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

1.2.22 · D3 · Physics › Newton's Laws & Dynamics › Gravitational potential energy — U = −GMm - r (not mgh)

Do quantities jo hum har ek problem mein use karenge — inhe ek baar define kar lete hain, clearly:

$U (r) = - \frac{GM m}{r}$ — do-mass system ki stored energy jab woh $r$ distance par hain. Hamesha $\leq 0$ . Pit ka bottom bahut negative hota hai; door ( $r \to \infty$ ) jaate-jaate yeh $0$ tak climb karta hai.
$K = \frac{1}{2} m v^{2}$ — kinetic energy, motion ki energy. Hamesha $\geq 0$ .
$E = K + U$ — total mechanical energy. Kyunki gravity ek conservative force hai, $E$ nahi badlti jab object coast karta hai (koi engine nahi, koi air nahi). Yeh akela fact — " $E$ before = $E$ after" — neeche ke almost sab kuch solve kar deta hai.

Parent dekho: (parent topic).

The scenario matrix

Gravity-energy ka har question inhi cells mein se ek hai. Columns batate hain kya fixed hai aur kya solve karna hai.

#	Cell class	Kya special hai	Solved in
A	Sign of $E$ negative — bound object	$E < 0$ : object escape nahi kar sakta	Ex 1
B	Sign of $E$ zero — exact escape	$E = 0$ : bas $\infty$ tak pahunch paata hai	Ex 2
C	Sign of $E$ positive — unbound / excess speed	$E > 0$ : $\infty$ par bhi move karta hua pahunchta hai	Ex 3
D	Small- $h$ limit — degenerate to $m g h$	check $- GM m / r \to m g h$ , $v \to 2 g h$	Ex 4
E	Large- $r$ / limiting value	$r \to \infty$ , $U, g$ ka kya hota hai?	Ex 5
F	Circular orbit energy split	$K, U, E$ relationship, virial	Ex 6
G	Changing orbit (radius $r_{1} \to r_{2}$ )	energy difference, work ka sign	Ex 7
H	Real-world word problem	messy numbers, satellite launch	Ex 8
I	Exam twist — two bodies / non-radial	conservation with a hidden trick	Ex 9
J	Degenerate input $r \to 0$ / $M = 0$	formula edges par kya kehta hai	Ex 10

Constants jo throughout use honge (yeh teen yaad karo): $G = 6.67 \times 1 0^{- 11} N\cdotm^{2} / kg^{2}, M_{\oplus} = 5.97 \times 1 0^{24} kg, R_{\oplus} = 6.37 \times 1 0^{6} m .$

Ex 1 — Cell A: ek bound object (negative total energy)

Worked example

Ek $1 kg$ probe Earth ki surface par baitha hai aur use seedha upar $v = 5000 m/s$ pe throw kiya jaata hai. Kya yeh bound hai (wapas girega) ya free? Iska total energy $E$ nikalo.

Forecast: $5 km/s$ escape speed se adhe se bhi kam hai. Aage padhne se pehle $E$ ka sign guess karo.

Step 1 — Surface par dono energies likho. Yeh step kyun? $E = K + U$ ; hume sirf ek moment ki state chahiye kyunki $E$ conserved hai. $K = \frac{1}{2} (1) (5000)^{2} = 1.25 \times 1 0^{7} J .$ $U = - \frac{G M _{\oplus} m}{R _{\oplus}} = - \frac{( 6.67 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} ) ( 1 )}{6.37 \times 1 0 ^{6}} = - 6.25 \times 1 0^{7} J .$

Step 2 — Dono ko add karo. Yeh step kyun? $E$ ka sign hi poora answer hai: $E < 0$ matlab "pit mein hai, wapas aayega." $E = 1.25 \times 1 0^{7} - 6.25 \times 1 0^{7} = - 5.00 \times 1 0^{7} J .$

Step 3 — Interpret karo. $E < 0 \Rightarrow$ bound. Probe upar jaata hai, rukta hai, wapas girta hai. Yeh Cell A hai.

