Escape velocity — derivation: Question bank — Hinglish deep dive

1.2.23 · D5 · Physics › Newton's Laws & Dynamics › Escape velocity — derivation

Yeh reminders answer karte waqt kaam aa sakte hain:

Escape speed: $v_{e} = \frac{2 GM}{R} = 2 g R$ .
Gravitational PE with the $U (\infty) = 0$ convention: $U (r) = - \frac{GM m}{r}$ (negative ⇒ bound).
Energy bookkeeping: $E = \frac{1}{2} m v^{2} - \frac{GM m}{r}$ , conserved once only gravity acts.

True or false — justify

Ek tennis ball aur ek battleship, same planet se launch kiye jayein, toh dono ko same escape speed chahiye.

True. Kinetic energy ki supply aur gravitational energy ka debt dono object ke mass

m

ke saath scale karte hain, isliye

m

cancel ho jaata hai

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

mein — threshold ek speed hai, energy nahi.

Escape velocity ek vector hai, isliye launch ka seedha upar hona matter karta hai.

False. Yeh actually ek energy condition hai, isliye ek speed hai. Air aur rotation ko ignore karein toh sirf magnitude

\frac{1}{2} m v^{2}

hi energy equation mein aati hai, toh koi bhi launch direction (zameen mein jaane ke alawa) same speed maangti hai.

Jo object exactly escape speed par launch kiya jaaye woh infinity par rest mein aa jaata hai aur phir wapas khincha jaata hai.

False. Gravity ki force

\propto 1/ r^{2} \to 0

jab

r \to \infty

, toh infinity par kuch bhi nahi bachta jo use wapas kheenche; infinity par

v = 0

ke saath pahunchna just-barely-forever case hai.

Ek just-escaping object ki total mechanical energy exactly zero hoti hai.

True. Infinity par

K = 0

aur

U (\infty) = 0

, isliye

E_{f} = 0

; kyunki energy conserved hai,

E

path mein har jagah zero hai, launch ke waqt bhi.

Ek bound object (jo kabhi escape nahi karta) ki total mechanical energy negative hoti hai.

True. Agar

E < 0

hai, toh

\frac{1}{2} m v^{2} = E + \frac{GM m}{r}

ek finite

r

par zero ho jaata hai (turning point), isliye object ruk jaata hai aur wapas aata hai — yahi "bound" ka matlab hai.

Escape velocity, orbital velocity se ek fixed ratio se zyada tez hoti hai, chahe planet koi bhi ho.

True.

v_{e} = 2 GM / R

aur orbital

v_{o} = GM / R

, isliye

v_{e} = 2 v_{o}

hamesha;

2

geometry-free aur planet-independent hai (dekho Orbital velocity & circular motion).

Kisi planet ka mass double karne par, fixed radius par, escape velocity double ho jaati hai.

False.

v_{e} \propto M

, toh

M

double karne par

v_{e}

sirf

2 \approx 1.41

se multiply hoti hai,

2

se nahi.

Do planets jinki same density hai lekin alag sizes hain, unki escape velocity same hogi.

False. Fixed density par,

M \propto R^{3}

, isliye

v_{e} = 2 GM / R \propto R^{3} / R = R

— escape speed radius ke directly proportional badhti hai, toh bada planet escape karna zyada mushkil hai.

Surface par, ek body jo exactly

v_{e}

par move kar rahi hai use hamesha ke liye jaane ke liye koi aur push nahi chahiye.

True. Jab uske paas

v_{e}

speed ho aur koi drag na ho, energy conservation guarantee karta hai ki woh infinity par

v \to 0

ke saath pahunchegi; uske baad koi engine ki zaroorat nahi.

Spot the error

"Kyunki

U = + \frac{GM m}{r}

\frac{1}{2} m v_{e}^{2} = U

set karne par escape speed milti hai."

Sign galat hai:

U (\infty) = 0

convention mein bound PE

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

hoti hai. Positive

U

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

milega, ek imaginary speed — yeh red flag hai ki minus drop ho gaya.

