Gravitational field intensity g = GM - r²: Worked examples

1.2.20 · D3 · Physics › Newton's Laws & Dynamics › Gravitational field intensity g = GM - r²

Kuch bhi shuru karne se pehle: yaad karo woh do tools jo hum baar baar use karte hain.

Scenario matrix

$g$ ke baare mein har question inhi boxes mein se ek mein aata hai. Neeche ke examples tagged hain us box ke saath jise woh fill karte hain.

Cell	Case class	Tricky kyon hai	Example
C1	Single mass, $r = R$ (surface par)	sahi distance choose karo	Ex 1
C2	Single mass, $r > R$ (altitude) ratio ke zariye	recompute mat karo — $1/ r^{2}$ se scale karo	Ex 2
C3	Do fields, same line, opposite	vectors subtract hote hain; sign matter karta hai	Ex 3
C4	Do fields, null point ( $g = 0$ )	degenerate: net field vanish ho jaata hai	Ex 4
C5	Do fields right angles par	Pythagoras, addition nahi	Ex 5
C6	Limiting behaviour: $r \to \infty$ aur $r \to 0$	field → 0 vs. field "blow up" ho jaati hai	Ex 6
C7	Inverse problem: $g$ diya hai, $M$ ya $r$ dhundho	formula rearrange karo	Ex 7
C8	Word problem (real world, altitude)	words translate karo → $r = R + h$	Ex 8
C9	Exam twist: naya planet, $M$ aur $R$ ke ratios	do scalings ek saath combine karo	Ex 9

Ab hum har cell fill karte hain.

Ex 1 — Surface field (Cell C1)

Worked example Mars ki surface par field

Mars: $M = 6.42 \times 1 0^{23} kg$ , radius $R = 3.39 \times 1 0^{6} m$ . Iski surface par $g$ find karo.

Forecast: Mars Earth se kaafi halka hai lekin chota bhi hai. Guess karo: Earth ke $9.8$ se bada hoga ya chota? (Apna guess likh lo.)

Step 1 — Distance choose karo. Surface par, Mars ke centre se tumhari distance uska radius hai, isliye $r = R = 3.39 \times 1 0^{6} m$ . Yeh step kyon? $GM / r^{2}$ mein $r$ hamesha mass ke centre se measure hota hai, aur "surface" ka matlab exactly ek radius bahar hai.

Step 2 — Tool A mein plug in karo. $g = \frac{( 6.67 \times 1 0 ^{- 11} ) ( 6.42 \times 1 0 ^{23} )}{( 3.39 \times 1 0 ^{6} ) ^{2}}$ Yeh step kyon? Ek mass, ek distance — yeh direct-substitution case hai, koi vectors ki zaroorat nahi.

Step 3 — Arithmetic. Numerator $= 4.28 \times 1 0^{13}$ ; denominator $= (3.39 \times 1 0^{6})^{2} = 1.149 \times 1 0^{13}$ . $g = \frac{4.28 \times 1 0 ^{13}}{1.149 \times 1 0 ^{13}} \approx 3.7 N kg^{- 1}$

Verify: Units: $\frac{N m ^{2} kg ^{- 2} \cdot kg}{m ^{2}} = N kg^{- 1}$ ✔. Magnitude ≈ $3.7$ , Earth ka lagbhag $0.38 \times$ — yeh jaani-maani baat se match karta hai ki "Martian gravity Earth ki roughly ek-tehayi hai." Tumhara forecast: chota tha, jo sahi hai.

Ex 2 — Altitude by ratio (Cell C2)

Worked example Teen Earth-radii door orbit karta ek satellite

Earth ka surface field $g_{0} = 9.8 N kg^{- 1}$ hai. Ek satellite Earth ke centre se $r = 3 R$ par orbit karta hai. Wahan $g$ find karo bina $GM$ dobara plug kiye.

Forecast: Distance teen guna ho gayi. Kya field $3$ se drop hogi, $6$ se, ya $9$ se?

