Newton's law of gravitation — universal, action at distance: Worked examples

1.2.19 · D3 · Physics › Newton's Laws & Dynamics › Newton's law of gravitation — universal, action at distance

Shuru karne se pehle, har symbol ko dobara samjhte hain taaki koi bhi naya reader confuse na ho.

Scenario matrix

Har gravitation problem in case classes mein se ek (ya mix) hoti hai. Har row ka kam se kam ek worked example neeche cover kiya gaya hai.

#	Case class	Isme kya alag hai	Covered by
A	Do ordinary masses, seedha plug-in	kuch nahi — yeh baseline hai	Ex 1
B	Surface gravity $g$ from $M, R$	kaun sa $r$ ? mass cancel ho jaata hai	Ex 2
C	Distance scaling (double/halve $r$ )	inverse-square, koi numbers nahi	Ex 3
D	Do forces as vectors (3 bodies)	arrows add karne padte hain, numbers nahi	Ex 4 (figure)
E	Degenerate: $r \to 0$	formula blow up karta hai — asli matlab kya?	Ex 5
F	Limiting: $r \to \infty$	force zero ki taraf fade ho jaati hai, exactly 0 kabhi nahi	Ex 5
G	Real-world word problem	prose se numbers nikaalte hain	Ex 6
H	Exam twist: change both mass & distance at once	combine A + C carefully	Ex 7
I	Ratio / "which is bigger" trap	Sun ki pull vs Moon ki pull tumpar	Ex 8

Examples

Worked example Ex 2 — Case B: surface gravity, mass cancel ho jaata hai

Earth ka $g$ estimate karo $M_{\oplus} = 5.97 \times 1 0^{24} kg$ , $R_{\oplus} = 6.37 \times 1 0^{6} m$ se.

Forecast: kaun sa $r$ use karoge — object ki height, ya Earth ka radius?

Surface par rakhi ek test mass $m$ feel karegi $F = \frac{G M _{\oplus} m}{R _{\oplus}^{2}}$ . Yeh step kyun? Shell theorem ke hisaab se, Earth aise pull karta hai jaise saari mass uske centre par ho, isliye $r = R_{\oplus}$ (centre-to-surface), zero nahi.
Lekin weight bhi $F = m g$ hoti hai. Dono equal set karo: $m g = \frac{G M _{\oplus} m}{R _{\oplus}^{2}}$ . Yeh step kyun? Ek hi neeche ki taraf lagte force ke do naam hain.
$m$ cancel ho jaata hai: $g = \frac{G M _{\oplus}}{R _{\oplus}^{2}}$ . Yeh step kyun? Isi wajah se har cheez same rate se girti hai — girne wali mass dono sides par aati hai aur vanish ho jaati hai.
Plug in karo: $g = \frac{( 6.674 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} )}{( 6.37 \times 1 0 ^{6} ) ^{2}} \approx 9.82 m/s^{2}$ .

Verify: Known value $\approx 9.8 m/s^{2}$ hai ✓. Units: $\frac{N m ^{2} k g ^{- 2} \cdot kg}{m ^{2}} = N/kg = m/ s^{2}$ ✓. Dekho Weight vs Mass.

Worked example Ex 4 — Case D: do pulls as arrows (kabhi plain numbers mein nahi)

Mass $M_{C} = 4 kg$ ek corner par rakhi hai. Mass $M_{A} = 3 kg$ usse $2 m$ west mein hai; mass $M_{B} = 3 kg$ usse $2 m$ south mein hai. $M_{C}$ par total gravitational force find karo.

Forecast: do equal pulls $9 0^{\circ}$ par — kya total $F + F$ hoga, ya kuch chhota?

$A$ ki taraf se $C$ par force: yeh west ki taraf point karti hai ( $A$ ki taraf), magnitude $F_{A} = \frac{( 6.674 \times 1 0 ^{- 11} ) ( 4 ) ( 3 )}{2 ^{2}} = 2.00 \times 1 0^{- 10} N$ . Yeh step kyun? Har pair ke liye same scalar formula use hota hai; direction line ke along pulling mass ki taraf hoti hai.
$B$ ki taraf se $C$ par force: south ki taraf point karti hai, aur symmetry se $F_{B} = F_{A} = 2.00 \times 1 0^{- 10} N$ . Yeh step kyun? Same masses, same distance ⇒ same magnitude, lekin alag direction.
Yeh perpendicular arrows hain, isliye Pythagoras se add karo (figure mein red resultant dekho): $F_{tot} = F_{A}^{2} + F_{B}^{2} = F_{A} 2 = 2.83 \times 1 0^{- 10} N$ . Yeh step kyun? Tum $F_{A} + F_{B}$ nahi jod sakte kyunki yeh alag directions mein point karte hain — forces vectors hain. Perpendicular vectors right triangle ki sides ki tarah combine hote hain.
Direction: west aur south ke beech exactly halfway, yaani $4 5^{\circ}$ south-of-west ( $A$ aur $B$ ke midpoint ki taraf point karta hai). Yeh step kyun? Equal legs ⇒ resultant right angle ko bisect karta hai.

