Gravitation: Class 11 Physics Chapter 7 Hinglish mein

Physics Part 1 · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 18 Oct 2026 · Updated 18 Oct 2026

90 min
1 simulation
26 practice questions
28 MCQs

Is chapter mein aap kya seekhenge

Kepler ke teen laws batao, law of areas ko conservation of angular momentum se derive karo, aur calculations mein T² ∝ a³ use karo
Newton ka law of gravitation scalar aur vector form mein batao, forces ko superposition se jodo, aur dono shell results use karo
Samjhao ki Cavendish ne G kaise measure kiya aur G, g aur Earth ki radius se Earth ka mass aur density kaise milte hain
Height aur depth ke saath g ke badalne ko derive karo aur apply karo
Gravitational potential energy −GMm/r derive karo, gravitational potential define karo aur samjhao ki mgh ek approximation kyun hai
Escape speed derive karo aur Earth, Moon aur doosre bodies ke liye use calculate karo
Satellite ki orbital speed, period aur energy derive karo aur ek orbit se doosre orbit mein jaane wale problems solve karo

NCERT Physics Class 11, Chapter 7, pp. 127–143 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

7.1 Introduction

Ek paka aam apni daal se girta hai. Nadi pahaad se neeche samudra ki taraf behti hai, kabhi upar ki taraf nahi. Do metre paar karne wala high-jumper ek second se kam mein wapas mat par hota hai. In teeno ke peeche ek hi kheench hai: Earth apne paas ki har cheez ko attract karti hai. Is chapter mein hum us kheench, gravitation (गुरुत्वाकर्षण), ko padhenge aur dekhenge ki yahi Moon aur planets ko bhi chalati hai.

Galileo Galilei pehle insaan the jinhone girti cheezon ko dhyaan se measure karke study kiya. Free fall itna tez hota hai ki aankh se time nahi kiya ja sakta, isliye unhone halki dhalaan par balls ko lurhka kar use dheema kar diya. Unka nateeja tha ki jab air resistance negligible ho, to har body, chahe uska mass kuch bhi ho, zameen ki taraf ek hi steady rate se speed badhati hai. Aaj hum is rate ko acceleration due to gravity kehte hain, $g \approx 9.8 m / s^{2} .$ (Leaning Tower of Pisa se do balls girane wali mashhoor kahani shayad kabhi hui hi nahi.)

Ek doosri paheli alag lagti thi. Raat dar raat taare apne patterns banaye rakhte hain; jo constellations tumhare dada-dadi ne dekhe the, wahi tum dekhte ho. Lekin kuch chamakdaar points aisa nahi karte: hafton mein woh constellations ke beech dheere-dheere khisakte hain, aur kabhi-kabhi kuch samay ke liye ulti disha mein chalne lagte hain. Greeks ne inhe planetai, yaani wanderers (bhatakne wale), kaha. Inke raaston ko samjhane mein lagbhag do hazaar saal ke models lage, jo Fig. 7.1 aur neeche ki timeline mein dikhaye gaye hain.

2/3Ulte loops ki nakal karne ke liye har planet ek chhote circle (epicycle) par chalta hai, jo ek bade circle par ghoomta hai.

Fig. 7.1Solar system ka centre rakhne ke do tareeke.NCERT p. 127

Aasmaan ki teen pictures. Pehle step mein Earth centre mein hai aur Sun aur ek planet uske chaaron taraf circles par chalte hain. Doosre step mein planet ek chhote circle, yaani epicycle, par chalta hai jiska centre bade circle par chalta hai, isliye planet ka path loops banata hai. Teesre step mein Sun centre mein hai aur Earth aur planet dono uske chaaron taraf circles par chalte hain.

Timeline

Bhatakti roshniyon se Kepler ke laws tak

c. 150 CE
Ptolemy ka Earth-centred model
Alexandria ke Claudius Ptolemy ne Earth ko centre mein rakha aur planets ke loops se match karne ke liye epicycles use kiye.
499 CE
Aryabhata ki ghoomti Earth
Aryabhatiya mein Aryabhata ne aasmaan ke roz ghoomne ko Earth ke apni axis par rotate karne se samjhaya; kuch scholars unke kaam mein ek Sun-centred scheme bhi dekhte hain.
1543
Copernicus ne Sun ko centre mein rakha
Nicolaus Copernicus ki kitaab, jo unki maut wale saal chhapi, mein planets Sun ke chaaron taraf ghoomte the.
1576–1597
Tycho Brahe ke measurements
Is Danish nobleman ne bina telescope ke planets ki positions measure ki, apne se pehle kisi se bhi zyada precisely.
1609–1619
Kepler ke teen laws
Tycho ke assistant Johannes Kepler ne Tycho ke data mein chhupe teen rules dhoondh nikaale.
1633
Galileo ka trial
Galileo, jinke telescope ne Sun-centred view ko support kiya tha, par Roman Inquisition ne trial chalaya aur unhe house arrest mein rakha.
1687
Newton ki Principia
Isaac Newton ne dikhaya ki gravitation ka ek hi law girti cheezon aur Kepler ke laws dono ko samjhata hai.

