Variation of g — with altitude, latitude, depth: Worked examples

1.2.21 · D3 · Physics › Newton's Laws & Dynamics › Variation of g — with altitude, latitude, depth

Kuch bhi shuru karne se pehle, yeh woh tools hain jo hum baar baar use karte rahenge. Sab kuch parent note mein anchored hai; hum yahan sirf apply kar rahe hain.

Scenario matrix

Is topic ka har problem in cells mein se ek (ya inka blend) hota hai. Agar tum har row ka ek example kar sako, to tumne poora topic cover kar liya.

Cell	Kya vary karta hai	Sign / regime	Trap ya edge
A	Altitude, small $h ≪ R$	$g$ decrease hota hai, linear use karo	factor 2 bhool jaana
B	Altitude, large $h$	$g$ decrease hota hai, exact use karna zaroori	linear formula toot jaata hai
C	Depth $d$	$g$ decrease hota hai, linear exact	galti se " $2$ " use karna
D	Degenerate: $d = R$ (center) aur $h \to \infty$	limiting values	$g \to 0$ dono taraf, alag reasons se
E	"Equal drop" compare — same $g$ up vs down	$h$ aur $d$ ka ratio	$2 \times$ factor decide karta hai
F	Latitude, endpoints $λ = 0, 9 0^{\circ}$	equator pe sabse chhota, pole pe sabse bada	$cos^{2}$ na ki $cos$
G	Latitude, general $λ$ (real-world word problem)	intermediate value	radians vs degrees
H	Exam twist: spin-up limit / weight change	$g_{eff} = 0$ kab hoga?	tum $g < 0$ nahi bana sakte

Ab hum har cell ko hit karte hain.

Cell A — Altitude, small height

Worked example A · 3.2 km par ek drone

$h = 3.2 km$ par hover karne wale drone ke liye $g$ kitna drop karta hai? Percentage mein batao.

Forecast: Pehle guess karo — kya drop $0.1%$ se zyada hoga ya kam? Apna guess likh lo.

Step 1. Regime check karo: $h = 3.2 km$ vs $R = 6400 km$ , to $h / R = 0.0005 ≪ 1$ . Yeh step kyun? Linear formula sirf tab valid hai jab $h ≪ R$ ; use karne se pehle confirm karna zaroori hai, warna exact form chahiye hogi.

Step 2. Apply karo $g_{h} \approx g (1 - \frac{2 h}{R})$ : $\frac{2 h}{R} = \frac{2 ( 3.2 )}{6400} = 0.001$ Yeh step kyun? Fractional drop exactly $2 h / R$ hai — "percentage change" ka matlab yahi hai, isliye hum yeh ratio directly compute karte hain.

Step 3. Percent mein convert karo: $0.001 = 0.1%$ . To $g_{h} \approx 0.999 g$ . Yeh step kyun? Fraction ko $100$ se multiply karo taaki use percent mein likha ja sake (sirf bare symbol mat likho, $%$ likho).

Verify: Units — $2 h / R$ (km)/(km) = dimensionless ✅, ratio ke liye sahi. Sanity: parent ne $32 km$ par $1%$ nikala tha; hum ek-tenth height par hain, to ek-tenth drop → $0.1%$ ✅.

Cell B — Altitude, large height (linear formula FORBIDDEN)

Worked example B · Geostationary-ish:

$h = R$ Ek satellite $h = R$ (ek Earth-radius upar) par baith gayi hai. $g_{h}$ ko $g$ ke fraction ke roop mein nikalo. Phir dikhao ki linear formula yahan kyon bekar hai.

Forecast: Naive linear formula deta hai $g (1 - 2 h / R) = g (1 - 2) = - g$ . Negative gravity?! Sahi value guess karo.

Step 1. Yahan $h / R = 1$ hai, jo not $≪ 1$ hai. Isliye hume exact form ZAROOR use karni hai. Yeh step kyun? Binomial approximation $(1 + x)^{- 2} \approx 1 - 2 x$ sirf small $x$ ke liye hoti hai. $x = 1$ par yeh bilkul galat hai — yeh negative bhi ho jaati hai, jo gravity ke liye impossible hai.