Verify: $∣ U ∣$ ("pit ki gehraai," $6.25 \times 1 0^{7}$ ) $K$ ( $1.25 \times 1 0^{7}$ ) se zyada hai, isliye probe bahar nahi nikal sakta — $E < 0$ ke saath consistent hai. Units: sab joules. ✔

Ex 2 — Cell B: exact escape (total energy zero)

Worked example

Kaun si launch speed Earth ki surface se exactly $E = 0$ deti hai? (Yeh escape speed hai, Cell B ko cleanly hit karne ke liye re-derived.)

Forecast: Kya answer mass $m$ par depend karega? Haan/nahi guess karo.

Step 1 — $E = 0$ set karo. Yeh step kyun? $E = 0$ bound ( $E < 0$ ) aur free ( $E > 0$ ) ke beech ki boundary hai: object $r = \infty$ par $v = 0$ ke saath pahunchta hai. $\frac{1}{2} m v_{esc}^{2} - \frac{G M _{\oplus} m}{R _{\oplus}} = 0.$

Step 2 — $m$ cancel karo, solve karo. Yeh step kyun? Har term mein $m$ ka ek factor hota hai, isliye woh divide ho jaata hai — issi liye escape speed mass-independent hoti hai. $v_{esc} = \frac{2 G M _{\oplus}}{R _{\oplus}} = \frac{2 ( 6.67 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} )}{6.37 \times 1 0 ^{6}} .$

Step 3 — Compute karo. $= 1.118 \times 1 0^{4} m/s \approx 11.2 km/s$ .

Verify: Ek pankh aur ek rocket dono ko same $11.2 km/s$ chahiye — $m$ cancel ho gaya, forecast "koi dependence nahi" confirm karta hai. Dekho Escape Velocity. Units: $m^{2} / s^{2} = m/s$ . ✔

Ex 3 — Cell C: unbound object (positive total energy, leftover speed)

Worked example

Ek meteoroid Earth ki surface se $v_{0} = 15 km/s$ (escape se tez) fire kiya jaata hai. Bahut door ( $r \to \infty$ ) yeh kitni tez chal raha hoga?

Forecast: Kya $v_{\infty}$ , $15 km/s$ se zyada hoga ya kam? Lagbhag kitna?

Step 1 — Energy conserve karo, surface se infinity tak. Yeh step kyun? $\infty$ par $U \to 0$ , isliye baaki sari energy kinetic hai — yeh "excess" escape ke upar ka leftover hai. $\frac{1}{2} m v_{0}^{2} - \frac{G M _{\oplus} m}{R _{\oplus}} = \frac{1}{2} m v_{\infty}^{2} + 0.$

Step 2 — $m$ cancel karo, $v_{\infty}$ isolate karo. $v_{\infty}^{2} = v_{0}^{2} - \frac{2 G M _{\oplus}}{R _{\oplus}} = v_{0}^{2} - v_{esc}^{2} .$ Yeh step kyun? Yeh neat identity hai $v_{\infty}^{2} = v_{0}^{2} - v_{esc}^{2}$ : escape speed exactly woh amount hai jo gravity pit "kha jaati" hai.

Step 3 — Plug in karo. $v_{0} = 1.5 \times 1 0^{4}$ , $v_{esc} = 1.118 \times 1 0^{4}$ : $v_{\infty} = (1.5 \times 1 0^{4})^{2} - (1.118 \times 1 0^{4})^{2} = 2.25 \times 1 0^{8} - 1.250 \times 1 0^{8} \approx 10.0 \times 1 0^{3} m/s .$

Verify: $v_{\infty} \approx 10.0 km/s < 15 km/s$ — bahar nikalte waqt yeh slow ho gaya, jaise forecast kiya tha. Aur $E = \frac{1}{2} m v_{\infty}^{2} > 0$ Cell C (unbound) confirm karta hai. ✔

Ex 4 — Cell D: degenerate small-height limit ( $m g h$ )

Worked example

Ek patthar $45 m$ ki cliff se giraya jaata hai. Uski landing speed do tarikон se nikalo — exact $- GM m / r$ se aur approximation $2 g h$ se — aur dikhao ki woh agree karte hain.

