Gravitational field intensity g = GM - r²: Question bank

1.2.20 · D5 · Physics › Newton's Laws & Dynamics › Gravitational field intensity g = GM - r²

True or false — justify karo

Ek heavier test mass ek stronger gravitational field mein baithti hai.

False. Field

g = GM / r^{2}

sirf source

M

aur distance

r

par depend karta hai; test mass

m

g = F / m

mein cancel ho jaata hai. Ek pankh aur ek pathar ek hi point par identical

g

feel karte hain.

g

ek scalar hai kyunki ye

9.8

jaisa ek single number nikalta hai.

False.

g

ek vector hai; "

9.8

" sirf iska magnitude hai. Direction matter karta hai — kisi planet ke paas ye inward point karta hai, aur fields ko combine karne ke liye vector addition chahiye, number addition nahi.

Ek uniform solid planet ke exact centre par,

g = 0

True. Symmetry ki wajah se har direction ka pull opposite side ke equal pull se balance ho jaata hai, isliye net field zero ho jaata hai — chahe tum mass ke andar kaafi gehre ho.

Field intensity aur free-fall acceleration do alag physical quantities hain jo ek hi number share karte hain.

False. Ye literally equal hain:

g = F / m

aur Newton's second law deta hai

a = F / m

, isliye

a = g

. Same quantity, do naam (field view vs. motion view).

Agar do planets ek point par opposite directions mein fields create karein, toh net field magnitude donon ka sum hai.

False. Opposite directions subtract karte hain:

g_{n e t} = ∣ g_{1} - g_{2} ∣

. Ye tabhi add hote hain jab same direction mein point karein; null point par ye zero tak cancel ho jaate hain.

Null point par

g

ka zero ho jaana matlab gravity wahan "switch off" ho gayi.

False. Har source abhi bhi apna poora pull exert karta hai; ye sirf vectors ke roop mein cancel hote hain. Wahan rakhi gayi ek mass koi net force feel nahi karti, lekin individual fields alive hain.

Units

N kg^{- 1}

aur

m s^{- 2}

ek hi cheez describe karte hain.

True.

N = kg \cdot m s^{- 2}

, isliye

N kg^{- 1} = m s^{- 2}

. Pehla form "force per mass" padhta hai (field view), doosra "acceleration" (motion view).

Source mass

M

ko double karne se field double hoti hai, lekin distance

r

ko double karne se ye aadhi ho jaati hai.

Doosre half par False. Field

M

mein linear hai (double

M

→ double

g

) lekin

r

mein inverse-square hai (double

r

→

g /4

g /2

nahi).

Error dhundho

"Altitude

h

par,

g = GM / h^{2}

use karo."

Error:

r

M

ke centre se measure kiya jaata hai, surface se nahi. Sahi hai

r = R + h

, isliye

g = GM / (R + h)^{2}

. Sirf

h

use karne se

h \to 0

par galat result aata hai.

"Moon ke liye field Earth se zyada strong hai kyunki Moon chota hai, isliye uski lines zyada crowded hain."

Error: Field strength

M / r^{2}

se set hoti hai, sirf size se nahi. Earth ka kaafi bada

M

uski surface field ko Moon ki surface field se lagbhag chhe guna bana deta hai, bade radius ke bawajood.

"Kyunki

g = - \frac{GM}{r ^{2}} \overset{r}{^}

mein minus sign hai, gravity kabhi kabhi outward push karti hai."

Error: Minus fixed hai, "kabhi kabhi" nahi.

\overset{r}{^}

away from

M

point karta hai; minus use flip karke toward

M

point karta hai — hamesha inward. Gravity yahan purely attractive hai.

"Kyunki

g = F / m

, ek rocket jisme koi test mass nahi (

m = 0

) uske liye field undefined hai."

Error:

g

ek location ki property hai, jo ek chote test mass ke liye

F / m

ke limit ke roop mein define hoti hai;

F \propto m

isliye ratio finite rehta hai. Field ke exist karne ke liye koi test mass actually present hona zaroori nahi.

"Moon ki surface par objects slow girte hain, isliye wahan

G

chhota hai."

Error:

G

ek universal constant hai, har jagah. Chhoti surface

g

Moon ke chhote

M

aur

R

se aati hai

GM / R^{2}

mein,

G

mein kisi bhi change se nahi.

"Field lines ka

4 π r^{2}

par spread hona explain karta hai kyun

g \propto r^{2}

Error: Area

r^{2}

ki tarah badhta hai, isliye line density (field) dilute hoti hai — ye

1/ r^{2}

ki tarah jaati hai,

r^{2}

nahi. Spreading distance ke saath field ko kamzor karti hai.

