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

1.2.19 · D4 · Physics › Newton's Laws & Dynamics › Newton's law of gravitation — universal, action at distance

Constants jo is poore page par use honge (inhe apne paper par ek baar likh lo):

Definition Sign & direction convention (har problem mein use hoti hai)

Gravity hamesha attractive hoti hai, isliye yeh dono masses ko hamesha ek doosre ki taraf kheenchti hai, unke centres ko milane wali line ke saath. Is page par:

Jab hum magnitude $F = G m_{1} m_{2} / r^{2}$ likhte hain, yeh ek positive number hota hai; direction words mein batayi jaati hai ("toward Earth", "inward").
Jab directions ko add karna hota hai (jaise balance-point problem mein), hum masses ko milane wali line ke saath ek positive axis chunte hain. $+$ direction ki taraf pull $+$ count hoti hai; doosri taraf ki pull $-$ count hoti hai. "Cancel" ka matlab hai signed sum $0$ hai; equivalently, dono magnitudes equal hain aur opposite direction mein point karti hain.
Gravity mein koi "repulsive" case nahi hota — koi minus sign kabhi ise push mein nahi badalta.

Neeche sab kuch ek geometric fact par tikha hai: do spheres ke liye, $r$ unke centres ke beech ki distance hai, surfaces ke nahi (yeh hai Shell Theorem). Har problem ke liye neeche di gayi picture dhyan mein rakho.

Figure — Newton's law of gravitation — universal, action at distance

Level 1 — Recognition

Kya tum formula mein values plug kar sakte ho aur units sahi se padh sakte ho?

L1.1 — Do shopping bags

Do $1 kg$ ke bags ek table par $0.5 m$ door rakhe hain. Unke beech gravitational force kya hai?

Recall Solution

KYA: $F = G m_{1} m_{2} / r^{2}$ mein seedha substitution. KYUN: dono masses aur separation di gayi hain — kuch rearrange nahi karna. $F = \frac{( 6.674 \times 1 0 ^{- 11} ) ( 1 ) ( 1 )}{( 0.5 ) ^{2}} = \frac{6.674 \times 1 0 ^{- 11}}{0.25}$ $F = 2.67 \times 1 0^{- 10} N$ Yeh magnitude hai; har bag doosre ki taraf kheencha jaata hai unhe milane wali line ke saath. Lesson: behadh chota — ek bacterium ke weight ka fraction. Isliye hum roz ke objects ke beech gravity kabhi feel nahi karte; sirf bhari-bhoori Earth hi kaafi heavy hai ki matter kare.

L1.2 — Konsi quantity "universal" hai?

Tum Earth par $g = 9.8 m/s^{2}$ aur Moon par $g = 1.6 m/s^{2}$ measure karte ho. Kya $G$ dono measurements ke beech badal gaya? Ek line mein explain karo.

Recall Solution

Nahi. $G$ ek universal constant hai — Earth par, Moon par, aur sabse door ki galaxy mein identical hai. Jo badla woh hai $g = GM / r^{2}$ , jo local body ke mass $M$ aur radius $r$ par depend karta hai. Moon ka $M$ bahut chhota hai, isliye uska $g$ chhota hai. Dekho Gravitational Field & Potential.

Level 2 — Application

Formula ko rearrange karo; ek mass cancel karo; bade astronomical numbers substitute karo.

L2.1 — Scratch se $g$ recover karo

Sirf $G$ , $M_{\oplus}$ , aur $R_{\oplus}$ use karke Earth ki surface gravity $g$ compute karo.

Recall Solution

KYA: gravity force ko weight ke equal set karo, $\frac{G M _{\oplus} m}{R _{\oplus}^{2}} = m g$ , phir $m$ cancel karo. $m$ kyun cancel karein? Test mass $m$ dono sides par appear hoti hai. Ise khatam karne se prove hota hai ki har object ko same $g$ milta hai — Galileo ka leaning-tower result seedha algebra se nikal aata hai. $g = \frac{G M _{\oplus}}{R _{\oplus}^{2}} = \frac{( 6.674 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} )}{( 6.37 \times 1 0 ^{6} ) ^{2}}$ $g = \frac{3.984 \times 1 0 ^{14}}{4.057 \times 1 0 ^{13}} \approx 9.82 m/s^{2} ✓$

L2.2 — Pahad par weight

$70 kg$ mass ka ek astronaut itna chadhta hai ki Earth ke centre se uski distance $r = 1.02 R_{\oplus}$ ho jaati hai. Wahan uska weight kya hai? (Uska mass unchanged hai — dekho Weight vs Mass.)