Notation pehle: maano $R = R_{\oplus} = 6.37 \times 1 0^{6} m$ Earth ka radius hai (Earth ke centre se ground tak ki distance), aur $h = 45 m$ girne ki height hai. Toh patthar $r = R + h$ se $r = R$ tak girta hai.

Forecast: Kya exact answer $2 g h$ se thoda bada hoga ya chota? (Socho: kya $g$ neeche thoda zyada strong hota hai?)

Step 1 — Exact energy conservation. Yeh step kyun? Hum test karte hain ki kya jab $h ≪ R$ ho toh full formula schoolbook wala collapse ho jaata hai. $v = 2 G M_{\oplus} (\frac{1}{R} - \frac{1}{R + h}) .$ $R = 6.37 \times 1 0^{6} m$ aur $h = 45 m$ ke saath: $v = 2 (3.98 \times 1 0^{14}) (\frac{1}{6.37 \times 1 0 ^{6}} - \frac{1}{6.37 \times 1 0 ^{6} + 45}) \approx 29.7 m/s .$

Step 2 — $g = 9.8$ se approximate karo. Yeh step kyun? $m g h$ tab aata hai jab $R + h \approx R$ ; yahaan $h / R \sim 7 \times 1 0^{- 6}$ bahut chota hai. $v \approx 2 (9.8) (45) = 882 \approx 29.7 m/s .$

Step 3 — Compare karo. Woh dikhaye gaye digits tak match karte hain; difference ek part in $1 0^{5}$ ka hai.

Verify: Exact value microscopically badi hai (gravity ground par $45 m$ se thodi zyada strong hoti hai), forecast ke saath match karta hai. Roz ke drops ke liye $g$ use karo. ✔

Ex 5 — Cell E: $r \to \infty$ aur $r$ double hone par limiting values

Worked example

(a) $r \to \infty$ hone par $U$ aur field $g (r) = GM / r^{2}$ kya hoti hain? (b) Ek $100 kg$ satellite ke liye $r = 2 R_{\oplus}$ vs $r = R_{\oplus}$ par $U$ compare karo.

Forecast: $r \to \infty$ hone par kya $U$ , $0$ ke upar se approach karta hai ya neeche se?

Step 1 — Limits. Yeh step kyun? End-behaviour check karna hi batata hai ki reference sensible hai. $lim_{r \to \infty} (- \frac{GM m}{r}) = 0^{-}, lim_{r \to \infty} \frac{GM}{r ^{2}} = 0.$ Dono vanish ho jaate hain — infinitely door, koi stored energy nahi aur koi pull nahi. $U$ , $0$ ke neeche se approach karta hai (hamesha negative).

Step 2 — $R_{\oplus}$ par $U$ . Yeh step kyun? Hume surface par ek concrete reference value chahiye taaki Step 3 ki "doubling" comparison mein kuch ho jo halve ho sake. $U (R) = - \frac{( 6.67 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} ) ( 100 )}{6.37 \times 1 0 ^{6}} = - 6.25 \times 1 0^{9} J .$

Step 3 — $2 R_{\oplus}$ par $U$ . Yeh step kyun? Kyunki $U \propto 1/ r$ , $r$ double karne se magnitude half ho jaani chahiye — yeh step abstract " $1/ r$ " scaling ko numerically visible banata hai. $U (2 R) = - \frac{G M _{\oplus} m}{2 R _{\oplus}} = - 3.13 \times 1 0^{9} J .$

Verify: $U (2 R) = \frac{1}{2} U (R)$ exactly (formula $\propto 1/ r$ hai). Upar jaane par ⇒ kam negative ⇒ bada $U$ , "upar jaana energy raise karta hai" se match karta hai. ✔

Ex 6 — Cell F: circular orbit ka energy split

Worked example

Ek $500 kg$ satellite radius $r = 7.0 \times 1 0^{6} m$ par Earth ke chakkar laga raha hai. $K$ , $U$ , aur $E$ nikalo, aur virial relations confirm karo.

