Visual walkthrough — Potential energy — definition, gravitational (mgh and −GMm - r), elastic (½kx²)

1.3.5 · D2 · Physics › Work, Energy & Power › Potential energy — definition, gravitational (mgh and −GMm -

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Step 0 — Teeno ke peeche ek hi idea

Kisi bhi formula se pehle, woh ek sentence pakad lo jo sab kuch generate karta hai:

Ek master rule (dekho Work done by a force):

$\underset{\text{stored energy mein change}}{\underbrace{\Delta U}}=-\underset{\text{force ne jo work kiya}}{\underbrace{W}}=-\underset{\text{i se f tak tiny works ka sum}}{\underbrace{{\int }_i^f\vec{F}\cdot d\vec{r}}}$

• $\Delta U$ — stored energy kitni badi ho gayi (final minus initial).
• $W$ — force ne khud kitna total work kiya.
• Minus sign — agar force motion mein help karta hai (positive $W$), toh woh store kharach kar raha hai, isliye $U$ girta hai. Minus ka poora reason yahi hai.
• $\int \dots d\vec{r}$ — symbol $\int$ ("integral") ka matlab hai "infinite tiny slices ko add karo." Hum isse isliye use karte hain kyunki object kai tiny steps se guzarta hai aur hume total chahiye.

Integral kyun, sirf $F\times d$ kyun nahi? Kyunki hamare teeno cases mein se do mein force object ke move karne ke saath badalti hai. Jab force constant hoti hai toh multiply kar sakte ho; jab woh badhti ya ghatati hai toh tumhe slice-and-sum karna padta hai. Yahi exactly integral karta hai — aur geometrically, integral force-vs-position graph ke neeche ka area hota hai. Yeh picture apne paas rakho; yahi is poore page ko sambhalta hai.

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Step 1 — Constant force: flat graph se $mgh$ milta hai

KYA. Ek object jiska mass $m$ hai, use seedha upar uthao, height $y=0$ se $y=h$ tak, zameen ke paas.

YE PEHLE KYUN. Earth ki surface ke paas gravity barely change hoti hai, isliye force constant hai. Constant force sabse simple case hai — uska graph ek flat horizontal line hai, aur "uske neeche ka area" bas ek rectangle hai. Abhi calculus ki zaroorat nahi.

PICTURE. Figure mein gravity neeche ki taraf $mg$ size ka arrow hai:

• $m$ — mass (kilograms), "kitna stuff hai."
• $g$ — zameen ke paas gravity ki strength, lagbhag $9.8$ (metres per second squared).
• $mg$ — inका product, actual downward pull (newtons mein).

Force graph $mg$ value par flat hai. Object $h$ distance upar jaata hai.

$\text{Rectangle ka Area}=\underset{\text{graph ki height}}{\underbrace{mg}}\times \underset{\text{graph ki width}}{\underbrace{h}}=mgh$

Kyunki gravity neeche point karti hai jabki lift upar jaati hai, force negative work karti hai, $W=-mgh$. Master rule lagao:

$\Delta U=-W=-(-mgh)=mgh$

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Step 2 — Varying force: triangle graph se $\frac{1}{2}k{x}^2$ milta hai

KYA. Ek spring ko dhire dhire uski natural length ($x=0$) se stretch $x$ tak kheencho.

YE YAHAN KYUN. Ab force constant nahi hai — jitna zyada kheencho, spring utna hi zyada wapas kheenchti hai (Hooke's Law). Toh graph ab flat nahi hai; yeh ek sloped line hai. Yeh haara pehla real "slanted graph ke neeche area" hai — ek triangle.

PICTURE. Hooke's law kehta hai spring force hai

$F=-kx$

• $x$ — spring natural length se kitni dur stretch hui hai (metres).
• $k$ — stiffness (newtons per metre): bada $k$ matlab stiff spring.
• minus — force $x=0$ ki taraf wapas point karti hai, tumhari stretch ka virodh karti hai. Yeh "restore" karti hai.

Tumhari applied force $kx$ ki size ko $x$ ke against plot karo: ek straight ramp jo $0$ se shuru hoti hai aur stretch $x$ par $kx$ tak jaati hai. Ek ramp ke neeche ka area triangle hai:

$\text{Triangle ka Area}=\frac{1}{2}\times \underset{\text{base}}{\underbrace{x}}\times \underset{\text{height}}{\underbrace{kx}}=\frac{1}{2}k{x}^2$

Spring ki apni force $W=-\frac{1}{2}k{x}^2$ work karti hai (woh stretch ka virodh karti hai), isliye:

$\Delta U=-W=\frac{1}{2}k{x}^2$

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Step 3 — Stretch aur squeeze dono ki cost same kyun hai

KYA. $x<0$ (compression) ka case bhi check karo saath mein $x>0$ (stretch) ke.