Step 2. Exact form with $r = R + h = 2 R$ : $g_{h} = \frac{GM}{( R + h ) ^{2}} = \frac{GM}{( 2 R ) ^{2}} = \frac{GM}{4 R ^{2}} = \frac{1}{4} g$ Yeh step kyun? Center se distance double → inverse-square se ek-chauthai pull milti hai. Figure dekho: field lines $2 \times$ radius par spread ho jaati hain, to unki density (strength) $2^{2} = 4$ se gir jaati hai.

Step 3. Compare karo: exact deta hai $+ 0.25 g$ ; linear ne diya tha $- 1.0 g$ . Linear wala sirf galat nahi — uska sign hi wrong hai. Yeh step kyun? Memory mein burn karne ke liye ki "small- $h$ " formulas ek baar $h$ aur $R$ comparable ho jaayein to jhooth bolta hai.

Verify: $g$ kabhi negative nahi ho sakta (gravity hamesha pull karta hai), isliye $- g$ turant galat hai; $\frac{1}{4} g$ positive hai aur $g$ se kam hai (hum upar gaye) ✅. $h = R$ par, $(1 + h / R)^{- 2} = (2)^{- 2} = 1/4$ ✅.

Cell C — Depth (exact linear, no factor 2)

Worked example C · Duniya ki sabse gehri mine

Ek mine $d = 4 km$ tak pahunchi. Neeche $g$ kitne fraction se change hoga (uniform-Earth model)?

Forecast: Kya change utni hi doori upar jaane se zyada hoga ya kam? Guess karo.

Step 1. Depth ke liye exact linear formula $g_{d} = g (1 - \frac{d}{R})$ use hoti hai. Yeh step kyun? Uniform sphere ke andar $g \propto r$ (parent note), isliye koi factor 2 nahi aur koi approximation bhi nahi — yeh exact hai.

Step 2. Plug in karo: $\frac{d}{R} = \frac{4}{6400} = 6.25 \times 1 0^{- 4} \approx 0.0625%$ To $g_{d} \approx 0.999375 g$ . Yeh step kyun? Fractional change sirf $d / R$ hai.

Step 3. Upar $4 km$ jaane se compare karo: woh hoga $2 d / R = 0.125%$ . Yeh step kyun? Same distance, up vs down — upar twice as fast drop hota hai. Cell C ka yahi to poora point hai.

Verify: Depth change $0.0625%$ exactly half altitude change $0.125%$ hai same $4 km$ ke liye ✅ (factor-2 rule confirm). $g_{d} < g$ (bahar wali shell ka would-be pull remove ho gaya) ✅.

Cell D — Degenerate limits (Earth ka center, infinite height)

Worked example D · Dono raaste zero par jaate hain — opposite reasons se

$g$ nikalo (a) Earth ke center $d = R$ par, aur (b) infinitely dur upar $h \to \infty$ par. Dono $0$ hain — explain karo ki reasons alag kyun hain.

Forecast: Kaun sa zero "smoothly aur linearly" pahunchta hai, kaun sa "asymptotically"?

Step 1. Center: $g_{d} = g (1 - \frac{d}{R})$ mein $d = R$ daalo: $g_{R} = g (1 - 1) = 0$ Yeh step kyun? Exact center par koi inner sphere nahi bachi — Earth ki SAARI mass tumhare chaaron taraf shells mein hai, aur shell theorem ke anusaar har shell zero pull karti hai. Figure dekho: symmetric pulls cancel ho jaate hain.

Step 2. Infinity: $g_{h} = \frac{GM}{( R + h ) ^{2}}$ ; jaise jaise $h \to \infty$ denominator $\to \infty$ , to $lim_{h \to \infty} g_{h} = 0$ Yeh step kyun? Distance bina bound ke badhti hai, aur $1/ r^{2} \to 0$ . Lekin finite height par yeh kabhi equals zero nahi hota — sirf approach karta hai.