Kepler ke rules batate hain ki planets kaise chalte hain; Newton ka law batata hai kyun. Hum bhi yahi raasta lenge: Kepler ke laws, Newton ka law, $G$ ka measurement, $g$ jagah-jagah kaise badalta hai, gravitational energy, planet se escape, aur satellites.

Quick check

2 mein se 1

Galileo ka nateeja tha ki air resistance negligible ho to ek bhaari patthar aur ek halka kankad, ek hi height se saath girane par

saath mein zameen par girte hainpatthar pehle girta haikankad pehle girta haiaise order mein girte hain jo height par depend karta hai

Answer dekho

Answer (A) saath mein zameen par girte hain

Dono ka acceleration same

g

hai, isliye woh saath girte hain. Bhaari patthar par force zyada lagta hai, lekin use accelerate karne ke liye mass bhi zyada hai: ratio same rehta hai.

7.2 Kepler's Laws

Kepler ke laws empirical hain: measurements se nichode gaye rules, jinke peeche tab tak koi theory nahi thi. Har ek law us force ka ek clue nikla jo planets ko thaame rakhta hai.

LawKepler ka first law (law of orbits)

Har planet ek ellipse (दीर्घवृत्त) par chalta hai, aur Sun us ellipse ke ek focus par hota hai, centre par nahi.

Ellipse ek khincha hua circle hai. Do fixed points chuno, yeh foci hain; ellipse un saare points ka set hai jinki dono foci se distances ka jod hamesha same total hota hai. Foci se guzarne wali line ellipse ko uske lambe axis ke dono siron par kaatti hai. Us lambe axis ka aadha semi-major axis $a$ hai, jo orbit ke size ka natural measure hai, aur chhote axis ka aadha semi-minor axis $b$ hai.

Planet ke liye lambe axis ka jo sira Sun ke sabse paas hai woh uska perihelion (उपसौर) hai aur door wala sira aphelion (अपसौर) hai. Agar $c$ centre se focus tak ki distance hai, to yeh distances $a - c$ aur $a + c$ hain, isliye $a$ bas inka average hai. Ratio $e = c / a$ , yaani eccentricity, batata hai ki ellipse kitna khincha hua hai: circle ke liye $e = 0$ , aur lambe patle ellipse ke liye 1 ke kareeb. Circle woh special ellipse hai jiske dono foci ek hi jagah hote hain, aur $a$ uski radius ke barabar hota hai. Planet ke orbit ka doosra focus khaali hota hai: wahan kuch nahi hota.

4/4Sun ek focus par hai, perihelion sabse paas wala point hai aur aphelion sabse door wala; $a$ aur $b$ semi-axes hain.

Fig. 7.2Dhaage ka loop aur do pins ek ellipse banate hain.NCERT Fig. 7.1 · p. 128

Ek board par do pins khadi hain aur unke chaaron taraf dhaage ka loop hai. Pehle step mein ek pencil loop ko kheench kar triangle jaisa tight karti hai. Doosre step mein pencil ghoom chuki hai aur poora ellipse ban gaya hai, triangle ek nayi position par hai. Teesre step mein pencil se pins tak ke do seedhe pieces par r1 aur r2 likha hai; inka jod same rehta hai. Aakhri step mein dhaaga hat gaya hai, Sun left focus par hai, lambe axis ke left sire par perihelion likha hai, right sire par aphelion, aur centre se arrows semi-major axis a aur semi-minor axis b dikhate hain.

Try karke dekho

Ghar par ek orbit banao

Tumhe chahiye

mote cardboard ki ek sheet
do drawing pins
dhaage ka ek $20 c m$ loop
ek pencil aur ek ruler

Pehle guess karo

Agar tum pins ko paas-paas kar do aur usi loop se phir se banao, to naya curve hoga

zyada khincha huazyada gol, circle ke kareebbilkul same

Cardboard mein do drawing pins $8 c m$ ki doori par gaado aur unke upar loop daalo.
Pencil ki nok se loop ko tight kheencho aur dhaage ko tana rakhte hue poora ghuma kar banao.
Apne ellipse ke lambe aur chhote axes measure karo aur likh lo.
Pins ko $4 c m$ ki doori par rakh kar dobara karo, aur phir dono pins ek hi chhed mein rakh kar.