KYUN. Contract kehta hai har sign cover karo. Jab spring ko andar push karo toh kya hota hai?

PICTURE. Kyunki $U=\frac{1}{2}k{x}^2$ aur squaring sign ko khatam kar deti hai, $U(-x)=U(+x)$. Energy curve ek symmetric parabola hai — ek valley jiska lowest point $x=0$ par hai. $0.1\text{m}$ compress karna exactly utna hi store karta hai jitna $0.1\text{m}$ stretch karna.

$x=0$ par (natural length) stored energy zero hai — valley ka bottom. Yeh degenerate case hai: koi stretch nahi, koi store nahi.

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Step 4 — Ek force jo distance ke saath khatam hoti hai: $-\frac{GMm}{r}$ setup karna

KYA. Ek mass $m$ ko $M$ mass wale planet ki taraf move karo, aur poochho ki separation $r$ ke change hone par stored energy kaise behave karti hai.

KYUN ek naya tool. Spring ki force linearly badhti thi. Deep space mein gravity iska ulta karti hai — woh door jaane par fade hoti hai, Gravitation — Newton's law follow karte hue:

$F=-\frac{GMm}{r^2}$

• $r$ — do centres ke beech distance (metres).
• $M,m$ — do masses; dono aate hain kyunki gravity ek mutual pull hai.
• $G$ — universal gravitational constant, ek fixed tiny number.
• $\frac{1}{r^2}$ — inverse-square shape: distance double karo, force quarter ho jaati hai.
• minus — pull andar ki taraf hai, planet ki taraf.

PICTURE. Force graph ab ek steep curve hai jo planet ke paas deep jaati hai aur door jaane par zero ki taraf flat hoti hai. Iska area rectangle ya triangle nahi hai — yeh ek curved $1/r^2$ shape ke neeche ka area hai. Yahi reason hai ki hume poora integral chahiye.

Zero infinity par kyun rakha? $mgh$ ke liye humne $U=0$ floor par rakha. Yahan koi natural floor nahi hai — lekin infinitely dur pull zero hai, isliye yeh jagah declare karna sabse clean hai ki "koi stored energy nahi." Hum sab kuch "infinity tak escape kar gaya" ke relative measure karte hain.

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Step 5 — Curved-area sum karna: formula appear hota hai

KYA. Tiny works add karo jaise $m$ infinity se distance $r$ tak girta hai.

KYUN. Yeh Step 4 ke graph ka payoff hai: integral exactly shaded curved area hai.