Step 1 — Dono fields ko ratios mein likho. $g_{0} = \frac{GM}{R ^{2}}, g = \frac{GM}{( 3 R ) ^{2}}$ Yeh step kyon? Dono mein same $GM$ aata hai. Agar hum dono equations ko divide karein, $GM$ cancel ho jaata hai aur hume kabhi uski value ki zaroorat nahi padti.

Step 2 — Divide karo. $\frac{g}{g _{0}} = \frac{R ^{2}}{( 3 R ) ^{2}} = \frac{1}{9}$ Yeh step kyon? Yeh inverse-square law action mein hai: distance $\times 3 \Rightarrow$ field $\div 3^{2}$ .

Step 3 — Solve karo. $g = \frac{9.8}{9} \approx 1.09 N kg^{- 1}$

Verify: $1.09 \times 9 = 9.8$ ✔. Forecast check: yeh $9$ se drop hui, $3$ se nahi — "square" hi poora point hai. Yeh 80/20 shortcut hai: ratios arithmetic khatam kar deti hain.

Ex 3 — Do fields, opposite directions (Cell C3)

Ab humhe Tool B chahiye — vectors. Picture dekho: do masses point $P$ ko ek line par opposite directions mein kheeenchte hain.

Figure — Gravitational field intensity g = GM - r²

Worked example Ek line par do masses ke beech net field

Mass $A$ point $P$ ke baayein baitha hai aur wahan uska field left (toward $A$ ) point karta hai, magnitude $g_{A} = 6.0 \times 1 0^{- 6} N kg^{- 1}$ . Mass $B$ daayein baitha hai; $P$ par uska field right point karta hai $g_{B} = 2.0 \times 1 0^{- 6} N kg^{- 1}$ ke saath. Net field find karo.

Forecast: Net arrow kis taraf point karega, aur kya yeh $6.0 \times 1 0^{- 6}$ se bada hoga ya chota?

Step 1 — Ek positive direction choose karo. Maano "right" $= +$ , toh "left" $= -$ . Yeh step kyon? Ek line par vector sirf ek signed number hai jab hum choose kar lein kaun sa direction positive hai. Isse arrow-addition ordinary $\pm$ arithmetic mein badal jaati hai.

Step 2 — Signs assign karo. $g_{A}$ left point karta hai $\Rightarrow - 6.0 \times 1 0^{- 6}$ . $g_{B}$ right point karta hai $\Rightarrow + 2.0 \times 1 0^{- 6}$ . Yeh step kyon? Figure dekho: do red arrows ek doosre se ladte hain, isliye unke opposite signs hone chahiye.

Step 3 — Add karo. $g_{net} = (- 6.0 + 2.0) \times 1 0^{- 6} = - 4.0 \times 1 0^{- 6} N kg^{- 1}$ Yeh step kyon? Signed numbers add karna hi arrows add karna hai. Negative result humein direction batata hai.

Step 4 — Sign interpret karo. Negative = left point karta hai, $A$ ki taraf.

Verify: Zyada strong field ( $A$ ) jeetti hai, isliye net $A$ ki taraf point karna chahiye — karta hai ✔. Magnitude $4.0 < 6.0$ kyunki $B$ ne use partially cancel kiya ✔.

Ex 4 — Null point (Cell C4, degenerate zero)

Worked example Net field kahan vanish hoti hai?

Do masses: Earth $M_{E} = 5.97 \times 1 0^{24} kg$ aur Moon $M_{m} = 7.35 \times 1 0^{22} kg$ , jinke beech $d = 3.84 \times 1 0^{8} m$ ka separation hai. Unhe join karne wali line par, Earth ke centre se distance $x$ find karo jahan net field zero hai.

Forecast: Null point Earth ke paas hoga ya Moon ke paas?

Step 1 — Dono magnitudes equal set karo. Null point par dono fields opposite ways point karti hain aur unki size equal hoti hai, isliye cancel ho jaati hain: $\frac{G M _{E}}{x ^{2}} = \frac{G M _{m}}{( d - x ) ^{2}}$ Yeh step kyon? "Net field zero" do opposite vectors ke liye matlab hai ki unki magnitudes exactly match karti hain — yahi degenerate case ki definition hai.