Table 7.1Tumhare ellipses

Pins ke beech ki doori	Lamba axis (cm)	Chhota axis (cm)
8 cm
4 cm
0 cm (ek chhed)

Tumhari entries apne aap save hoti hain.

LawKepler ka second law (law of areas)

Barabar time mein Sun se planet tak ki line barabar areas sweep karti hai.

Isliye planet ek steady speed se nahi chal sakta. Perihelion ke paas Sun–planet line chhoti hoti hai, to ek mahine ka area sweep karne ke liye planet ko orbit par lambi doori tay karni padti hai. Aphelion ke paas line lambi hoti hai, aur orbit ka chhota hissa hi kaafi hota hai (Fig. 7.3). Planet perihelion par sabse tez aur aphelion par sabse dheema hota hai. Jaise Earth January ki shuruaat mein, jab woh Sun ke sabse paas hoti hai, lagbhag $30.3 k m / s$ se chalti hai, aur July ki shuruaat mein lagbhag $29.3 k m / s$ se.

3/3Dono wedges ke areas barabar hain, isliye planet Sun ke paas tez aur door dheere chalta hai.

Fig. 7.3Barabar time mein barabar areas (orbit kisi bhi planet ke orbit se zyada khincha hua dikhaya gaya hai).NCERT Fig. 7.2 · p. 128

Ek elliptical orbit jiske left focus par Sun hai. Pehle step mein Sun ke paas ek chhota, chauda wedge shaded hai; planet ek mahine mein orbit ka lamba arc tay karta hai. Doosre step mein door wali taraf ek lamba, patla wedge shaded hai; ek mahine mein planet sirf chhota arc tay karta hai. Aakhri step mein dono wedges saath dikhte hain, equal areas, equal times shabdon ke saath, aur labels Sun ke paas fast aur door slow.

LawKepler ka third law (law of periods)

Planet ka orbital period $T$ aur uske orbit ka semi-major axis $a$ , $T^{2} \propto a^{3}$ se jude hain: period ka square size ke cube ke saath badhta hai.

$T$ ko years mein aur $a$ ko astronomical units mein naapo, jahan $1 A U = 1.496 \times 10^{11} m$ Earth–Sun ki mean distance hai. Tab Earth ke liye $T = 1$ aur $a = 1$ , isliye law har planet ke liye $T^{2} / a^{3} = 1$ predict karta hai. Table 7.2 dikhata hai ki yeh Mercury se Neptune tak sahi hai, aur Halley's comet ke lambe, patle orbit ke liye bhi.

Table 7.2AU aur years mein check kiya gaya Kepler ka third lawNCERT Table 7.1 · p. 129

Body	$a$ (AU)	$T$ (years)	$T^{2} / a^{3}$
Mercury	0.387	0.241	1.002
Venus	0.723	0.615	1.001
Earth	1.000	1.000	1.000
Mars	1.524	1.881	1.000
Jupiter	5.203	11.86	0.999
Saturn	9.537	29.46	1.001
Uranus	19.19	84.01	0.999
Neptune	30.07	164.8	0.999
Halley's comet	17.83	75.32	1.001

Ek hi constant saare planets par fit hota hai kyunki, jaisa tum Section 7.9 mein prove karoge, yeh sirf Sun ke mass par depend karta hai. Kisi aur cheez ke chaaron taraf ghoomne wali bodies bhi $T^{2} \propto a^{3}$ follow karti hain, lekin alag constant ke saath: Jupiter ke moons ka ek value hai, Earth ke satellites ka doosra.

SimulationAndar ke planets, scale ke saath

Sun ke saath Mercury, Venus, Earth aur Mars apne orbits par, sahi scale par bane hue. Play dabane par time bees din per second ki speed se aage chalta hai; andar ke planets bahar walon se kaafi tez ghoomte hain, aur scale toggle sahi aur sikudi hui distances ke beech switch karta hai.

Fig. 7.4Play dabao aur dekho kaise andar ke planets bahar walon se aage nikal jaate hain.

Try karo

Jab tak Earth ek chakkar lagaati hai, Mercury ke chakkar gino. Kya $a = 0.387 A U$ ke saath $T = a^{3 / 2}$ lagbhag chaar predict karta hai?
Mars ke ek chakkar mein kitne Earth years lagte hain? Ise ${1.52}^{3 / 2}$ se check karo.

