Gravitational potential energy — U = −GMm - r (not mgh): Visual walkthrough

1.2.22 · D2 · Physics › Newton's Laws & Dynamics › Gravitational potential energy — U = −GMm - r (not mgh)

Step 1 — Do masses, ek pull

KYA. Ek badi mass $M$ (socho: ek planet) ko ek fixed point par rakho jise hum origin kehte hain — yeh haari picture ka centre hai. Ek chhoti mass $m$ (socho: ek ball) ko $r$ distance par rakho.

KYUN. Jab tak hum energy ki baat nahi kar sakte, pehle force jaanni padegi. Energy "stored-up work" hai, aur work force se aata hai. Isliye hum pull se shuru karte hain.

PICTURE. Figure dekho. $M$ se $m$ tak seedhi line distance $r$ hai. Chhota pale-yellow arrow $\overset{r}{^}$ ("r-hat") ek unit vector hai: exactly $1$ length ka arrow jo sirf direction batata hai — yeh $M$ se bahar ki taraf, outward point karta hai. Pink arrow actual gravitational force $F$ hai, aur yeh doosri direction mein, andar ki taraf, $M$ ki taraf point karta hai.

Figure — Gravitational potential energy — U = −GMm - r (not mgh)

Step 2 — Force ko uske minus sign ke saath likho

KYA. Newton's Law of Universal Gravitation humein batata hai ki pull ki size $\frac{GM m}{r ^{2}}$ hai, aur direction inward hai. Milaakar: $F = inward - how strong \frac{GM m}{r ^{2}} which way (outward) \overset{r}{^} .$

MINUS KYUN? $\overset{r}{^}$ outward point karta hai, lekin gravity inward pull karti hai — bilkul ulta. $- 1$ se multiply karne par outward arrow palat ke inward point karne lagta hai. Minus sign literally " $\overset{r}{^}$ ki direction reverse karo" hai.

$1/ r^{2}$ KYUN? Distance double karo toh pull quarter ho jaati hai. Yeh "inverse-square" fall-off gravity ki pehchaan hai, aur yahi baat final energy formula ko woh shape deti hai jo uski hai.

PICTURE. Figure mein $m$ ke teen snapshots dikhaye gaye hain $r$ , $2 r$ , $3 r$ par. Pink force arrow tezi se chhota hota hai — $2 r$ par yeh $1/4$ jitna lamba hai, $3 r$ par $1/9$ jitna. Woh shrinking $1/ r^{2}$ ka visible roop hai.

Step 3 — "Potential energy" ka matlab aakhir kya hai

KYA. Hum potential energy mein change ko minus the work the force does define karte hain: $Δ U = U_{B} - U_{A} = - work done by gravity \int_{A}^{B} F \cdot d r .$

YEH DEFINITION KYUN? Kyunki gravity ek conservative force hai — jo work yeh karta hai woh sirf start aur end points par depend karta hai, kabhi bhi tedhey-medhey path par nahi. Isse hum work ko ek single number $U$ mein "bottle" kar sakte hain jo sirf position par depend karta hai. Minus sign ek bookkeeping choice hai: jab gravity positive work kare ( $m$ ko andar kheenche), toh hum chahte hain ki stored energy neeche jaaye, jaise ek spent battery.

$\int$ (integral) KYUN, sirf force $\times$ distance kyun nahi? Kyunki force change karti hai jab $m$ move karta hai (Step 2 ne dikhaya tha ki yeh chhoti hoti hai). Tum ek changing force ko distance se multiply nahi kar sakte. Integral woh tool hai jo force-times-tiny-step ko ek path par add karta hai jahan force hamesha change hoti rehti hai — yeh chhote chhote dhakkon ka running total hai.

$F \cdot d r$ KYA HAI? Dot ( $\cdot$ ) dot product hai: yeh sirf force ka woh hissa leta hai jo motion ki direction ke saath lie karta ho. Agar tum motion ke sideways push karo, koi work nahi hota. Yahan hum seedha $\overset{r}{^}$ ke along move karenge, toh dot product next step mein nicely simplify ho jaayega.

PICTURE. Figure ek curvy path dikhata hai $A$ se $B$ tak jo tiny straight steps $d r$ mein kaata gaya hai. Har step par hum $F$ ka woh tukda lete hain jo us step ke along point karta hai (blue step par pink arrow ki chhaya) aur unhe sab add kar dete hain.

