Variation of g — with altitude, latitude, depth: Question bank

1.2.21 · D5 · Physics › Newton's Laws & Dynamics › Variation of g — with altitude, latitude, depth

Ek single expression jo har effect ko fold karta hai (altitude/depth se pure gravity, phir effective $g$ pane ke liye spin correction) neeche dikhaya gaya hai aur aage ke diagrams mein case-by-case unpack kiya gaya hai.

Figure — Variation of g — with altitude, latitude, depth

True or false — justify

Both altitude aur depth formulas surface ke paas

g

ko reduce karte hain, toh 1 km climb aur 1 km dig se

g

utni hi amount se change hoti hai

False. Altitude inverse-square law use karta hai jisse

- 2 h / R

milta hai, lekin depth linear hai jisse

- d / R

milta hai; equal distance ke liye upar jaana neeche jaane se do guna tezi se

g

change karta hai (s01 figure dekho).

Formula

g_{h} = g (1 - 2 h / R)

kisi bhi height ke liye kaam karta hai

False. Yeh ek binomial approximation hai jo sirf $h ≪ R$ ke liye valid hai; badi

h

(satellites) ke liye exact

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

use karna zaroori hai.

Mine mein neeche jaane se hamesha

g

weak hota hai

Sirf uniform-density Earth ke liye. Clean formula

g_{d} = g (1 - d / R)

constant

ρ

assume karta hai; real Earth mein dense iron core hai, isliye depth ke pehle ~2900 km mein

g

thoda badhta bhi hai, phir center par zero ho jaata hai.

Earth ke exact center par

g = 0

hota hai

True. Har jagah mass ab tumhare around shells mein hai, aur Shell Theorem ke mutabik ek shell apne andar kisi cheez par zero net force lagaata hai, isliye sab directions se pulls cancel ho jaate hain (s02 figure dekho).

Equator par

g

sabse zyada hota hai kyunki equator Sun ke pull ke sabse kareeb hota hai

False. Sun local

g

se irrelevant hai; effective

g

actually equator par sabse kam hota hai kyunki wahan rotation tumhe baahir fenk ta hai aur equatorial bulge tumhe Earth ke center se aur door rakhta hai.

Rotation term

ω^{2} R cos^{2} λ

poles par zero ho jaata hai

True. Poles par

λ = 9 0^{\circ}

hota hai toh

cos^{2} 9 0^{\circ} = 0

; axis par ek point zero radius ka circle trace karta hai aur koi centripetal force nahi chahiye, isliye wahan koi gravity "kharach" nahi hoti.

Agar Earth ghoomna band kar de, toh effective

g

instantly har jagah same ho jaayega

False. Rotation correction ek dum gayab ho jaayegi, lekin oblate shape (Oblate Spheroid Earth) bani rehti hai — aur woh sirf geological timescales (laakhon saal, agar ho bhi toh) par relax hoti hai, isliye spin band hone ke kaafi baad bhi shape-driven

g

-variation bani rahegi.

Orbit mein satellite zero gravity experience karta hai

False. Orbital height par gravity abhi bhi badi hoti hai (

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

); astronauts weightless isliye feel karte hain kyunki woh free fall mein hain,

g

zero hai isliye nahi — Weight vs Mass dekho.

Equator se pole jaane par mass change hota hai

False. Mass matter ki quantity hai aur location ke saath kabhi change nahi hoti; sirf weight (

m g_{λ}

) change hota hai kyunki effective

g

change hoti hai. Yahi core distinction hai Weight vs Mass mein.

Spot the error

g_{h} = g (1 - 2 h / R)

exact hai, isliye main isse

h = R

par satellite ke liye use kar sakta hoon."

Yeh formula ek binomial approximation hai jo sirf $h ≪ R$ ke liye valid hai.

h = R

par exact

g_{h} = GM / (R + h)^{2} = g /4

use karna hoga, linear wala nahi jo absurdly

g (1 - 2) = - g

dega.

"Aur neeche jaane par,

g_{d} = g (1 - d / R)

, toh

d = 2 R

par gravity

- g

ho jaayegi (upar ki taraf point karti hai)."

Yeh formula sirf

0 \leq d \leq R

(Earth ke andar) ke liye apply hota hai;

d

positive downward measure hota hai aur kabhi

R

se zyada nahi ho sakta. Center ke baad koi Earth hi nahi "below" mein, isliye derivation collapse ho jaati hai.

"Latitude formula mein

cos^{2} λ

isliye hai kyunki humne radius ko square kiya."

Galat reason hai. Ek

cos λ

circle radius

r = R cos λ

se aata hai, aur doosra horizontal centripetal vector ko local vertical par project karne se — do alag geometric steps hain, squared radius nahi (s03 figure dekho).

"Equator par poora centripetal term

ω^{2} R

reduce karta hai

g

ko; latitude

4 5^{\circ}

par reduction exactly half hai."

Reduction

ω^{2} R cos^{2} λ

hai, aur

cos^{2} 4 5^{\circ} = \frac{1}{2}

hai, toh wahan half hoti hai — lekin dependence

cos^{2} λ

hai,

λ

mein linear nahi, isliye "half latitude" ka matlab "half effect" nahi hota generally.

"Effective gravity hamesha seedha Earth ke center ki taraf point karti hai."

Sirf equator aur poles par. Baki jagah centripetal correction ka ek sideways component hota hai, isliye plumb line thodi si true center-direction se off hang karti hai — jo "down" tum feel karte ho woh effective gravity

g_{λ}

hai, equator ki taraf thoda tilt hua.