Step 2 — $G$ cancel karo aur square root lo. $\frac{M _{E}}{x ^{2}} = \frac{M _{m}}{( d - x ) ^{2}} \Rightarrow \frac{d - x}{x} = \frac{M _{m}}{M _{E}}$ Yeh step kyon? $G$ dono mein common hai aur cancel ho jaata hai; square root awkward squares ko ek clean linear ratio mein badal deta hai.

Step 3 — Mass ratio compute karo. $\frac{7.35 \times 1 0 ^{22}}{5.97 \times 1 0 ^{24}} = 0.01231 = 0.1109$

Step 4 — $x$ ke liye solve karo. $d - x = 0.1109 x$ se: $d = 1.1109 x \Rightarrow x = \frac{3.84 \times 1 0 ^{8}}{1.1109} \approx 3.457 \times 1 0^{8} m$

Verify: $x \approx 3.46 \times 1 0^{8}$ m out of $3.84 \times 1 0^{8}$ — yeh Moon tak ke raste ka $90%$ hai, yaani Moon ke paas kyunki Moon kaafi halka hai aur Earth ko tabhi balance kar sakta hai jab tum almost uske upar khade ho ✔. Forecast: Moon ke paas sahi tha.

Ex 5 — Fields at right angles (Cell C5)

Do masses ab $P$ se ek line par nahi hain. Figure dekho: arrows $9 0^{\circ}$ par milte hain, isliye hum Pythagoras use karte hain.

Worked example Perpendicular fields

Point $P$ par, mass $A$ ek field $g_{A} = 3.0 \times 1 0^{- 6} N kg^{- 1}$ produce karta hai jo east point karta hai, aur mass $B$ $g_{B} = 4.0 \times 1 0^{- 6} N kg^{- 1}$ produce karta hai jo north point karta hai. Net field ki magnitude aur direction find karo.

Forecast: Answer $7.0 \times 1 0^{- 6}$ nahi hai. Guess karo ki yeh $5.0 \times 1 0^{- 6}$ se bada hoga ya chota.

Step 1 — Recognize karo ki inhe directly add nahi kar sakte. East aur north perpendicular hain. Yeh step kyon? Sirf same line par wale fields numbers ki tarah add ya subtract hote hain. $9 0^{\circ}$ par arrows ek right triangle ki do legs form karte hain (figure dekho), aur net hypotenuse hai.

Step 2 — Pythagoras apply karo. $g_{net} = g_{A}^{2} + g_{B}^{2} = (3.0)^{2} + (4.0)^{2} \times 1 0^{- 6}$ Yeh step kyon? Ek right triangle ki hypotenuse ki length $leg_{1}^{2} + leg_{2}^{2}$ hoti hai — exactly aise hi perpendicular vectors combine hote hain.

Step 3 — Compute karo. $g_{net} = 9 + 16 \times 1 0^{- 6} = 25 \times 1 0^{- 6} = 5.0 \times 1 0^{- 6} N kg^{- 1}$

Step 4 — Direction. North-of-east angle: $θ = arctan (\frac{g _{B}}{g _{A}}) = arctan (\frac{4.0}{3.0}) = 53. 1^{\circ}$ $arctan$ kyon? $arctan$ jawab deta hai "is rise-over-run ka angle kaun sa hai?" — yahan north leg over east leg.

Verify: $3^{2} + 4^{2} = 5$ (classic 3-4-5 triangle) ✔, aur $5.0 < 7.0$ isliye direct addition over-count karta ✔. $tan 53. 1^{\circ} = 1.33 = 4/3$ ✔.

Ex 6 — Limiting behaviour (Cell C6)

Worked example

$r \to \infty$ aur $r \to 0$ par kya hota hai? $g = \frac{GM}{r ^{2}}$ use karke, bahut door aur ek point mass ke bahut paas field describe karo. Phir far limit numerically dikhane ke liye Earth ke liye $r = 10 R$ par $g$ compute karo ( $g_{0} = 9.8$ ).

Forecast: Kya field kabhi exact zero finite distance par reach karti hai?

