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

1.2.22 · D5 · Physics › Newton's Laws & Dynamics › Gravitational potential energy — U = −GMm - r (not mgh)

Shuru karne se pehle, actual shape dekho jis par neeche saare traps refer karte hain — gravitational potential well:

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

Recall The single rule behind all these traps

$U = - \frac{GM m}{r}$ negative hai, aur bada hona matlab zero ki taraf jaana hai (well curve par rightward aur upar). Toh "more energy" = "less negative" = "bigger $r$ ". Well par higher energy kaunsi taraf hai? ::: Rightward aur upar — $U = 0$ ceiling ki taraf. Planet se door jaana well se bahar climb karna hai, energy raise karta hai.

True or false — justify karo

Har item ek statement hai. True/false decide karo, phir reason do — sirf T/F se kuch nahi milega.

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

real masses ke liye finite

r

par kabhi positive nahi ho sakta.

True.

G, M, m, r

sab positive hain, isliye

- GM m / r < 0

hamesha; yeh

0

sirf

r \to \infty

ki limit mein reach karta hai, kabhi upar cross nahi karta.

Agar satellite A

r

par hai aur satellite B

2 r

par, toh B ka gravitational PE zyada (less negative) hai.

True.

U_{B} = - GM m /2 r

sirf aadha magnitude hai

U_{A} = - GM m / r

ka, aur negative number ka aadha less negative hota hai, isliye

U_{B} > U_{A}

. Higher orbit = higher potential energy. Neeche two-orbit figure dekho.

Surface se height

h

ko double karne par PE gain hamesha double ho jaati hai.

False. Sirf

m g h

(constant-

g

) approximation mein. Exact

Δ U = GM m (\frac{1}{R} - \frac{1}{R + h})

h

mein nonlinear hai; chadhte waqt har metre par increase shrink hoti hai.

Gravitational PE ka zero ek law of nature hai jo infinity par fixed hai.

False. Zero ek choice hai. Hum

U (\infty) = 0

isliye choose karte hain kyunki yeh ek inverse-square force ke liye ek saath sab

r

ke liye kaam karne wala ek hi reference hai, lekin koi bhi constant add kiya ja sakta hai — sirf

Δ U

physical hai.

Ek bound circular orbit ke liye, total mechanical energy negative hoti hai.

True.

E = - GM m /2 r < 0

. Negative hona (

r = \infty

ke zero se neeche) exactly yehi hai jiska matlab "bound" hai — freedom reach karne ke liye energy add karni padegi.

Escape speed escaping object ke mass par depend karti hai.

False.

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

mein

m

cancel ho jaata hai,

v_{esc} = 2 GM / R

deta hai. Ek pebble aur ek probe ko same launch speed chahiye.

m g h

aur

- GM m / r

do rival, competing formulas hain.

False.

m g h

asal mein

- GM m / r

hi hai

h ≪ R

ke liye expand kiya gaya

g = GM / R^{2}

ke saath. Yeh dono zoom levels par same physics hain, rivals nahi.

Kyunki planet ke paas

U < 0

hai, paas wala object door wale se less energy mein hai.

True (PE ke liye). Paas matlab zyada negative matlab less PE. Lekin total energy mein kinetic energy bhi shamil hai, isliye "less PE" ≠ "less total energy" bina speed check kiye.

Force wahan sabse strong hoti hai jahan

U

sabse zyada negative hoti hai.

True.

U

chhote

r

par sabse zyada negative hoti hai, aur

F = GM m / r^{2}

bhi chhote

r

par sabse bada hota hai. Well mein gehraai mein,

U

ka slope aur force magnitude dono sabse bade hote hain.

Spot the error

Har line mein ek flawed statement ya step hai. Kya galat hai yeh batao aur usse correct karo.

"Upar jaane se

r

badhta hai, aur

- GM m / r

aur negative ho jaata hai, isliye upar jaane se energy decrease hoti hai."

Error: bada

r

- GM m / r

ko zero ki taraf le jaata hai, yani less negative. Upar jaane se energy badhti hai. Galti signed value ki jagah magnitude se confuse karna hai.

g = 6.67 \times 1 0^{- 11}

Error: yeh

G

hai, universal constant. Local field

g = GM / R^{2} \approx 9.8 m/s^{2}

hai.

G

universal hai;

g

derived aur planet-specific hai.

"Infinity par potential energy apne minimum par hai kyunki object sabse door hai."

Error: infinity par

U = 0

hai, jo

U

ka maximum value hai. Minimum sabse chhote

r

par hai, well mein sabse gehraai mein.

"Satellite ki PE find karne ke liye hum

m g h

use karte hain jahan

h

uski orbital altitude hai."

Error:

m g h

ko constant

g

chahiye, valid sirf

h ≪ R

ke liye. Orbital altitudes par

g

kaafi kam ho jaati hai; tumhe

- GM m / r

use karna hi hoga.

U = - GM m / r

, toh agar main

r

ko zero kar dun toh energy zero ho jaayegi."

Error: jaise

r \to 0

U \to - \infty

, zero nahi. Formula blow up ho jaata hai (idealised point mass);

U = 0

r \to \infty

par hota hai, opposite end par.

"Escape ka matlab infinity par kuch leftover speed ke saath pahunchna hai, toh

E > 0

set karo."

Error: minimum escape ka matlab infinity par zero speed ke saath pahunchna hai, isliye

E = 0

(

\frac{1}{2} m v^{2} + U = 0

E > 0

minimum se tez escape deta hai, threshold nahi.