Verify: $F_{A} 2 = 2.00 \times 1 0^{- 10} \times 1.414 = 2.83 \times 1 0^{- 10}$ ✓. Total ( $2.83$ ) naive sum ( $4.00$ ) se kam hai — sahi hai, kyunki do pulls ek doosre ke across partly kaam karti hain. Dekho Newton's Third Law janne ke liye ki har pull ka $A$ aur $B$ par ek matching reaction kyun hota hai.

Worked example Ex 5 — Cases E & F: do dangerous ends,

$r \to 0$ aur $r \to \infty$ $F = \frac{G m _{1} m _{2}}{r ^{2}}$ kya predict karta hai jab masses (a) touch karein, aur (b) infinitely door chali jaayein?

Forecast: kya formula "break" ho jaata hai, ya sirf kuch extreme bata deta hai?

Jab $r \to 0$ : denominator $r^{2} \to 0$ , toh $F \to \infty$ . Yeh step kyun? Ek fixed top ko shrinking bottom se divide karne par result explode karta hai (figure mein left spike dekho).
Lekin yeh ek degenerate case hai: formula sirf point masses ke liye ya solid spheres ke baahri region ke liye valid hai. Real spheres ke liye kabhi bhi centres ko $r = 0$ tak nahi la sakte bina unhe overlap karaye — ek baar jab tum sphere ke andar jaate ho, toh shell theorem kehta hai ki tumhare radius ke bahar wali mass pull karna band kar deti hai, isliye centre par $F$ actually wapas $0$ ho jaati hai. Yeh step kyun? Hamesha poochho "kya formula ki assumption yahan bhi sach hai?" Real bodies ke liye $r = 0$ par yeh sach nahi hai.
Jab $r \to \infty$ : $r^{2} \to \infty$ , toh $F \to 0$ . Yeh step kyun? Ek fixed top bade hote bottom par shrink karke zero ki taraf jaata hai (figure ka right tail).
Lekin $F$ kisi bhi finite $r$ ke liye exactly zero kabhi nahi hoti: gravity ki infinite reach hoti hai, bas vanishingly weak ho jaati hai. Yeh step kyun? $1/ r^{2}$ zero ki taraf jaata hai lekin usse kabhi touch nahi karta — iska asli matlab yahi hai "universal, action at a distance."

Verify: Numerically, $r$ ko chaar baar halve karne par ( $r \to r /16$ ) $F$ , $1 6^{2} = 256$ guna ho jaati hai — blow-up trend confirm hota hai. $r$ ko das baar double karne par $F$ , $2^{20} \approx 1 0^{6}$ guna shrink ho jaati hai — fade confirm hota hai ✓. Dekho Shell Theorem aur Gravitational Field & Potential.

Worked example Ex 6 — Case G: real-world word problem

Ek $70 kg$ astronaut $400 km$ altitude (ISS height) par orbit kar rahi hai. Earth uske upar kitna gravitational force lagata hai? Use karo $M_{\oplus} = 5.97 \times 1 0^{24} kg$ , $R_{\oplus} = 6.37 \times 1 0^{6} m$ .

Forecast: woh "float" karti hai, toh kya force (a) zero hai, ya (b) ground jitna hi almost strong?

Sahi centre-to-centre distance nikalo: $r = R_{\oplus} + h = 6.37 \times 1 0^{6} + 0.40 \times 1 0^{6} = 6.77 \times 1 0^{6} m$ . Yeh step kyun? $r$ Earth ke centre se measure hoti hai, isliye altitude ko radius mein add karo — yeh classic word-problem trap hai.
Substitute karo: $F = \frac{( 6.674 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} ) ( 70 )}{( 6.77 \times 1 0 ^{6} ) ^{2}}$ . Yeh step kyun? Corrected $r$ ke saath standard plug-in.
Compute karo: numerator $= 2.789 \times 1 0^{16}$ ; denominator $= 4.583 \times 1 0^{13}$ ; $F \approx 608 N$ .

Verify: Ground par uska weight $70 \times 9.8 = 686 N$ hota. ISS height par $608 N$ iska 89% hai — woh weightless nahi hai kyunki gravity khatam ho gayi; woh float karti hai kyunki woh Earth ke around free fall mein hai. Ratio check: $(R / r)^{2} = (6.37/6.77)^{2} = 0.885 \Rightarrow 686 \times 0.885 = 607 N$ ✓. Dekho Circular Motion & Centripetal Force.