Step 3. Dono curves ka contrast:

Depth: surface par $g$ se center par $0$ tak straight line.
Altitude: ek curve jo fast dive karta hai phir flatten ho jaata hai, $0$ sirf infinity par touch karta hai.

Verify: Depth zero finite $d = R$ (center) par reach hota hai ✅. Altitude zero ek limit hai, kabhi finite nahi ✅. Dono har jagah $\geq 0$ — gravity kabhi reverse nahi hoti ✅.

Cell E — Equal-drop comparison

Cell F — Latitude endpoints (equator vs pole)

Worked example F · Sabse bhaari aur sabse halke spots

Sirf rotation effect use karke, $g_{pole} - g_{eq}$ ko $m/s^{2}$ mein nikalo, aur ek $70 kg$ insaan ke liye fractional weight change batao.

Forecast: Apparent weight difference kuch grams hogi ya kuch kilograms? Guess karo.

Step 1. $g_{λ} = g - ω^{2} R cos^{2} λ$ ke endpoints:

Pole $λ = 9 0^{\circ}$ : $cos^{2} 9 0^{\circ} = 0 \Rightarrow g_{pole} = g$ .
Equator $λ = 0^{\circ}$ : $cos^{2} 0^{\circ} = 1 \Rightarrow g_{eq} = g - ω^{2} R$ .

Yeh step kyun? $cos^{2}$ (na ki $cos$ ) kyunki ek factor spin-circle radius $r = R cos λ$ se aata hai aur ek outward push ko local vertical par project karne se (parent note).

Step 2. Difference poora rotation term hai: $g_{pole} - g_{eq} = ω^{2} R = (7.27 \times 1 0^{- 5})^{2} (6.4 \times 1 0^{6}) \approx 0.0338 m/s^{2}$ Yeh step kyun? Subtract karne se $g$ cancel ho jaata hai, pure spin contribution bachti hai.

Step 3. $m = 70 kg$ ke liye weight change: $Δ W = m Δ g = 70 \times 0.0338 \approx 2.37 N$ , yaani lagbhag $0.24 kg$ apparent weight (kyunki $2.37/9.8 \approx 0.24$ ). Yeh step kyun? Weight $= m g_{eff}$ ; $g$ mein jo drop hai woh scale pe lost weight ke roop mein dikhta hai.

Verify: $0.0338 m/s^{2}$ parent ke $\approx 0.034$ se match karta hai ✅. $70 \times 0.0338/9.8 \approx 0.24 kg$ — couple hundred grams, bilkul waise jaise parent ne big-picture mein claim kiya tha ✅.

Cell G — Latitude, general angle (word problem)

Worked example G ·

$λ = 4 0^{\circ}$ N par ek city Ek physics lab latitude $4 0^{\circ}$ par hai. Effective $g$ nikalo (sirf rotation), $g = 9.8 m/s^{2}$ lo.

Forecast: Equator value ke zyada close hoga ya pole value ke? Guess karo.

Step 1. Humein $cos^{2} 4 0^{\circ}$ chahiye. Radians mein, $4 0^{\circ} = 40 \times π /180 = 0.698 rad$ , aur $cos 4 0^{\circ} \approx 0.766$ . Yeh step kyun? Calculator/sympy par trig functions radians use karte hain; conversion bhool jaana cell G mein #1 error hai. To $cos^{2} 4 0^{\circ} \approx 0.587$ .

Step 2. Formula apply karo: $g_{λ} = g - ω^{2} R cos^{2} λ = 9.8 - (0.0338) (0.587) \approx 9.8 - 0.0198$ $g_{λ} \approx 9.780 m/s^{2}$ Yeh step kyun? Cell F mein humne already $ω^{2} R = 0.0338$ compute kiya tha; ise reuse karo, $cos^{2} λ$ se scale karo.

Verify: $9.780$ , $g_{eq} = 9.766$ aur $g_{pole} = 9.8$ ke beech mein hai ✅. Pole ke zyada close hai ( $cos^{2}$ ke terms mein $4 0^{\circ}$ pole ki taraf halfway se aage hai) ✅. Units: poore mein $m/s^{2}$ ✅.