Recall Solution

KYA: naye $r$ par $g$ compute karo, phir $W = m g$ . 9.8 kyun use nahi karein? Kyunki $g$ $1/ r^{2}$ ki tarah shrink hota hai; zyada altitude par field weaker hoti hai. $g^{'} = \frac{G M _{\oplus}}{( 1.02 R _{\oplus} ) ^{2}} = \frac{g}{1.0 2 ^{2}} = \frac{9.82}{1.0404} = 9.44 m/s^{2}$ $W = m g^{'} = 70 \times 9.44 = 661 N$ Uska mass abhi bhi $70 kg$ hai; sirf pull (weight) lagbhag $4%$ gira.

L2.3 — Earth aur Moon ke beech force

$M_{Moon} = 7.35 \times 1 0^{22} kg$ , Earth–Moon centre distance $r = 3.84 \times 1 0^{8} m$ . Pull find karo.

Recall Solution

KYA: $m_{1} = M_{\oplus}$ , $m_{2} = M_{Moon}$ ke saath seedha substitution. $F = \frac{( 6.674 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} ) ( 7.35 \times 1 0 ^{22} )}{( 3.84 \times 1 0 ^{8} ) ^{2}}$ Numerator $= 2.929 \times 1 0^{37}$ . Denominator $= 1.475 \times 1 0^{17}$ . $F \approx 1.99 \times 1 0^{20} N$ Bahut bada — kyunki dono masses astronomical hain, chahe woh door hain.

Level 3 — Analysis

Raw plugging ki jagah ratios aur scaling ke baare mein reason karo.

L3.1 — Girta hua Moon (Newton ka apna test)

Moon $r \approx 60 R_{\oplus}$ par orbit karta hai. (a) $g$ ki inverse-square scaling se uska acceleration predict karo. (b) Centripetal acceleration $a = \frac{4 π ^{2} r}{T ^{2}}$ se compare karo, jahan $T$ orbital period hai — ek poora chakkar lagane ka time — $T = 27.3 days$ aur $r = 3.84 \times 1 0^{8} m$ use karke.

Recall Solution

(a) KYA: wahi law, $1/ r^{2}$ se scale kiya gaya. $60$ Earth-radii par field $6 0^{2} = 3600$ guna weaker hai. $a_{grav} = \frac{g}{6 0 ^{2}} = \frac{9.8}{3600} = 2.72 \times 1 0^{- 3} m/s^{2}$ (b) Centripetal kyun? Moon near-circle travel karta hai; circular path ke liye inward acceleration $a = 4 π^{2} r / T^{2}$ zaroori hai, jahan $T$ (the period) ek orbit kitne time mein poori hoti hai (Circular Motion & Centripetal Force). $T$ ko seconds mein convert karo: $27.3 \times 86400 = 2.359 \times 1 0^{6} s$ . $a_{circ} = \frac{4 π ^{2} ( 3.84 \times 1 0 ^{8} )}{( 2.359 \times 1 0 ^{6} ) ^{2}} = \frac{1.516 \times 1 0 ^{10}}{5.564 \times 1 0 ^{12}} = 2.72 \times 1 0^{- 3} m/s^{2}$ Dono match karte hain. Jo force apple ko neeche kheenchti hai wahi force Moon ko steer karti hai — yeh agreement universality ka empirical proof hai.

L3.2 — Doubling aur halving

Do masses ke beech force $F_{0}$ act karti hai. Tum $m_{1}$ triple karte ho, $m_{2}$ half karte ho, aur $r$ double karte ho. Force kis factor se badlegi?