Step 1 — Far limit, $r \to \infty$ . Jaise $r$ badhta hai, $r^{2}$ bina bound ke badhta hai, isliye $\frac{GM}{r ^{2}} \to 0$ . Yeh step kyon? Ek fixed numerator over ek ever-larger denominator zero ki taraf shrink hota hai — lekin sirf reach karta hai zero ko infinity par, kisi bhi finite $r$ par kabhi nahi.

Step 2 — Far limit ka numerical taste. $r = 10 R$ par: $g = \frac{g _{0}}{1 0 ^{2}} = \frac{9.8}{100} = 0.098 N kg^{- 1}$ Yeh step kyon? Ratio trick phir se: distance $\times 10 \Rightarrow$ field $\div 100$ . Chota lekin non-zero — Step 1 confirm karta hai.

Step 3 — Close limit, $r \to 0$ . Jaise $r \to 0$ , $r^{2} \to 0$ aur $\frac{GM}{r ^{2}} \to \infty$ : formula "blow up" ho jaata hai. Yeh step kyon? Ek vanishing number se divide karna unbounded result deta hai — mathematically field diverge ho jaati hai ek idealised point mass ke liye.

Step 4 — Physics caveat. Ek real planet ke liye tum actually $r = 0$ par khade nahi ho sakte; ek baar jab tum andar jaate ho, sirf tumhare neeche wala mass tumhe kheenchta hai, aur $g$ actually centre par zero tak girta hai (dekho Variation of g with Altitude and Depth). $1/ r^{2}$ blow-up sirf external point mass ke liye hai.

Verify: $0.098 \times 100 = 9.8$ ✔. Door jaao → tiny lekin positive; field sirf limit mein zero hai ✔.

Ex 7 — Inverse problem (Cell C7)

Worked example Measured field se mass find karo

Ek unknown planet ke centre se $r = 2.0 \times 1 0^{7} m$ par ek field probe $g = 5.0 N kg^{- 1}$ read karta hai. Planet ka mass $M$ find karo.

Forecast: Earth ke $5.97 \times 1 0^{24}$ kg se bada hoga ya chota?

Step 1 — Tool A ko $M$ ke liye rearrange karo. $g = \frac{GM}{r ^{2}}$ se: $M = \frac{g r ^{2}}{G}$ Yeh step kyon? Hume $g$ , $r$ , $G$ pata hai aur $M$ chahiye — dono taraf $r^{2}$ se multiply karo aur $G$ se divide karo unknown isolate karne ke liye.

Step 2 — Substitute karo. $M = \frac{( 5.0 ) ( 2.0 \times 1 0 ^{7} ) ^{2}}{6.67 \times 1 0 ^{- 11}}$ Yeh step kyon? Direct plug-in; ek hi trap hai ki distance pehle square karo.

Step 3 — Arithmetic. $r^{2} = 4.0 \times 1 0^{14}$ ; numerator $= 5.0 \times 4.0 \times 1 0^{14} = 2.0 \times 1 0^{15}$ . $M = \frac{2.0 \times 1 0 ^{15}}{6.67 \times 1 0 ^{- 11}} \approx 3.0 \times 1 0^{25} kg$

Verify: Plug back karo: $g = \frac{( 6.67 \times 1 0 ^{- 11} ) ( 3.0 \times 1 0 ^{25} )}{( 2.0 \times 1 0 ^{7} ) ^{2}} = \frac{2.0 \times 1 0 ^{15}}{4.0 \times 1 0 ^{14}} = 5.0$ ✔. Earth se bada — ek heavier planet, forecast tumhare guess par dependent hai.

Ex 8 — Word problem, altitude (Cell C8)

Worked example Ek mountaintop observatory

Ek observatory Earth ki surface se $h = 4.8 km$ upar ek peak par baitha hai. Surface $g_{0} = 9.80 N kg^{- 1}$ aur $R = 6.37 \times 1 0^{6} m$ diye hue hain, observatory par $g$ find karo.

Forecast: $g$ kabhi roughly kitna drop karega — kuch parts per thousand, ya kuch percent?