"Jaise mass andar girta hai gravity ka work negative hota hai, kyunki gravity ek negative force hai."

Error: andar girte waqt, force aur displacement dono ek hi taraf point karte hain (dono andar), isliye gravity positive work karti hai.

F = - \frac{GM m}{r ^{2}} \overset{r}{^}

mein minus sirf

\overset{r}{^}

ke relative direction encode karta hai, work ke sign ko nahi. Neeche vector figure mein dono arrows ek saath lined up dikhte hain.

Why questions

Answer the reasoning, not just the result.

Gravitational PE ka reference infinity par kyun choose kiya jaata hai, surface par kyun nahi?

Infinity ek hi aisi distance hai jahan inverse-square force ek saath har planet ke liye vanish ho jaati hai, ek universal, natural zero deta hai. Surface reference arbitrary hogi aur har body ke liye alag hogi.

Potential-energy formula mein aage minus sign kyun hona zaroori hai?

Kyunki

Δ U = - \int F \cdot d r

: PE change gravity ke kaam ka minus hota hai. Gravity cheezein andar kheenchte waqt positive work karti hai, isliye PE andar jaate waqt drop hoti hai — minus isko build-in karta hai.

Total orbital energy negative hone se orbit bound kyun guarantee hoti hai?

Negative matlab

r = \infty

ke zero se neeche. Freedom reach karne ke liye (jahan

E \geq 0

) tumhe energy add karni padegi, isliye object khud escape nahi kar sakta — yeh trapped hai.

Gravity ke liye ek single-number

U (r)

define kyun kar sakte hain?

Kyunki gravity ek conservative force hai: kaam sirf start aur end points par depend karta hai, path par nahi. Yahi path-independence ek number ko position encode karne deta hai.

m g h

ek positive number kyun deta hai jabki

- GM m / r

hamesha negative hai?

m g h

ek change measure karta hai apne local zero se jo tum

h = 0

par rakhte ho, isliye yeh positive ho sakta hai.

- GM m / r

infinity zero ke against ek absolute value hai, isliye negative rehta hai. Alag references hain, alag physics nahi.

Escape speed launch angle par independent kyun hai (planet ki surface/air ignore karte hue)?

Energy conservation sirf speed (through

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

) aur

r

ki parwah karta hai, direction ki nahi. Koi bhi direction jo eventually infinity reach kare usse same

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

chahiye.

Edge cases

Woh scenarios jo naive intuition tod dete hain. Har limit cover karo.

Surface par

U

exactly kya hai,

r = R

par?

U (R) = - GM m / R

, ek finite negative number — planet ke bahar wali sabse gehri PE jo ek object ho sakti hai. Yeh surface launches ke liye "bottom of the well" hai, escape-speed calculations ka starting point.

r \to \infty

jaise

U

ka kya hota hai?

U \to 0

neeche se. Yeh chosen zero aur well ki ceiling hai; yahan ka object exactly free/bound boundary par hai.

Idealised point mass ke liye

r \to 0

jaise

U

ka kya hota hai?

U \to - \infty

. Formula diverge ho jaata hai kyunki ek true point mass infinite depth concentrate karta hai; real bodies ka finite radius hota hai, isliye

- GM m / r

branch sirf

r \geq R

ke liye apply hoti hai.

Radius

R

wale real planet ke liye, body ke andar (

r < R

)

U (r)

kya karta hai?

Yeh

- \infty

ki taraf nahi jaata. Ek uniform sphere ke andar enclosed mass

r^{3}

ki tarah shrink hoti hai,

U (r) = - \frac{GM m}{2 R ^{3}} (3 R^{2} - r^{2})

deta hai — ek smooth parabola jo centre par ek finite value

- \frac{3 GM m}{2 R}

par bottom out hoti hai. Point-mass formula sirf outside branch hai; finite-bottom well figure dekho.

Agar total energy

E

exactly

0

hai, toh trajectory kya hai?

Threshold case: object barely escape karta hai, infinity par zero speed ke saath pahunchta hai.

E < 0

bound hai (wapas girta hai),

E > 0

unbound hai leftover speed ke saath — Escape Velocity dekho.

Do identical objects: ek

r = R

par rest mein, ek

r = R

par orbiting. Kiska total energy zyada hai?

Orbiting wale ka — uska same

U

hai lekin extra kinetic energy

\frac{1}{2} m v^{2} > 0

hai. Same PE, zyada KE, isliye higher total energy.

Kya do point masses ke liye gravitational PE finite distance par exactly zero ho sakti hai?

Nahi.

- GM m / r = 0

ke liye

r \to \infty

chahiye. Kisi bhi finite separation par real positive masses ke saath,

U

strictly negative hai.

Kya

m g h

kabhi

Δ U

ke exactly equal hoti hai, ya sirf approximately?

Sirf approximately, aur sirf

h \to 0

ki limit mein. Exact

Δ U = GM m h / [R (R + h)]

m g h

ke barabar exactly tab hota hai jab

R + h = R

, jo

h > 0

ke liye kabhi truly nahi hota — yeh ek leading-order match hai.

Connections

Parent topic — woh derivations jo yeh traps test karte hain.
Conservative Forces and Potential Energy — kyun ek single-number $U$ exist karta hai.
Work-Energy Theorem — work ke sign ke traps.
Escape Velocity — $E = 0$ threshold edge cases.
Kepler's Laws & Orbital Energy — "negative = bound" reasoning.
Acceleration due to gravity g — $g$ vs $G$ error.
Newton's Law of Universal Gravitation — minus sign ke peeche ki force.