Recall Solution

KYA: har proportionality alag se track karo, kyunki $F \propto m_{1} m_{2} / r^{2}$ .

$m_{1}$ triple: $\times 3$ .
$m_{2}$ half: $\times \frac{1}{2}$ .
$r$ double: $2^{2} = 4$ se divide, isliye $\times \frac{1}{4}$ . $factor = 3 \times \frac{1}{2} \times \frac{1}{4} = \frac{3}{8} = 0.375$ Nayi force $0.375 F_{0}$ hai — distance-squared term sabse zyada kaam karti hai.

Level 4 — Synthesis

Gravitation ko ek doosre idea (equilibrium, orbits) ke saath combine karo.

L4.1 — Earth aur Moon ke beech balance point

Earth–Moon line par kahin ek point hai jahan ek chhoti probe ko zero net gravity feel hoti hai (pulls cancel ho jaati hain). Agar $M_{\oplus} / M_{Moon} = 81.3$ aur centre distance $d = 3.84 \times 1 0^{8} m$ hai, toh yeh point Earth ke centre se kitna door hai?

Recall Solution

KYA: probe ko Earth se $x$ door aur Moon se $d - x$ door rakho. Dono pulls equal set karo (figure mein ek red probe har taraf se ek black arrow feel karti hai). Equal kyun, sum nahi? Hamari sign convention use karke, Earth ki taraf pull $-$ hai aur Moon ki taraf pull $+$ hai (axis ke saath opposite directions). "Net zero" matlab $(+) F_{Moon} + (-) F_{Earth} = 0$ , yaani dono magnitudes equal hain. $\frac{G M _{\oplus} m}{x ^{2}} = \frac{G M _{Moon} m}{( d - x ) ^{2}}$ $G$ aur probe mass $m$ cancel karo. Yahan $m$ kyun cancel karein? Probe ki apni mass $m$ dono sides par hai — yeh ek common factor hai, isliye divide ho jaati hai. Physically iska matlab hai balance point ek pankh ya ek boulder ke liye same hai; probe ki mass kabhi nahi decide karti ki equilibrium kahan hai. $\frac{M _{\oplus}}{x ^{2}} = \frac{M _{Moon}}{( d - x ) ^{2}} \Rightarrow \frac{( d - x ) ^{2}}{x ^{2}} = \frac{M _{Moon}}{M _{\oplus}} = \frac{1}{81.3}$ Square root lo (positive, kyunki dono lengths positive hain): $\frac{d - x}{x} = \frac{1}{81.3} = \frac{1}{9.017}$ $d - x = \frac{x}{9.017} \Rightarrow d = x (1 + \frac{1}{9.017}) = 1.1109 x$ $x = \frac{d}{1.1109} = \frac{3.84 \times 1 0 ^{8}}{1.1109} = 3.457 \times 1 0^{8} m$ Toh null point Earth ke centre se lagbhag $3.46 \times 1 0^{8} m$ door hai — Moon ke raaste ka roughly $90%$ tak, kyunki Earth bahut zyada heavy hai aur zyaadatar gap par dominate karti hai. Negative root kyun reject kiya: $\cdot$ allow bhi karta hai $(d - x) / x = - 1/9.017$ , jo $x > d$ deta hai (probe Moon se aage). Wahan pulls same direction mein point karti hain, isliye add hoti hain (dono same sign), kabhi cancel nahi hoti — physically impossible, isliye hum ise drop karte hain.

L4.2 — Gravity = centripetal se orbital speed

Earth ke around $r$ radius ki circular orbit mein ek satellite ke liye, gravity hi centripetal force hai. Orbital speed $v$ derive karo, phir $r = R_{\oplus} + 4.0 \times 1 0^{5} m$ par low orbit ke liye compute karo (≈ 400 km upar, ISS).