"Escape ke liye itni kinetic energy chahiye ki gravity cancel ho, isliye

\frac{1}{2} m v_{e}^{2} = F R = \frac{GM m}{R ^{2}} \cdot R

Force aur energy mein confusion hai. Tumhe work integral

\int_{R}^{\infty} F d r = \frac{GM m}{R}

pay karni hogi, force

\times

ek radius nahi; sahi condition hai

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

, jisse root ke andar extra factor of

2

milta hai.

"Air resistance matter nahi karta, isliye real rocket ko bhi bas

11.2

km/s ground par chahiye."

Vacuum mein energy threshold

11.2

km/s hai, lekin real rockets continuous thrust ke saath climb karte waqt drag aur gravity se ladte hain, isliye unhe kabhi ek saath woh single ground speed nahi chahiye — woh problem ko sustained acceleration se replace karte hain.

"Moon par atmosphere nahi hai kyunki uske paas gravity nahi hai."

Moon par gravity hai (

g \approx 1.62

m/s²); asli wajah yeh hai ki uski escape speed (

\approx 2.4

km/s) itni kam hai ki gas molecules routinely isse zyada ho jaate hain aur leak ho jaate hain (dekho Why the Moon has no atmosphere).

v_{e}

par object orbit mein hai, bas bahut zyada ऊँcha."

Escape speed par object ek unbound parabolic path follow karta hai aur kabhi wapas nahi aata; orbit ek bound closed ellipse hoti hai jahan

E < 0

hota hai. Line

E = 0

dono ko separate karti hai — orbit iske neeche hai, escape iske upar.

"Kyunki

v_{e} = 2 g R

, aur

g

Earth ki surface par har jagah same hai, escape speed launch altitude se affect nahi hoti."

Do jagah galat hai: height

h

se launch karne par

R

ki jagah

R + h

aata hai (aur

g

ghatta hai), isliye escape speed altitude badhne par ghatti hai. Galti yeh hai ki

R

ko fixed maana gaya chahe tum kahaan se actually start karo.

"Ek black hole ke liye

v_{e} = c

set karna bekar hai kyunki derivation mein light ka kuch tha hi nahi."

Newtonian estimate

v_{e} = c

sach mein sahi Schwarzschild radius

R_{s} = 2 GM / c^{2}

numerically deta hai (ek lucky coincidence), isliye yeh Black holes — Schwarzschild radius ke liye standard heuristic hai — halanki asli justification relativistic hai.

Why questions

Escaping object ka mass

m

final formula se cancel kyun ho jaata hai?

Kyunki

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

ki dono sides par

m

ka ek factor hota hai; kinetic supply aur gravitational debt saath scale karte hain, sirf speed par ek condition chhodke.

Hum

U (\infty) = 0

kyun choose karte hain na ki

U (surface) = 0

Infinity natural zero hai — wahan masses interact nahi karte, isliye

U \to 0

automatically. Isse "escape" ka matlab hota hai "energy ceiling

E = 0

tak pahunchna," jo sabse clean boundary condition hai.

Minimum escape case mein final speed zero kyun honi chahiye, koi small value nahi?

Infinity par koi bhi leftover speed matlab hai ki requirement se zyada kinetic energy supply ki gayi, isliye launch speed minimum se zyada thi;

v_{f ina l} = 0

exact break-even hai jo threshold define karta hai.

v_{e}

force ke ek single point ki bajaye force ke integral se kyun banta hai?

Gravity distance ke saath weaken hoti hai, isliye

R

se infinity tak climb karne ka work kai shrinking contributions ka sum hai —

\int_{R}^{\infty} \frac{GM m}{r ^{2}} d r

— jo sirf integration se capture hota hai (dekho Newton's law of universal gravitation).

Escape speed exactly

2

times orbital speed kyun hai, koi aur factor kyun nahi?

