < 0$ , yani ek imaginary speed — yeh sign chhoot jaane ka signal hai.

Verify: Dono limits monotonically behave karti hain: $v_{e}$ badhta hai jab $R$ ghatta hai aur badhta hai jab $M$ badhta hai, toh $\partial v_{e} / \partial R < 0$ aur $\partial v_{e} / \partial M > 0$ . $M = 0$ test karne par exactly $0$ milta hai, aur trap negative radicand deta hai. ✓ Parent ke "sign error" warning se consistent.

Cell I — exam twist: same density, different size

Worked example Ex 9 — Do planets, same density (Cell I)

Planet B same stuff se bana hai jaise planet A (same density $ρ$ ) lekin 3× the radius hai. $v_{e}$ kaise change hota hai?

Forecast: Bada planet, lekin same material. Factor of $3$ ? $3$ ka? Ya kuch aur?

Step 1. Density se mass likho: $M = ρ \cdot \frac{4}{3} π R^{3}$ , toh fixed $ρ$ par $M \propto R^{3}$ . Yeh step kyun? Is twist mein density lock hai, mass nahi — toh mass free nahi hai; woh volume ke saath badhta hai.

Step 2. $v_{e} = 2 GM / R$ mein substitute karo: $v_{e} \propto \frac{M}{R} \propto \frac{R ^{3}}{R} = R^{2} = R .$ Yeh step kyun? $M$ ko $\propto R^{3}$ se replace karna saari dependence ko ek clean $v_{e} \propto R$ mein collapse kar deta hai — naive feeling se kaafi zyada strong growth.

Step 3. Isliye $v_{e}$ waise hi scale karta hai jaise $R$ : $\times 3$ .

Verify: Numbers se check karo — maan lo A ka $R = 1, v_{e} = 1$ (arbitrary units). B ka $R = 3$ hai, toh $M \propto 27$ , aur $27/3 = 9 = 3$ . ✓ Escape speed fixed density par size ke saath linearly badhta hai — exam-favourite surprise.

Active recall

Escape from

r = 2 R

vs surface

v = v_{e, surf} / 2 \approx 7.92

km/s.

Leftover speed at infinity

v_{\infty} = v_{0}^{2} - v_{e}^{2}

; yahaan

1 5^{2} - 11. 2^{2} \approx 9.98

km/s.

Compress Sun to

v_{e} = c

R = 2 GM / c^{2} \approx 2.95

km (Schwarzschild radius).

Same density,

3 \times

radius

v_{e} \propto R

, toh

v_{e}

triple ho jaata hai.

Degenerate

M \to 0

v_{e} \to 0

(koi gravity nahi, koi well nahi).

Degenerate

R \to 0

v_{e} \to \infty

(bottomless well).

Gravitational PE used all through

U (r) = - GM m / r

, negative kyunki mass bound hai.

Connections

Escape velocity — derivation (Hinglish)
Gravitational potential energy
Newton's law of universal gravitation
Conservation of mechanical energy
Orbital velocity & circular motion
Black holes — Schwarzschild radius
Why the Moon has no atmosphere