Barabar time mein barabar areas kyun sweep hone chahiye? Iska jawab ek conservation law hai jo tumne Chapter 6 mein padha tha.

DerivationLaw of areas kyun sahi hai

Sun ko origin par rakho. Maan lo $m$ mass ka planet position $𝐫$ par hai aur uski velocity $𝐯$ hai. Chhote time $Δ t$ mein woh $𝐯 Δ t,$ chalta hai, aur jo patla triangle sweep hota hai uska area hai

Δ 𝐀 = \frac{1}{2} 𝐫 \times 𝐯 Δ t

(7.1)

kyunki cross product ka size ek parallelogram ka area hota hai, aur triangle uska aadha hai. $𝐩 = m 𝐯$ aur angular momentum $𝐋 = 𝐫 \times 𝐩,$ ke saath

\frac{Δ 𝐀}{Δ t} = \frac{𝐫 \times 𝐩}{2 m} = \frac{𝐋}{2 m}

(7.2)

Sun ki kheench $- 𝐫$ ki direction mein, seedhe Sun ki taraf hoti hai. Position vector ki direction wala force origin ke around koi torque nahi lagata, kyunki $𝐫 \times 𝐅 = 0,$ isliye $𝐋$ constant rehta hai aur area sweep hone ka rate bhi constant rehta hai. Yahi second law hai. Aisa force, jo hamesha ek fixed point ki line mein ho, central force kehlata hai. Constant $𝐋$ $𝐫$ aur $𝐯$ ko ek fixed plane mein bhi rakhta hai, isliye har orbit flat hota hai.

Example 7.1NCERT Example 7.1 · p. 129

Mercury sabse paas aur sabse door

Perihelion par Mercury Sun se $4.60 \times 10^{10} m$ tak aa jaata hai, jahan uski speed $59.0 k m / s$ hoti hai, aur aphelion par $6.98 \times 10^{10} m$ tak door chala jaata hai. (a) Aphelion par uski speed nikalo. (b) Uske orbit ka semi-major axis nikalo. (c) Sun se guzarne wali, lambe axis ke perpendicular ek line orbit ko perihelion side aur aphelion side mein baant deti hai. Kya Mercury apne saal ka aadhe se zyada ya kam hissa perihelion side par bitata hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Ek comet ki aphelion distance uski perihelion distance ki 9 guna hai. Perihelion par uski speed hai

aphelion wali speed ki 3 gunaaphelion wali speed ki 81 gunaaphelion wali speed ki 9 gunaaphelion wali speed ka nauvaan hissa

Answer dekho

Answer (C) aphelion wali speed ki 9 guna

Angular momentum

m r v

dono siron par same hai, isliye

v_{p} / v_{a} = r_{a} / r_{p} = 9.

Square root (3) lene ya square (81) karne ka yahan koi aadhaar nahi; nauvaan hissa ratio ko ulta kar deta hai.

7.3 Universal Law of Gravitation

Kahani hai ki girte seb ne Newton ko sochne par majboor kiya. Wajah jo bhi ho, unka khaas idea bahut bold tha: jo force seb ko neeche kheenchta hai, shayad wahi Moon tak pahunch kar use uske orbit mein thaame rakhta hai, bas kamzor hota hai kyunki Moon bahut door hai.

Woh ise apne zamaane ke jaane-maane numbers se check kar sakte the. Moon ke orbit ko $R_{m} \approx 3.84 \times 10^{8} m,$ radius ka circle maano, jo $T \approx 27.3$ din mein poora hota hai. Tab uska centripetal acceleration hai

a_{m} = \frac{v^{2}}{R_{m}} = \frac{4 π^{2} R_{m}}{T^{2}}

(7.3)

jo $a_{m} \approx 2.72 \times 10^{- 3} m / s^{2} .$ aata hai. Kya inverse-square kheench yahi predict karti hai? Seb centre se ek Earth radius par hai aur Moon lagbhag saath par, isliye $1 / r^{2}$ ki tarah ghatne wali kheench Moon par $60^{2}$ guna kamzor honi chahiye:

\frac{g}{a_{m}} = \frac{R_{m}^{2}}{R_{E}^{2}} \approx 3600

(7.4)

Aur sach mein $9.8 / 0.00272 \approx 3600.$ Fig. 7.5 ise doosre tareeke se dikhata hai: ek second mein zameen ke paas seb $4.9 m$ girta hai, jabki Moon us seedhi line se sirf lagbhag $1.4 m m$ neeche girta hai jis par woh gravity na hone par chalta.