Step 4 — Seedha bahar jaao, toh dot product collapse ho jaata hai

KYA. Sabse simple possible path choose karo: $m$ ko $M$ se guzarne wali line ke seedha bahar slide karo. Tab har tiny step $d r = d r \overset{r}{^}$ hai — outward direction mein $d r$ length ka step. Isse force mein feed karo: $F \cdot d r = (- \frac{GM m}{r ^{2}} \overset{r}{^}) \cdot (d r \overset{r}{^}) = - \frac{GM m}{r ^{2}} = 1 (\overset{r}{^} \cdot \overset{r}{^}) d r = - \frac{GM m}{r ^{2}} d r .$

HUM YEH PATH CHOOSE KARNE KE ALLOWED KYUN HAIN? Kyunki gravity conservative hai (Step 3), answer path par depend nahi karta. Isliye hum sabse aasaan path choose karte hain — ek seedhi radial line — aur messy dot product ordinary numbers mein badal jaata hai.

$\overset{r}{^} \cdot \overset{r}{^} = 1$ KYUN HOTA HAI? Kisi arrow ka khud ke saath dot product (length) $^{2}$ hota hai. Yahan length $1$ hai, isliye $\overset{r}{^} \cdot \overset{r}{^} = 1$ . Dono unit vectors "cancel" ho jaate hain, ek plain scalar bachta hai.

PICTURE. Figure force arrow $F$ (inward, pink) aur step $d r$ (outward, blue) ko same line par dikhata hai lekin opposite ways point karte hue. Same line ⇒ dot product sirf signed lengths ka product hai; opposite ways ⇒ result negative hai.

Step 5 — Energy ka zero infinity par pin karo

KYA. Potential energy hamesha sirf difference ( $Δ U$ ) ke roop mein aati hai, isliye hum choose kar sakte hain ki " $U = 0$ " kahan hoga. Hum choose karte hain: $U (\infty) = 0.$ Tab kisi bhi distance $r$ par energy hai $U (r) = U (r) - = 0 U (\infty) = - \int_{\infty}^{r} F \cdot d r .$

INFINITY KYUN? Inverse-square force ke liye, infinity single natural landmark hai — yeh woh jagah hai jahan pull poori tarah se zero ho gayi ho. Koi bhi doosra choice arbitrary hoga aur clean formula kharab kar dega. "Pull se mukta ⇒ zero energy" — yeh sabse clean possible reference hai.

PICTURE. Figure ek horizontal energy line hai. Ek dum right (bada $r$ ) par level $0$ par baitha hai. Jab hum baayi taraf $M$ ki taraf chalte hain toh hum neeche chalte hain, ek valley mein — woh valley next step compute karta hai.

Step 6 — Integral karo

KYA. Step 4 ka result Step 5 mein substitute karo aur integrate karo. (Main running distance ko $r^{'}$ rename karta hoon taaki yeh endpoint $r$ se clash na kare.) $U (r) = - \int_{\infty}^{r} (- \frac{GM m}{r ^{'2}}) d r^{'} = GM m \int_{\infty}^{r} \frac{d r ^{'}}{r ^{'2}} .$

Dono minus signs (ek $U$ ki definition se, ek inward force se) cancel ho jaate hain, ek clean positive $GM m$ front mein bachta hai. Ab yeh fact use karo ki $\frac{1}{r ^{'2}}$ ka antiderivative $- \frac{1}{r ^{'}}$ hai: $U (r) = GM m [- \frac{1}{r ^{'}}]_{\infty}^{r} = GM m at r^{'} = r - \frac{1}{r} - = 0 (- \frac{1}{\infty}) = - \frac{GM m}{r} .$

$\infty$ WALA TERM VANISH KYUN HOTA HAI? $- 1/\infty$ yaani $- 1/ (bahut bada) = 0$ . Yahi wajah hai ki infinity smart reference tha: yeh poora ek term kill kar deta hai.

PICTURE. Figure $U (r) = - GM m / r$ plot karta hai. Yeh ek curve hai jo $r \to 0$ par $- \infty$ ki taraf giraata hai (well ke deep mein, $M$ ke ekdum paas) aur $r$ badhne par dheere dheere $0$ line ki taraf uthta hai. Poora curve $0$ ke neeche rehta hai.

Step 7 — Edge case: $r \to 0$ aur $r \to \infty$ par kya hota hai?

KYA. Curve ki dono extremes check karo taaki koi reader kabhi surprised na ho.

Jab $r \to \infty$ : $U = - \frac{GM m}{r} \to 0^{-}$ ( $0$ ke neeche se approach karta hai). Masses essentially free hain.
Jab $r \to 0^{+}$ : $U = - \frac{GM m}{r} \to - \infty$ . Ideal point-mass model mein well bottomless hai.