Forecast: $E$ ka sign guess karo aur yeh bhi ki $∣ K ∣$ bada hoga ya $∣ U ∣$ ?

Step 1 — "Gravity = centripetal" use karo. Yeh step kyun? Circular orbit ke liye inward pull exactly centripetal force supply karta hai, jo $v$ (aur isliye $K$ ) ko sirf $r$ se fix kar deta hai. $\frac{GM m}{r ^{2}} = \frac{m v ^{2}}{r} \Rightarrow K = \frac{1}{2} m v^{2} = \frac{GM m}{2 r} .$ $K = \frac{( 6.67 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} ) ( 500 )}{2 ( 7.0 \times 1 0 ^{6} )} = 1.42 \times 1 0^{10} J .$

Step 2 — $U$ compute karo. Yeh step kyun? Hume $U$ alag se chahiye taaki use $K$ mein add karke total ka sign dekh sakein. $U = - \frac{GM m}{r} = - 2.84 \times 1 0^{10} J .$

Step 3 — $E$ ke liye add karo. Yeh step kyun? Total ka sign orbit ko bound classify karta hai, aur uski value virial pattern expose karti hai. $E = K + U = 1.42 \times 1 0^{10} - 2.84 \times 1 0^{10} = - 1.42 \times 1 0^{10} J = - \frac{GM m}{2 r} .$

Verify: $E = - 1.42 \times 1 0^{10}$ , $\frac{1}{2} U$ ke barabar hai aur $- K$ ke barabar hai — virial relations $E = \frac{1}{2} U = - K$ . $E < 0$ (bound). Dekho Kepler's Laws & Orbital Energy. ✔

Neeche ka figure $r$ mein curves ke roop mein yeh split plot karta hai: cyan potential $U$ zero se neeche baitha hai, amber kinetic $K$ uske upar hai, aur dashed white total $E = - GM m /2 r$ exactly beech mein thread karta hai. Orbit radius (marked) padhlo: us vertical line par dashed $E$ curve, cyan $U$ value ki half par hai aur amber $K$ value ka negative hai — $E = \frac{1}{2} U = - K$ ki picture.

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

Ex 7 — Cell G: ek orbit ko raise karna (energy difference aur uska sign)

Worked example

Ex 6 wale $500 kg$ satellite ko $r_{1} = 7.0 \times 1 0^{6} m$ se upar $r_{2} = 1.4 \times 1 0^{7} m$ par le jaao. Thrusters ko kitni energy add karni padegi?

Forecast: Kyunki higher orbits higher-energy hote hain, kya $Δ E$ positive hoga? Aur kya yeh poore $∣ E_{1} ∣$ se zyada hoga ya kam?

Step 1 — Har radius par $E = - \frac{GM m}{2 r}$ use karo. Yeh step kyun? Sirf total energies ka difference matter karta hai; jo work supply karni hai woh hai $W = E_{2} - E_{1}$ . $E_{1} = - \frac{GM m}{2 r _{1}} = - 1.42 \times 1 0^{10} J, E_{2} = - \frac{GM m}{2 r _{2}} = - 7.11 \times 1 0^{9} J .$

Step 2 — Subtract karo. Yeh step kyun? $E_{2} - E_{1}$ ka sign batata hai ki engine energy add kar raha hai ya remove. $W = E_{2} - E_{1} = (- 7.11 \times 1 0^{9}) - (- 1.42 \times 1 0^{10}) = + 7.11 \times 1 0^{9} J .$

Step 3 — Interpret karo. Positive ⇒ tumhe energy add karni padegi (engine fire karo), Work-Energy Theorem ke saath consistent.