Why questions

Test mass

m

final formula se kyun gayab ho jaata hai?

Kyunki gravitational force

m

ke proportional hoti hai (

F = GM m / r^{2}

g = F / m

mein

m

se divide karne par ye cancel ho jaata hai, ek aisi quantity bach jaati hai jo sirf source aur location describe karti hai.

Hum ek field define karne ki kyon zahmatt lete hain instead of har baar force compute karne ke?

Field space ko hi ek baar aur sab ke liye describe karta hai; phir kisi bhi visiting mass ki force sirf

F = m g

hai. Ye "source space ko kya karta hai" ko "visitor kaise respond karta hai" se alag karta hai.

Gravity inverse-square law kyun hai aur, maan lo, inverse-cube kyun nahi?

Geometrically, ek fixed number of field lines har surrounding sphere ko pierce karti hain, aur ek sphere ka area

4 π r^{2}

ki tarah badhta hai; line density (field) isliye

1/ r^{2}

ki tarah girta hai. Yahi Gauss's Law for Gravity ka content hai.

g

source ki taraf kyun point karta hai jabki unit vector

\overset{r}{^}

away point karta hai?

Hum

\overset{r}{^}

ko convention se outward point karne ke liye choose karte hain, isliye ek attractive pull ko opposite sign chahiye — isliye

g = - \frac{GM}{r ^{2}} \overset{r}{^}

, inward.

Surface value itni special kyun hai ki use apna symbol

g \approx 9.8

mil jaata hai?

Kyunki surface par centre se teri distance exactly planet ka radius

R

hoti hai, jo ek fixed number deta hai jo hum rozana experience karte hain.

r \neq = R

ke liye Variation of g with Altitude and Depth dekho.

Electric field

E = k Q / r^{2}

g = GM / r^{2}

se identical kyun lagta hai?

Dono "source strength over distance squared" fields hain jisme same inverse-square geometry hai; mass charge ki role play karta hai. Compare karo Electric Field Intensity E=kQ/r² mein.

Edge cases

r \to \infty

par

g

kya hai?

Ye zero ki taraf jaata hai:

GM / r^{2} \to 0

. Gravity kabhi truly vanish nahi hoti lekin bina kisi limit ke fade hoti jaati hai — iska infinite range hai lekin ever-weaker strength.

Ek point mass ke liye

r \to 0

par

g = GM / r^{2}

ka kya hota hai?

Ye infinity tak diverge ho jaata hai — ek mathematical singularity. Real bodies points nahi hain, isliye jab tum inside jaate ho, sirf chhote radius par mass count hota hai aur

g

uski jagah drop ho jaata hai (dekho Variation of g with Altitude and Depth).

Ek uniform planet ke andar gehre, kya

g

centre ki taraf badhta rehta hai?

Nahi. Sirf tumhare neeche wale sphere ki mass pull karti hai (outer shell zero net field contribute karta hai), isliye

g

linearly decrease hota hai centre par zero tak.

Do equal masses ke beech, null point kahan hai, aur kya ye stable hai?

Exactly beech mein, jahan do fields equal aur opposite hain isliye

g_{n e t} = 0

. Line ke saath nudge ki gayi ek mass ek restoring pull feel karti hai sirf tabhi — do equal point masses ke liye midpoint line ke saath unstable hai.

Agar ek test mass ki mass zero hai, toh kya

g

ka concept meaningless hai?

Nahi.

g

ek vanishingly small test mass ke liye

F / m

ke limit ke roop mein define hoti hai; kyunki

F \propto m

, ratio well-defined rehta hai aur field describe karta hai chahe koi visitor present na ho.

Kya

g

kisi point par non-zero ho sakta hai jahan gravitational potential zero ho?

Haan. Potential aur field is tarah se independent hain —

g = - d V / d r

, potential ki slope hai, uski value nahi. Flat-but-nonzero potential zero field deta hai; slope par zero potential nonzero field deta hai. Dekho Gravitational Potential and Potential Energy.

Connections

Parent topic — full derivation
Newton's Law of Universal Gravitation — yahan har trap ke peeche force law
Newton's Second Law (F=ma) — kyun field = free-fall acceleration
Gravitational Potential and Potential Energy — "zero potential ≠ zero field" edge case
Gauss's Law for Gravity — inverse-square aur shell result ka formal reason
Variation of g with Altitude and Depth — $r \to 0$ aur inside-planet edge cases
Electric Field Intensity E=kQ/r² — identical-structure comparison