Orbit ko KE chahiye well ki depth ke half ke equal (

\frac{1}{2} m v_{o}^{2} = \frac{1}{2} \frac{GM m}{R}

), jabki escape ko KE chahiye puri depth ke equal; KE double karne par speed

2

se multiply hoti hai.

Ek shallower "gravity valley" (small

g

, small

R

) chhotai escape velocity kyun deta hai?

Escape speed measure karti hai ki well se bahar nikalne ke liye per unit mass kitni kinetic energy chahiye; ek shallow well (

v_{e} = 2 g R

chote

g

aur

R

ke saath) mein kam energy lagti hai, isliye ek modest speed kaafi hai.

Hum

v_{e}

ke liye ordinary projectile problems ki tarah

\frac{1}{2} m v^{2} = m g h

kyun use nahi kar sakte?

Woh formula maanta hai ki poori rise mein

g

constant hai, lekin

g

infinity tak ki badi climb mein

1/ r^{2}

ke roop mein girti hai; constant

g

use karne par infinite

h

ke liye infinite energy maangni padegi aur answer galat aayega.

Edge cases

Sab mass se infinitely dur kisi point par escape velocity kya hai?

Effectively zero: jab

R \to \infty

v_{e} = 2 GM / R \to 0

. Infinitely door tum pehle se hi well ke top par ho, toh free rehne ke liye koi speed nahi chahiye.

Agar exactly

v_{e}

par launch kiya jaaye, toh escape mein kitna time lagta hai?

Infinitely zyada.

r = \infty

tak pahunchne mein unbounded time lagta hai chahe total energy finite ho, kyunki object

v = 0

ki taraf badhte hue dheemai hoti rehti hai climb karte waqt.

Agar launch speed

v_{e}

se sirf

1 m/s

kam ho toh kya hoga?

Object bound hai (

E < 0

): woh bahut badi lekin finite maximum radius tak climb karta hai, rukta hai, aur wapas girta hai. Koi gradual failure nahi hoti —

E = 0

cross karna ek sharp threshold hai.

Kya surface ke neeche (planet ke andar) kisi object ke liye escape velocity ka matlab banta hai?

Surface formula ab apply nahi hota; andar, sirf tumhari radius ke interior wala mass tumhe pull karta hai (shell theorem), isliye enclosed

M

aur potential badal jaata hai, aur bahar nikalne ke liye energy bhi badal jaati hai.

Kya planet ka apna rotation required launch speed ko change karta hai?

Haan, thoda, idealized problem mein hum ise usually ignore karte hain: equator par eastward launch karne par tumhare paas pehle se surface ki spin speed hai, isliye rocket ko

v_{e}

tak pahunchne ke liye kam additional speed supply karni padti hai — lekin infinity tak ki energy threshold khud nahi badlti.

Agar do bodies (jaise Earth aur Moon) dono ek saath object ko pull karein toh kya hoga?

Single-body formula kaam nahi karta; tumhe dono potentials

U = - \frac{G M _{1} m}{r _{1}} - \frac{G M _{2} m}{r _{2}}

add karne honge aur "escape" matlab combined well clear karna hai, jisme saddle points hote hain, ek clean

v_{e}

nahi.

Ek black hole ke liye

v_{e} = c

set karna physically kya claim karta hai?

Ki Schwarzschild radius

R_{s} = 2 GM / c^{2}

par light-speed bhi escape ke liye kaafi nahi — kuch bhi bahar nahi nikalta — jo Black holes — Schwarzschild radius mein event horizon ki defining property hai.

Recall Traps ka one-line summary

Escape velocity ek speed hai (vector nahi, energy nahi), condition $E = 0$ se set hoti hai, work integral se pay hoti hai negative bound PE ke saath, $m$ se independent hai, $M$ aur $R \cdot g$ ke roop mein scale karti hai — aur yeh ek sharp threshold hai, gradual nahi.

Question bank — Escape velocity — derivation

True or false — justify

Spot the error

Why questions

Edge cases

Connections