Verify: $W > 0$ jaise forecast kiya; aur kyunki $r_{2} = 2 r_{1}$ , $E_{2} = \frac{1}{2} E_{1}$ , isliye $W = - \frac{1}{2} E_{1} = 7.11 \times 1 0^{9} J$ . Interesting baat yeh hai ki satellite ki speed ghatti hai even though energy badhi — pit $K$ ke girne se tez shallow ho gayi. ✔

Ex 8 — Cell H: real-world word problem (satellite launch budget)

Worked example

Ek telecom company $1200 kg$ ka satellite circular orbit mein altitude $800 km$ ( $r = R_{\oplus} + 8.0 \times 1 0^{5} = 7.17 \times 1 0^{6} m$ ) par chahti hai. Earth ki surface par rest se start karte hue, launch ko minimum kitni energy deliver karni padegi? (Earth ka rotation aur air ignore karo.)

Forecast: Kya zyada energy lifting mein jaayegi ( $U$ raise karne mein) ya speeding up mein (orbital $K$ dene mein)?

Step 1 — Launch pad par energy. Yeh step kyun? Start state: rest par ( $K_{i} = 0$ ) surface par. $E_{i} = 0 - \frac{G M _{\oplus} m}{R _{\oplus}} = - \frac{( 6.67 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} ) ( 1200 )}{6.37 \times 1 0 ^{6}} = - 7.50 \times 1 0^{10} J .$

Step 2 — Final orbit mein energy. Yeh step kyun? Circular orbit total energy $- GM m /2 r$ hoti hai. $E_{f} = - \frac{G M _{\oplus} m}{2 r} = - \frac{( 6.67 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} ) ( 1200 )}{2 ( 7.17 \times 1 0 ^{6} )} = - 3.33 \times 1 0^{10} J .$

Step 3 — Required energy = difference. $W = E_{f} - E_{i} = (- 3.33 \times 1 0^{10}) - (- 7.50 \times 1 0^{10}) = + 4.17 \times 1 0^{10} J .$

Verify: Positive (tumhe supply karni padegi). Split: lifting cost $Δ U = U_{f} - U_{i} \approx + 8.3 \times 1 0^{9} J$ ; orbital $K_{f} \approx 3.33 \times 1 0^{10} J$ . Toh zyaatar energy orbital speed ban jaati hai, height nahi — zyaatar physics students ke liye surprising, forecast ke saath match karta hai. Units: throughout joules. ✔

Ex 9 — Cell I: exam twist (do comparable masses, non-radial launch)

Worked example

Do identical asteroids, har ek mass $m = 2.0 \times 1 0^{12} kg$ , deep space mein $d = 1.0 \times 1 0^{5} m$ door rest se release kiye jaate hain. Jab woh aadhi us distance par hon, $d /2$ , toh har ek kitni tez chal raha hoga?

Forecast: Trap: $U = - GM m / r$ mein kaun sa $M$ jaata hai jab dono masses move kar rahe hain? Speed guess karo — kya dono kinetic energy equally share karenge?

Step 1 — Shared potential energy likho. Yeh step kyun? Pair ki EK potential energy hai $U = - G m_{1} m_{2} / r$ jahan $r =$ unka separation — koi "central body" nahi. $U_{i} = - \frac{G m ^{2}}{d}, U_{f} = - \frac{G m ^{2}}{d /2} = - \frac{2 G m ^{2}}{d} .$

Step 2 — Energy conserve karo (rest se start). Yeh step kyun? $E$ conserved: initial $K = 0$ , isliye released potential dono bodies ki kinetic ban jaati hai. $0 + U_{i} = dono asteroids 2 \cdot \frac{1}{2} m v^{2} + U_{f} \Rightarrow m v^{2} = U_{i} - U_{f} = \frac{G m ^{2}}{d} .$ Symmetry se har ek ki same speed $v$ hai; isliye $K_{total} = 2 \cdot \frac{1}{2} m v^{2} = m v^{2}$ .

