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

1.2.22 · Physics › Newton's Laws & Dynamics

WHY do we even define potential energy?

HOW to derive $U = - GM m / r$ from scratch

Step 1 — Force likhein. Newton's law of gravitation: origin par rakha mass $M$ , distance $r$ par rakhe mass $m$ ko kheenchta hai $F = - \frac{GM m}{r ^{2}} \overset{r}{^} .$ Minus sign kyun? Force andar ki taraf point karta hai ( $M$ ki taraf), outward radial unit vector $\overset{r}{^}$ ke opposite.

Step 2 — $m$ ko purely radially move karein kisi distance $r$ se infinity tak. Tab $d r = d r \overset{r}{^}$ aur $F \cdot d r = - \frac{GM m}{r ^{2}} d r .$

Step 3 — Reference set karein. Choose karein $U (\infty) = 0$ . Tab $U (r) = U (r) - U (\infty) = - \int_{\infty}^{r} F \cdot d r = - \int_{\infty}^{r} (- \frac{GM m}{r ^{'2}}) d r^{'} .$

Step 4 — Integral karein. $U (r) = GM m \int_{\infty}^{r} \frac{d r ^{'}}{r ^{'2}} = GM m [- \frac{1}{r ^{'}}]_{\infty}^{r} = GM m (- \frac{1}{r} + 0) .$

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

WHY $m g h$ is just the local approximation

Earth ki surface ke paas, $r = R + h$ jahan $h ≪ R$ ( $R =$ Earth ki radius). Surface se height $h$ tak PE mein change: $Δ U = - \frac{GM m}{R + h} - (- \frac{GM m}{R}) = GM m (\frac{1}{R} - \frac{1}{R + h}) = GM m \cdot \frac{h}{R ( R + h )} .$

Kyunki $h ≪ R$ , $R + h \approx R$ , isliye $Δ U \approx GM m \cdot \frac{h}{R ^{2}} = m = g (\frac{GM}{R ^{2}}) h = m g h .$

Worked examples

Common mistakes (Steel-man + fix)

Forecast-then-Verify

Recall Dekhne se pehle predict karein

Forecast: Agar aap orbital radius $r$ ko double kar dein, toh total orbital energy $E = - GM m /2 r$ kaise change hogi? Verify: $E$ half as negative ho jaata hai (0 ke karib), yaani energy badhti hai. Higher orbits higher-energy hain — consistent hai kyunki orbit ko raise karne ke liye kaam karna padta hai. ✔

Flashcards

Gravitational PE negative kyun hoti hai?

Humne

U (\infty) = 0

set kiya; har finite

r

ek bound state hai reference se kam energy ke saath, isliye

U < 0

Force se

U (r)

derive karein.

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

m g h

surface ke paas kyun kaam karta hai?

Yeh

- GM m / r

ka Taylor/small-

h

limit hai jahan

g = GM / R^{2}

aur

R + h \approx R

g

aur

G

ka relation?

g = GM / R^{2}

— universal constant se local field.

Escape speed formula aur mass-independent kyun?

v_{esc} = 2 GM / R

;

m

cancel ho jaata hai

\frac{1}{2} m v^{2} = GM m / R

mein.

Circular orbit ki total energy?

E = - GM m /2 r

(aur

E = \frac{1}{2} U = - K

Jaise

r

badhta hai,

U

badhti hai ya ghatti hai?

Badhti hai (0 ki taraf); magnitude shrink hoti hai.

Potential energy exist karne ke liye force kaisi honi chahiye?

Ek conservative force (path-independent work).

Recall Feynman: 12-saal ke bachche ko explain karein

Ek ball ko ek gehri khai mein imagine karein. Use bahar nikalne ke liye aapko poore raaste upar dhakka dena padta hai — isme mehnat (energy) lagti hai. Space mein bhi aisa hi hota hai: planets invisible "gravity pits" mein baithe hain. Hum kehte hain ki khai ka bottom "zero" se neeche hai, isliye yeh ek negative number hai, aur khai ka top (space mein bahut door) zero hai. Jitna gehra khai mein (planet ke karib), utna zyada negative. $m g h$ tab use karte hain jab khai itni shallow ho ki flat ramp jaisi lage — hills aur buildings ke liye kaam aata hai, Moon jaane wale rockets ke liye nahi.

Connections

Newton's Law of Universal Gravitation — woh force jise humne integrate kiya.
Conservative Forces and Potential Energy — $U$ exist kyun karta hai.
Work-Energy Theorem — $W$ aur $Δ K E$ ko $Δ U$ se link karta hai.
Escape Velocity — direct application $E = 0$ set karke.
Kepler's Laws & Orbital Energy — $E = - GM m /2 r$ use karta hai.
Acceleration due to gravity g — local limit $g = GM / R^{2}$ .