$U \to - \infty$ physics kyun nahi todta? Real bodies ki finite radius hoti hai; tum actually $r = 0$ tak nahi pahunch sakte kyunki pehle planet ki surface hit hogi. Bottomless well point masses ke liye ek idealisation hai — physically tum $r = R$ (surface) par ruk jaate ho, ek finite, sensible number.

APPROACH KE SIGN KI PARWAH KYUN KAREIN? Kyunki ek common error (neeche dekho) yeh sochna hai ki bada $r$ matlab "aur zyaada negative." Figure isse kill karta hai: jab $r$ badhta hai, curve $0$ ki taraf chadhta hai — $U$ increase karta hai.

PICTURE. Step 6 jaisa hi curve, ab dono limits flag kiye gaye hain: left par $- \infty$ mein girta pink arrow, right par $0$ line par flatten hota blue arrow, aur ek shaded band jo physically reachable region $r \geq R$ mark karta hai.

Step 8 — Sanity check: well hi hai $m g h$ close up mein

KYA. Surface ke paas curve ke ek tiny sliver mein zoom karo, $r = R + h$ jahan $h ≪ R$ . Surface se height $h$ tak $U$ mein change: $Δ U = - \frac{GM m}{R + h} - (- \frac{GM m}{R}) = GM m \cdot \frac{h}{R ( R + h )} \approx GM m \cdot \frac{h}{R ^{2}} = m = g \frac{GM}{R ^{2}} h = m g h .$

"DERIVATION IN PICTURES" KE LIYE YEH KYUN MATTER KARTA HAI? Yeh dikhata hai ki schoolbook $m g h$ koi rival law nahi hai — yeh haare curve ka straight-line tangent hai $r = R$ ke paas. $g = GM / R^{2}$ ke liye Acceleration due to gravity g dekho.

PICTURE. Figure well ko $r = R$ ke paas zoom karta hai. Ek chhote span par curve ek seedhi ramp jaisi lagti hai; uski constant slope $m g$ hai, isliye height $h$ chadhne par $m g h$ cost aata hai. Surface se door curve vapas apne asli shape mein bend ho jaata hai — jahan $m g h$ fail karta hai aur tumhe $- GM m / r$ use karna hi padta hai.

Ek-picture summary

KYA. Ek figure jisme poori story hai: inward force (pink), infinity tak outward path (blue), shaded work jo integrate ho raha hai, reference $U (\infty) = 0$ , aur resulting energy well $U = - GM m / r$ apne $m g h$ tangent ke saath surface ke paas.

Recall Feynman: poora walkthrough ek 12-saal ke bacche ko retell karo

Ek planet space mein khodi gayi ek gehri gol khai ke mooh jaisi hai. Planet ke paas ek ball us khai mein kuch gehraai tak gir chuki hai. "Khai mein kitne gehre hain" yeh describe karne ke liye pehle poochhte hain: khai kitna zor se kheenchti hai? Jawab — yeh inward kheenchti hai, aur pull jitna door jaao utna tezi se kamzor hota jaata hai (do guna door, ek-chauthaai pull). "Depth energy" nikaalane ke liye hum imagine karte hain ki ball ko khai se poora baahar kheeench kar infinitely door le jaayein, raaste mein pull ke har chhote jhatke ko add karte huye — woh grand total energy hai. Hum decide karte hain ki khai se poori tarah baahar hona (infinitely door) zero depth count karta hai. Kyunki khai ke andar kahin bhi us se neeche hai, number negative aata hai: $U = - GM m / r$ . Bilkul top rim par, agar tum thoda sa upar hop karo, toh khai ki deewar seedhi ramp jaisi lagti hai, aur height $h$ tak hop karne par $m g h$ lagta hai — yeh flat-Earth shortcut hai. Lekin rockets ke liye jo poori khai se nikalte hain, tumhe real curved-well formula chahiye.

Connections

Newton's Law of Universal Gravitation — woh force jo humne Steps 2–4 mein integrate kiya.
Conservative Forces and Potential Energy — isliye lazy straight path pick karna (Step 4) legal hai.
Work-Energy Theorem — woh work jo humne $U$ mein bottle kiya.
Escape Velocity — is well ko use karke total energy ko $0$ set karo.
Kepler's Laws & Orbital Energy — isi curve se $E = - GM m /2 r$ padhta hai.
Acceleration due to gravity g — Step 8 mein well ka tangent slope.

Visual walkthrough — Gravitational potential energy — U = −GMm - r (not mgh)