Recall Solution

KYA: gravitational pull ko required centripetal force $m v^{2} / r$ ke equal set karo. KYUN: circle mein available only inward force gravity hai, isliye ise exactly $m v^{2} / r$ supply karna hoga (Circular Motion & Centripetal Force). $\frac{G M _{\oplus} m}{r ^{2}} = \frac{m v ^{2}}{r} \Rightarrow v^{2} = \frac{G M _{\oplus}}{r} \Rightarrow v = \frac{G M _{\oplus}}{r}$ Satellite mass $m$ cancel ho jaati hai — orbital speed sirf is par depend karti hai ki tum kitna door ho. Ab $r = 6.37 \times 1 0^{6} + 4.0 \times 1 0^{5} = 6.77 \times 1 0^{6} m$ : $v = \frac{( 6.674 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} )}{6.77 \times 1 0 ^{6}} = 5.885 \times 1 0^{7}$ $v \approx 7.67 \times 1 0^{3} m/s \approx 7.67 km/s$ Yeh lagbhag $27, 600 km/h$ hai — real ISS se match karta hai. Yahi $v^{2} = GM / r$ Kepler's Laws ka seed hai.

Level 5 — Mastery

Poore inverse-problems aur limiting behaviour — copy karne ke liye koi template nahi.

L5.1 — Orbit se planet ko weigh karo

Ek moon ek unknown planet ke around $r = 2.0 \times 1 0^{8} m$ radius par period $T = 5.0 days$ ke saath orbit karta hai ( $T$ = ek poori orbit ka time). Planet ka mass $M$ find karo.

Recall Solution

KYA: "gravity = centripetal" ko $v = 2 π r / T$ (speed = circumference over period) ke saath combine karo. YEH ROUTE KYUN? Hum $M$ directly measure nahi kar sakte, lekin hum ek orbit ko time kar sakte hain. Formula orbit ko mass report karne deta hai — literally aise hi astronomers planets aur stars ko weigh karte hain. $\frac{GM m}{r ^{2}} = \frac{m v ^{2}}{r}$ se shuru karo aur $v = \frac{2 π r}{T}$ substitute karo: Chhoti mass $m$ kyun cancel karein? Orbit karne wale moon ki mass $m$ dono sides par common factor ki tarah hai, isliye divide ho jaati hai — orbit ka size aur timing sirf central mass $M$ par depend karta hai, orbit karne wale body ki weight par nahi. Isliye hum planet ko weigh kar sakte hain bina uske moon ka mass jaane. $\frac{GM}{r} = v^{2} = \frac{4 π ^{2} r ^{2}}{T ^{2}} \Rightarrow M = \frac{4 π ^{2} r ^{3}}{G T ^{2}}$ $T = 5.0 \times 86400 = 4.32 \times 1 0^{5} s$ convert karo. $M = \frac{4 π ^{2} ( 2.0 \times 1 0 ^{8} ) ^{3}}{( 6.674 \times 1 0 ^{- 11} ) ( 4.32 \times 1 0 ^{5} ) ^{2}}$ Numerator $= 4 π^{2} (8.0 \times 1 0^{24}) = 3.158 \times 1 0^{26}$ . Denominator $= (6.674 \times 1 0^{- 11}) (1.866 \times 1 0^{11}) = 1.246 \times 1 0^{1}$ . $M = \frac{3.158 \times 1 0 ^{26}}{12.46} \approx 2.53 \times 1 0^{25} kg$ Note karo $M \propto r^{3} / T^{2}$ — woh ratio exactly Kepler's Third Law hai.

L5.2 — Limiting behaviour: kya gravity kabhi zero hoti hai?

Jab $r \to \infty$ , $F$ ke saath kya hota hai? Aur do point masses ke liye $r \to 0$ ke saath formula kya kehta hai — aur spheres ke liye yeh real crisis kyun nahi hai?