Cell H — Exam twist: the spin-up limit

Worked example H · Kitna fast spin tumhe weightless banata hai

Agar Earth itni fast ghoomti ki equatorial $g_{eff} = 0$ ho jaaye, to ek din kitne lamba hota? Dikhao ki $g$ negative nahi ho sakta.

Forecast: Din ki length guess karo — minutes? hours? days?

Step 1. Equatorial effective gravity zero set karo: $g - ω^{2} R = 0 ⟹ ω^{2} R = g ⟹ ω = \frac{g}{R}$ Yeh step kyun? Equator par $cos^{2} λ = 1$ , isliye poora spin term $ω^{2} R$ subtract ho jaata hai. Weightlessness ka matlab hai outward requirement poori pull ke equal hai.

Step 2. $ω$ compute karo: $ω = \frac{9.8}{6.4 \times 1 0 ^{6}} \approx 1.24 \times 1 0^{- 3} rad/s$ Yeh step kyun? Yeh critical spin rate hai jahan equator par objects ground level par orbit mein chale jaate hain.

Step 3. Ek din $T = \frac{2 π}{ω}$ mein convert karo: $T = \frac{2 π}{1.24 \times 1 0 ^{- 3}} \approx 5063 s \approx 1.41 hours$ Yeh step kyun? Period $2 π / ω$ hai; seconds ko hours mein convert karo $/3600$ se.

Step 4. Iss speed se aage, kya $g$ negative ho jaayega? Nahi — jab $ω^{2} R > g$ hoga, to equator par objects simply ground chhod dete hain (unhe gravity se zyada inward force chahiye hogi, to woh fly off ho jaate hain). Gravity khud, $GM / R^{2}$ , unchanged rehti hai; sirf apparent $g$ zero tak pahunchti hai aur rukk jaati hai. Yeh step kyun? Yeh misconception khatam karne ke liye ki spinning gravity ko invert kar sakta hai.

Verify: $T \approx 1.41 h$ parent ke "about 1.4 hours" se match karta hai ✅. $ω^{2} R = (1.24 \times 1 0^{- 3})^{2} (6.4 \times 1 0^{6}) \approx 9.8 = g$ ✅ (weightless condition satisfied).

Recall Har formula kis cell ke liye hai? (self-test)

Small altitude, percent drop chahiye — kaun sa formula? ::: linear $g (1 - 2 h / R)$ (Cell A) Height $R$ ke comparable — kaun sa formula? ::: exact $GM / (R + h)^{2}$ , linear kabhi nahi (Cell B) Depth $d$ — koi factor 2 hai? ::: Nahi — depth ke liye $g (1 - d / R)$ hai, exact aur linear (Cell C) Earth ke center par, $g = ?$ aur kyun ::: $0$ , kyunki saari mass tumhare upar shells mein hai (Cell D) Same $g$ reduction: depth vs height ::: depth height se double honi chahiye, $d = 2 h$ (Cell E) Pole minus equator $g$ (sirf rotation) ::: $ω^{2} R \approx 0.034 m/s^{2}$ (Cell F) Latitude $4 0^{\circ}$ : $cos$ mein radians ya degrees? ::: pehle radians mein convert karo (Cell G) Kya spinning $g$ ko negative bana sakta hai? ::: Nahi — apparent $g$ $0$ par bottom out karta hai; objects phir fly off ho jaate hain (Cell H)

Connections

Parent topic — woh char formulas jo yeh examples apply karte hain.
Newton's Law of Universal Gravitation — cells A, B, D ke peeche $1/ r^{2}$ .
Shell Theorem — center par $g = 0$ kyun (cell D) aur depth formula (cell C).
Centripetal Force & Circular Motion — cells F, G, H mein spin term.
Weight vs Mass — cell F ka apparent-weight change.
Oblate Spheroid Earth — shape effect jo equator par rotation mein add hota hai.
Escape Velocity & Orbital Mechanics — cell H ground-level orbit hai.