Worked example Ex 7 — Case H: exam twist, mass AUR distance dono ek saath badlein

Ek planet ki mass $2 \times$ Earth ki hai aur radius $2 \times$ Earth ki hai. Uski surface gravity $g_{p}$ Earth ke $g$ ke comparison mein kya hogi?

Forecast: mass double (zyada strong) lekin radius double (zyada weak) — kya yeh exactly cancel ho jaate hain?

Surface gravity: $g = \frac{GM}{R ^{2}}$ . Planet ke liye, $M_{p} = 2 M$ , $R_{p} = 2 R$ . Yeh step kyun? Field form use karo taaki girne wali object ki mass already cancel ho jaaye.
Substitute karo: $g_{p} = \frac{G ( 2 M )}{( 2 R ) ^{2}} = \frac{2 GM}{4 R ^{2}} = \frac{1}{2} \cdot \frac{GM}{R ^{2}} = \frac{1}{2} g$ . Yeh step kyun? Mass factor $\times 2$ deta hai; radius factor $\times \frac{1}{4}$ deta hai (squared!). Net $\times \frac{1}{2}$ .

Verify: Yeh cancel nahi hote — squared radius jeetta hai. $g_{p} = 0.5 \times 9.8 = 4.9 m/s^{2}$ . Ek common galat guess hai "same $g$ " (sochte hain $2/2 = 1$ ); trap yeh hai ki $R$ par square bhool jaate hain ✓.

Worked example Ex 8 — Case I: "kaun wali pull badi hai" trap

Tumhare $70 kg$ body par, kaun si zyada strong hai: Sun ki pull ya Moon ki pull? Sun: $M_{S} = 1.99 \times 1 0^{30} kg$ , $r_{S} = 1.50 \times 1 0^{11} m$ . Moon: $M_{M} = 7.35 \times 1 0^{22} kg$ , $r_{M} = 3.84 \times 1 0^{8} m$ .

Forecast: Moon itni zyada paas hai — surely woh jeette ga?

Sun ki pull: $F_{S} = \frac{( 6.674 \times 1 0 ^{- 11} ) ( 1.99 \times 1 0 ^{30} ) ( 70 )}{( 1.50 \times 1 0 ^{11} ) ^{2}} \approx 0.413 N$ . Yeh step kyun? Sun ke numbers ke saath direct plug-in.
Moon ki pull: $F_{M} = \frac{( 6.674 \times 1 0 ^{- 11} ) ( 7.35 \times 1 0 ^{22} ) ( 70 )}{( 3.84 \times 1 0 ^{8} ) ^{2}} \approx 0.00233 N$ . Yeh step kyun? Same formula, Moon ke numbers.
Ratio: $\frac{F _{S}}{F _{M}} \approx \frac{0.413}{0.00233} \approx 177$ . Yeh step kyun? Sun ki enormous mass ( $1 0^{7}$ times Moon se zyada) uski zyada distance ko beat kar deti hai, distance square hone ke baad bhi.

Verify: Sun tumhe Moon se taqreeban 180× zyada pull karta hai ✓ — toh Moon nahi jeetta. (Tides Moon ki wajah se hote hain, uski total pull zyada hone ki wajah se nahi, balki isliye kyunki uski pull Earth ke diameter ke across zyada vary karti hai — yeh Gravitational Field & Potential ka ek alag, gradient effect hai.)

Active Recall

Recall Answers cover kar lo

Jab $r$ double karo, force ban jaati hai... ::: ek chauthai (inverse-square: $1/ 2^{2}$ ). Do equal perpendicular pulls $F$ milke banti hain... ::: $F 2$ , unke beech $4 5^{\circ}$ par (Pythagoras, $2 F$ nahi). $400$ km upar astronaut ke liye $r$ kya hai? ::: $R_{\oplus} + 400 km$ , Earth ke centre se. Jab $r \to \infty$ , kya $F$ kabhi exactly $0$ hoti hai? ::: Nahi — $0$ ki taraf fade hoti hai lekin kabhi zero nahi hoti (infinite reach). Planet with $2 M$ aur $2 R$ : surface $g$ hai... ::: Earth ka aadha, kyunki radius squared hoti hai. Sun vs Moon pull tumpar — winner kaun? ::: Sun, taqreeban $180 \times$ se (mass distance ko beat karta hai).

Connections

Yeh Hinglish mein padho →
Newton's Third Law — Ex 4 ki har pull ka ek equal, opposite partner hota hai.
Shell Theorem — centre-to-centre $r$ aur $r \to 0$ resolution ko justify karta hai.
Gravitational Field & Potential — Ex 8 mein tide/gradient idea.
Circular Motion & Centripetal Force — Ex 6 ki astronaut kyun float karti hai.
Weight vs Mass — Ex 2 mein $m g = GM m / R^{2}$ cancellation.