Step 1 — Words ko ek distance mein translate karo. "Surface ke upar" matlab altitude hai, isliye centre se distance hai $r = R + h = 6.37 \times 1 0^{6} + 4.8 \times 1 0^{3} = 6.3748 \times 1 0^{6} m$ Yeh step kyon? Exam mein sabse common galti $h$ akela use karna hai. $r$ centre se measure hota hai — parent note ki "radius from the centre" warning dekho.

Step 2 — Ratio use karo. $g = g_{0} (\frac{R}{R + h})^{2} = 9.80 (\frac{6.37 \times 1 0 ^{6}}{6.3748 \times 1 0 ^{6}})^{2}$ Yeh step kyon? $GM$ surface aur peak dono mein common hai; ratio kabhi bhi $M$ ki zaroorat avoid karta hai.

Step 3 — Compute karo. $\frac{6.37}{6.3748} = 0.999247$ ; squared $= 0.998495$ . $g = 9.80 \times 0.998495 \approx 9.785 N kg^{- 1}$

Verify: Drop $= 9.80 - 9.785 = 0.015 N kg^{- 1}$ , lagbhag $0.15%$ — "kuch parts per thousand," jo is baat se match karta hai ki $4.8$ km Earth ke $6370$ km radius ke samne tiny hai ✔.

Ex 9 — Exam twist: ek naya planet (Cell C9)

Worked example Do scalings ek saath

Planet Zorp ki mass Earth ki do guni hai aur radius Earth ki aadhi hai. Earth ki surface $g_{0} = 9.8 N kg^{- 1}$ . Zorp ki surface gravity find karo.

Forecast: Zyada mass aur chota radius dono gravity ko strong banate hain. Multiple guess karo: ×2? ×4? ×8?

Step 1 — Zorp ka field Earth ke terms mein likho. $g_{Z} = \frac{G ( 2 M )}{( R /2 ) ^{2}}$ Yeh step kyon? Do changes Tool A mein substitute karo taaki hum Earth ka $\frac{GM}{R ^{2}} = g_{0}$ factor out kar sakein.

Step 2 — Do scalings alag karo. $g_{Z} = \frac{2 GM}{R ^{2} /4} = 2 \times 4 \times \frac{GM}{R ^{2}} = 8 g_{0}$ Yeh step kyon? Mass doubling $2$ se multiply karta hai; radius ko half karne se $(1/2)^{2} = 1/4$ se divide hota hai, yaani $4$ se multiply. Dono effects multiply hote hain, isliye $8$ milta hai.

Step 3 — Number. $g_{Z} = 8 \times 9.8 = 78.4 N kg^{- 1}$

Verify: Independent check — factors $2/ (1/2)^{2} = 2/0.25 = 8$ combine hote hain ✔. Toh $78.4 N kg^{- 1}$ . Forecast: ×8, kyunki dono effects pile up hote hain aur radius wala squared hai.

Active recall

Recall Kaun sa cell? Question ko uski method se match karo

"Field at $r = 5 R$ from surface $g_{0}$ " ::: C2 — ratio, $5^{2} = 25$ se divide karo
"Woh point jahan Earth aur Moon ke fields cancel hote hain" ::: C4 — magnitudes equal set karo, null point
"Do fields $9 0^{\circ}$ par" ::: C5 — hypotenuse par Pythagoras
" $g$ aur $r$ diya hai, $M$ find karo" ::: C7 — $M = g r^{2} / G$ ke liye rearrange karo
" $h$ height par ek peak par $g$ " ::: C8 — $r = R + h$ use karo, sirf $h$ nahi

Connections

Parent topic note — woh definition aur derivation jise yeh examples exercise karte hain
Newton's Law of Universal Gravitation — woh force law jo har $g$ ke peeche hai
Newton's Second Law (F=ma) — isiliye $g$ free-fall acceleration ke roop mein bhi kaam karta hai
Variation of g with Altitude and Depth — Ex 6 & Ex 8 ko generalize kiya $r \neq = R$ ke liye
Gauss's Law for Gravity — woh gehri wajah ki field lines $1/ r^{2}$ deti hain
Gravitational Potential and Potential Energy — inhi fields ka energy view
Electric Field Intensity E=kQ/r² — yahan har vector-addition trick verbatim transfer hoti hai