Step 3 — $v$ isolate karo aur compute karo. Yeh step kyun? Last equation ko $m$ se divide karo taaki $v^{2} = G m / d$ mile, phir square root lo — yeh final numerical answer hai. $v = \frac{G m}{d} = \frac{( 6.67 \times 1 0 ^{- 11} ) ( 2.0 \times 1 0 ^{12} )}{1.0 \times 1 0 ^{5}} = 1.334 \times 1 0^{- 3} \approx 3.65 \times 1 0^{- 2} m/s .$

Verify: Momentum throughout zero hai (rest se start kiya, symmetric), isliye dono velocities equal aur opposite hain — har ek $3.65 \times 1 0^{- 2} m/s$ . Agar tumne galti se ek asteroid fixed maana hota, toh KE double ho jaati aur galat $v$ milta. Dekho Newton's Law of Universal Gravitation. ✔

Ex 10 — Cell J: degenerate inputs ( $r \to 0$ , $M \to 0$ )

Worked example

$U = - GM m / r$ kya predict karta hai jab (a) $r \to 0$ aur (b) $M \to 0$ ? Kya yeh physical hain?

Forecast: Guess karo ki jab do masses touch karein toh $U \to 0$ hoga ya $U \to - \infty$ ?

Step 1 — $r \to 0$ karo. Yeh step kyun? Edge cases test karte hain kahan ek formula pe trust karna band hota hai — $r$ ko zero push karna "masses touch" limit probe karta hai. $lim_{r \to 0^{+}} (- \frac{GM m}{r}) = - \infty.$ Pit infinitely deep ho jaati hai. Interpretation: real masses ki finite size $R$ hoti hai, isliye $r$ actually $0$ tak kabhi nahi pahunchta; formula point masses / distances $\geq$ surface ke liye hai. Surface ke neeche ek alag expression apply hota hai, isliye $- \infty$ ek modelling artefact hai, koi real infinite energy nahi.

Step 2 — $M \to 0$ karo. Yeh step kyun? Source mass ko zero send karna "no gravity" limit probe karta hai aur check karta hai ki formula gracefully degrade karta hai. $lim_{M \to 0} (- \frac{GM m}{r}) = 0.$ Interpretation: koi central mass nahi ⇒ koi gravity nahi ⇒ koi stored energy nahi. Bilkul sensible — pit kuch bhi nahi ho jaati.

Step 3 — Final interpretation. Cell J "edges ki sanity" check hai: $M \to 0$ limit physically clean hai ( $U \to 0$ ), jabki $r \to 0$ limit ( $U \to - \infty$ ) ek warning flag hai ki point-mass model apni valid range chod chuka hai. $U = - GM m / r$ sirf tab trust karo jab $r \geq$ body ka radius aur $M > 0$ ho.

Verify: $M = 0$ par answer exactly $0$ hai (checkable); $r \to 0$ par magnitude unboundedly badhti hai. Dono $U \propto M$ aur $U \propto 1/ r$ ke saath consistent hain. ✔

Recall Matrix ke across quick self-test

Har question kaun sa cell hai? "Satellite 400 km se 300 km tak girta hai — energy change?" ::: Cell G (changing orbit). "Patthar 20 km/s par throw kiya — infinity par speed?" ::: Cell C (unbound, $E > 0$ ). "Ball 10 m ki roof se — landing speed?" ::: Cell D (small- $h$ , use $2 g h$ ).

Connections

Conservative Forces and Potential Energy — kyun ek number $E$ fixed rehta hai.
Work-Energy Theorem — thruster work $= Δ E$ Ex 7–8 mein.
Escape Velocity — Cells B aur C.
Kepler's Laws & Orbital Energy — Cell F virial split.
Newton's Law of Universal Gravitation — $U$ ke peeche force, Ex 9.
Acceleration due to gravity g — local limit $g = GM / R^{2}$ , Cell D.