= -GMm\left[-\frac{1}{r'}\right]_{\infty}^{r} = -GMm\left(-\frac{1}{r}+0\right) = \frac{GMm}{r}$$ Term by term padhte hain: - $r'$ — ek dummy stand-in varying distance ke liye jab hum $\infty$ se $r$ tak sweep karte hain. - $\left[-\dfrac1{r'}\right]$ — "antiderivative": woh function jiska slope $\dfrac1{r'^2}$ hai. Hum ise isliye pehchante hain kyunki $\dfrac{d}{dr'}\!\left(-\dfrac1{r'}\right)=\dfrac1{r'^2}$. - $\infty$ plug in karne par $0$ milta hai; $r$ plug in karne par $-\dfrac1r$ milta hai. Gravity ne **positive** work kiya $+\dfrac{GMm}{r}$ (usne mass ko girne mein *help* ki), toh master rule se: $$U(r) = -W = -\frac{GMm}{r}$$ ![[deepdives/dd-physics-1.3.05-d2-s06.png]] > [!formula] Universal gravitational PE > $$U(r) = -\frac{GMm}{r} \qquad (U=0 \text{ at } r=\infty)$$ > **Negative** value ka matlab hai "infinity line ke neeche" — mass ek **well** mein baitha hai. Wapas infinity tak climb karne ke liye tumhe energy *add* karni padegi (yahi [[Escape velocity|escape energy]] hai). > [!intuition] Yeh *zaroori* negative kyun hai > Infinity se shuru karo jahan $U=0$ hai. Andar giro — gravity positive work karti hai, toh $U$ *girna* chahiye, zero se neeche jaana chahiye. Zyada gehra girna (chhota $r$) matlab aur zyada negative $U$. Dekho [[Conservative vs non-conservative forces]] ki kyun yeh single-number $U$ exist bhi karta hai. --- ## Step 6 — Do gravity formulas actually ek hi formula hain **KYA.** Dikhao ki $-\dfrac{GMm}{r}$ thodi si movement par $mgh$ ban jaata hai. **KYUN.** Ek smart reader sochega: "gravity ke liye do alag formulas?" Chhoti climbs ke liye inhe agree karna chahiye. **PICTURE.** Deep gravity curve ko surface $r=R$ par zoom karo. Ek tiny height $h$ par, curve ek *straight ramp* jaisi dikhti hai — aur ek straight ramp exactly Step 1 ki flat-force duniya hai. $$\Delta U = -GMm\left(\frac{1}{R+h}-\frac{1}{R}\right)\approx GMm\,\frac{h}{R^2}=\left(\frac{GM}{R^2}\right)mh = mgh$$ $g=\dfrac{GM}{R^2}$ use karte hue. - $R$ — Earth ka radius (jahan tum khade ho). - $h$ — chhoti climb, $R$ ke compare mein tiny. - approximation $\approx$ — valid sirf isliye kyunki $h\ll R$ curve ko flat kar deta hai. ![[deepdives/dd-physics-1.3.05-d2-s07.png]] Toh $mgh$ bas bade gravity well ka **pehla tiny slice** hai, itna close dekha gaya ki curve flat lagti hai. --- ## Step 7 — Energy landscape se force wapas padhna **KYA.** Kisi bhi $U$ se slope use karke force recover karo (dekho [[Force from potential — F = -dU/dx]]). **KYUN.** Humne $U$ force se banaya; reverse bhi kaam karna chahiye — aur yeh hamare answers double-check bhi karta hai. $$F_x = -\frac{dU}{dx}$$ - $\dfrac{dU}{dx}$ — energy curve ka **slope** (move karne par $U$ kitni tezi se badhti hai). - **minus** — force *downhill* point karti hai, lower energy ki taraf, jaise ek ball valley mein girta hai. **PICTURE.** Spring parabola par: $x>0$ par curve *upar* slope karti hai, slope positive hai, toh $F=-kx$ wapas left point karti hai. $x<0$ par neeche slope karti hai, force right point karti hai. Har jagah arrow valley ke bottom ki taraf point karta hai. $$F=-\frac{d}{dx}\!\left(\tfrac12 kx^2\right)=-kx\;\checkmark$$ ![[deepdives/dd-physics-1.3.05-d2-s08.png]] Landscape par arrows exactly wahi Hooke's restoring force hain jo humne shuru ki thi — loop close ho gaya. Yeh [[Conservation of mechanical energy]] visible ho gaya: energy curve ki height ($U$) aur speed ($KE$) ke beech sloshe karti hai, kabhi nahi khoti. --- ## Ek-picture summary ![[deepdives/dd-physics-1.3.05-d2-s09.png]] Teen forces, teen graph shapes, teen areas: | Case | Force graph | Area = $U$ | |---|---|---| | Near-Earth gravity | flat line $mg$ | rectangle → $mgh$ | | Spring | ramp $kx$ | triangle → $\tfrac12kx^2$ | | Deep-space gravity | curve $\dfrac{GMm}{r^2}$ | curved area → $-\dfrac{GMm}{r}$ | > [!recall]- Feynman retelling — poora walkthrough plain words mein > Potential energy bas "wo work hai jo tumne bank kiya." Ye jaanne ke liye ki tumne kitna bank kiya, force ka graph banao us distance ke against jitni cheez move hui, aur neeche ka area measure karo — kyunki us graph par area *hi* total work hai. Ek flat force (zameen ke paas uthana) ek plain rectangle deta hai, aur uska area $mgh$ hai. Spring jitni zyada kheencho utna hard pull karti hai, toh uska graph ek slanted ramp hai, aur ramp ka area ek triangle hai — woh triangle $\tfrac12kx^2$ hai, aur chhota $\tfrac12$ bas "half base times height" hai. Deep-space gravity iska ulta karti hai: woh distance ke saath ek swooping $1/r^2$ curve mein fade hoti hai, toh uska area ek curvy piece hai jise hum integral se add karte hain, $-GMm/r$ deta hai — negative isliye kyunki hum "infinity tak escape" ko zero point kehte hain, toh planet ke paas trapped rehna ek hole mein baithne jaisa count hota hai. Us hole ko surface par zoom karo aur curve phir straight lagti hai — woh straight bit exactly $mgh$ hai, toh do gravity formulas hamesha se ek hi thi. Finally, force wapas paane ke liye bas dekho energy curve kitni *steeply* rise karti hai: force hamesha downhill point karti hai, valley ke bottom ki taraf. > [!mnemonic] Shape formula batati hai > **Rectangle → $mgh$. Triangle → $\tfrac12kx^2$. Curve → $-GMm/r$.** Aur reverse ke liye: **slope down = force ka rasta** ($F=-dU/dx$). ## Connections - [[1.3.05 Potential energy — definition, gravitational (mgh and −GMm - r), elastic (½kx²) (Hinglish)]] - [[Work done by a force]] - [[Hooke's Law]] - [[Gravitation — Newton's law]] - [[Force from potential — F = -dU/dx]] - [[Conservation of mechanical energy]] - [[Conservative vs non-conservative forces]] - [[Escape velocity]]