Recall Solution

Bada $r$ (KYA): $F = G m_{1} m_{2} / r^{2} \to 0$ jab $r \to \infty$ , lekin kisi finite distance par yeh kabhi exactly zero nahi hoti. Gravity ki infinite range hai — sabse door ki galaxy abhi bhi tumhein kheenchti hai, bas immeasurably faintly. Isliye "universal" literal hai. Point masses ke liye chhota $r$ (KYA): jab $r \to 0$ , $F \to \infty$ — formula diverge karta hai. Yeh alarming lagta hai. YEH REAL CRISIS KYUN NAHI: real objects points nahi hain. Ek sphere ke andar jaane par, Shell Theorem kehta hai sirf $r$ radius ke andar wala mass tumhein kheenchta hai, aur woh mass $r^{3}$ ki tarah shrink karta hai. Isliye ek uniform sphere ke andar pull actually $F \propto r$ ki tarah jaati hai aur centre par smoothly zero ho jaati hai — koi infinity nahi. Blow-up sirf idealized points ke liye hai jo exactly ek doosre ke upar baithe hों, jo kabhi nahi hota. (True point singularities General Relativity mein milti hain.)

L5.3 — Escape speed

Dikhao ki planet ki surface se hamesha ke liye nikalne ki minimum speed $v_{esc} = 2 GM / R$ hai, phir Earth ke liye evaluate karo. (Energy use karo: kinetic energy ko gravitational potential energy $U = - GM m / R$ poori tarah pay off karni hogi; dekho Gravitational Field & Potential.)

Recall Solution

KYA: total mechanical energy ko zero set karo — "just barely escapes" matlab infinity par pahunche tab zero speed bachti hai. ZERO KYUN? Infinity par $U \to 0$ ; minimum launch mein koi leftover kinetic energy bhi nahi hoti, isliye total (jo conserved hai) shuru se $0$ hona chahiye. $\frac{1}{2} m v_{esc}^{2} + (- \frac{GM m}{R}) = 0 \Rightarrow \frac{1}{2} m v_{esc}^{2} = \frac{GM m}{R}$ Launch mass $m$ kyun cancel karein? Yeh dono sides par common factor ki tarah hai, isliye divide ho jaati hai — ek pebble aur ek rocket ko same escape speed chahiye. $\frac{1}{2} v_{esc}^{2} = \frac{GM}{R} \Rightarrow v_{esc}^{2} = \frac{2 GM}{R} \Rightarrow v_{esc} = \frac{2 GM}{R}$ Ab Earth ke liye evaluate karo ( $M = 5.97 \times 1 0^{24} kg$ , $R = 6.37 \times 1 0^{6} m$ ): $v_{esc} = \frac{2 ( 6.674 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} )}{6.37 \times 1 0 ^{6}} = 1.251 \times 1 0^{8}$ $v_{esc} \approx 1.12 \times 1 0^{4} m/s \approx 11.2 km/s$ Note karo yeh L4.2 ki low-orbit speed se exactly $2$ guna hai — ek neat sanity check.

Active Recall

Recall Ek-line answers (inhe cover karo)

0.5 m par do 1 kg bags ke beech force? ::: $2.67 \times 1 0^{- 10} N$
Universal kya hai, $g$ ya $G$ ? ::: $G$ ; $g = GM / r^{2}$ local hai.
Orbital speed formula? ::: $v = GM / r$ (satellite mass cancel hoti hai).
Orbit se planet weigh karo? ::: $M = 4 π^{2} r^{3} / (G T^{2})$ — Kepler's third law.
Escape speed? ::: $v_{esc} = 2 GM / R$ , Earth ke liye lagbhag $11.2 km/s$ .
Kya gravity ki range end hoti hai? ::: Nahi — yeh $1/ r^{2}$ ki tarah fade hoti hai lekin finite $r$ par kabhi zero nahi hoti.

Connections

Parent: Universal Gravitation — woh formula jise yeh exercises drill karte hain.
Circular Motion & Centripetal Force — falling-Moon aur orbital-speed problems.
Kepler's Laws — L5.1 ka $M \propto r^{3} / T^{2}$ Kepler's third law hai.
Shell Theorem — kyun $r$ centre-to-centre hai aur centre par gravity zero kyun hai.
Gravitational Field & Potential — escape speed ke peeche potential energy $U = - GM m / r$ .
Weight vs Mass — weight $g$ ke saath badlta hai; mass kabhi nahi badlta.
General Relativity — $r \to 0$ singularity ka asli anjaam.