"Kyunki tumhare upar wale shells zero contribute karte hain, isliye jitna aur neeche jaao

g

utni hi tezi se drop honi chahiye."

Uniform density ke liye rate constant hoti hai:

g_{d} = g (1 - d / R)

linear hai, isliye

g

steady rate

- g / R

per metre se girta hai, tezi se nahi.

Why questions

g

altitude ke saath ghatta kyun hai?

Newton's law center se distance

r

mein inverse-square hai; height

h

par tum

r = R + h > R

par ho, isliye

1/ r^{2}

pull weaker hoti hai — yeh seedha Newton's Law of Universal Gravitation hai.

Depth ka formula exact kyun hai jabki altitude ka sirf approximate hai?

Uniform sphere ke andar

g \propto r

perfectly linear hai, isliye

g_{d} = g (1 - d / R)

ko koi approximation nahi chahiye; bahar,

g \propto 1/ r^{2}

nonlinear hai, isliye tidy

1 - 2 h / R

paane ke liye binomial-expand karna padta hai (assuming

h ≪ R

Spinning Earth tumhe halka kyun karta hai, instead of ground ka zyada push karne ke?

Pure gravity ka ek hissa centripetal force (Centripetal Force & Circular Motion) supply karne mein divert ho jaata hai jo tumhe circle mein chalata hai; sirf bacha hua — effective

g_{λ}

— tumhe neeche rokta hai, isliye scale kam read karta hai.

Rotation effect equator par sabse zyada kyun hoti hai?

Equator spin axis se sabse door hota hai, isliye woh sabse bada, sabse tez circle mein move karta hai aur sabse zyada centripetal force chahiye (

ω^{2} R cos^{2} λ

λ = 0

par peak karta hai), sabse zyada gravity divert karta hai.

Oblate shape aur rotation dono polar

g

ko kyun largest banate hain?

Poles par tum Earth ke center ke sabse kareeb hote ho (polar radius choti hoti hai) aur rotation effect zero hoti hai — dono factors ek hi direction mein push karte hain, isliye effective

g

wahan maximum hoti hai.

Earth ko tezi se ghuma kar effective

g

negative kyun nahi ho sakta?

Equatorial reduction

ω^{2} R

par capped hai; jab yeh

g

ke barabar ho jaata hai, effective gravity exactly zero ho jaati hai aur objects orbit/float off karne lagte hain press down hone ki jagah — tum

g_{λ} = 0

tak pahunchte ho, usse neeche kabhi nahi. Yeh Escape Velocity & Orbital Mechanics ki boundary hai.

Edge cases

Atmosphere ke top par aur mantle ke andar deep mein

g

roughly kitna hota hai?

Zyada altitude par

g

smoothly

1/ (R + h)^{2}

ki tarah decrease karta hai; real (non-uniform) Earth ke andar

g

dense core ki wajah se briefly badh sakta hai phir drop hota hai — uniform model sirf first approximation hai.

Latitude formula exactly

λ = 0

aur

λ = 9 0^{\circ}

par kya hota hai?

λ = 0

(equator) par

cos^{2} 0 = 1

maximum reduction

g - ω^{2} R

deta hai;

λ = 9 0^{\circ}

(pole) par

cos^{2} 9 0^{\circ} = 0

koi reduction nahi deta,

g_{λ} = g

— poore formula ke do extremes.

d = R

(center) par depth formula aur shell argument dono agree karne chahiye — kya karte hain?

Haan.

g_{d} = g (1 - R / R) = 0

, aur shell theorem independently kehta hai ab saara mass tumhare baahir hai, zero net force deta hai — do routes, same answer.

Agar

h = d

ho (upar aur neeche same distance), kaunse point par zyada

g

hoga?

Neeche wala point:

g (1 - d / R) > g (1 - 2 h / R)

jab

h = d

ho, kyunki depth surface ke paas altitude ki half rate se gravity loose karta hai.

Equatorial gravity vanish hone se pehle sabse chota possible "day length" kya hai?

ω^{2} R = g

set karne par

ω = g / R

milta hai, roughly 1.4 hours ka period; isse tez spin karo aur equator objects ko neeche nahi rok sakta.

Kya depth formula Moon ya gas planet par kaam karta hai?

Linear result

g_{d} = g (1 - d / R)

ko uniform density chahiye; differentiated Moon ya huge density gradient wale gas giant mein bahut deviation hoga, isliye yeh formula ek model hai, law nahi.

Recall Har trap ki ek-line summary

UP, DOWN se do guna tezi se $g$ lose karta hai (aur sirf $h ≪ R$ ke liye); DOWN sirf uniform density ke liye aur sirf $0 \leq d \leq R$ ke liye kaam karta hai; SPIN pure $g$ ko effective $g_{λ}$ mein convert karta hai, equator par sabse zyada churaata hai, aur zyada se zyada $g_{λ}$ ko zero tak drive kar sakta hai, kabhi negative nahi; mass kabhi nahi badlta, sirf weight badlta hai; aur oblate shape spin se zyada time tak rehti hai kyunki woh sirf geological time par relax hoti hai.

Connections

Parent topic — woh derivations jinhe ye traps test karte hain.
Shell Theorem — har depth aur center-of-Earth question ke peeche.
Weight vs Mass — mass-doesn't-change traps yahan rehte hain.
Oblate Spheroid Earth — "shape vs spin" separation.
Escape Velocity & Orbital Mechanics — jahan $g \to 0$ spin limit le jaati hai.