3/3Moon Earth ke centre se 60 guna door hai aur $60^{2} = 3600$ guna kam girta hai.

Fig. 7.5Wahi kheench, Moon par 3600 guna kamzor (gap bada karke dikhaya gaya).NCERT p. 130

Do panels. Pehle step mein ek seb ped se girta hai, ek arrow dikhata hai ki woh ek second mein 4.9 metre girta hai. Doosre step mein right panel Earth aur Moon ke orbit ka ek hissa dikhata hai; Moon se ek dashed seedhi line nikalti hai, aur Moon ka asli path uske neeche mudta hai, ek chhota arrow dikhata hai ki woh kitna girta hai. Aakhri step mein ek banner par likha hai 60 guna door, 3600 guna kamzor.

LawNewton ka law of universal gravitation

Koi bhi do particles ek doosre ko attract karte hain. Force unhe jodne wali line ke along lagta hai; uska size dono masses mein se har ek ke proportional hota hai aur unki doori ke inverse square ki tarah ghatta hai.

Law of gravitation

F = G \frac{m_{1} m_{2}}{r^{2}}

(7.5)

jahan

$G$: universal gravitational constant
$r$: dono particles ke beech ki distance

Newton ne yeh law 1687 mein apni Philosophiae Naturalis Principia Mathematica (Mathematical Principles of Natural Philosophy) mein publish kiya. Sabse bold shabd hai universal: ek hi law, ek hi constant $G$ ke saath, seb, moons, planets aur, jahan tak hum jaante hain, galaxies ke liye.

Vector form mein, maan lo $𝐫$ , $m_{1}$ ke relative $m_{2}$ ki position hai, aur unit vector $\hat{𝐫} = 𝐫 / r$ $m_{1}$ se $m_{2}$ ki taraf point karta hai. $m_{1}$ ki wajah se $m_{2}$ par force wapas $m_{1}$ ki taraf point karta hai:

𝐅_{21} = - G \frac{m_{1} m_{2}}{r^{2}} \hat{𝐫}

Newton ke third law se $m_{2}$ ki wajah se $m_{1}$ par force $𝐅_{12} = - 𝐅_{21} .$ hai. Earth $1 k g$ ke aam ko lagbhag $9.8 N$ se kheenchti hai, aur aam bhi poori Earth ko $9.8 N$ se kheenchta hai; Earth ke bahut bade mass par yeh force itna chhota acceleration deta hai ki pata hi nahi chalta.

PrinciplePrinciple of superposition

Kisi particle par net gravitational force un sab forces ka vector sum hai jo baaki har particle us par lagata hai, aur har pair aise interact karta hai jaise baaki maujood hi na hon.

𝐅_{1} = \sum_{i = 2}^{n} \frac{G m_{1} m_{i}}{r_{1 i}^{2}} {\hat{𝐫}}_{1 i}

jahan

${\hat{𝐫}}_{1 i}$: $m_{1}$ se $m_{i}$ ki taraf unit vector

Example 7.2NCERT Example 7.2 · p. 130

Gravitational forces ko jodna

(a) Ek table par teen chhote lead spheres rakhe hain: A ( $10 k g$ ) ek right angle ke corner par, B ( $10 k g$ ) A se $x$ -axis ke along $0.40 m$ par, aur C ( $20 k g$ ) A se $y$ -axis ke along $0.30 m$ par. A par net gravitational force nikalo. (b) Ek regular hexagon ke corners par chheh barabar masses $M$ rakhe hain, har ek centre se $d$ distance par, aur centre par $m$ mass ka ek particle hai. Ek corner wala mass hata diya jaata hai. Ab $m$ par net force nikalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Newton ka law particles ke baare mein hai. Ek asli body particles ki bheed hai, har ek apni direction mein kheenchta hai, isliye forces ko vectors ki tarah jodna padta hai, jo integral calculus ka kaam hai. Ek uniform spherical shell ke liye result kamaal ka simple hai.

TheoremShell theorem

(1) Ek patle uniform spherical shell ke bahar ke particle ke liye, shell bilkul ek point mass ki tarah kheenchta hai, jiska mass shell ke mass ke barabar hai aur jo uske centre par rakha hai. (2) Aise shell ke andar kahin bhi rakhe particle ke liye, shell ki net kheench zero hai.

Bahar. Shell ke har chhote patch ko particle se centre tak ki line ke doosri taraf uske mirror image se jodo. Unki kheench ke sideways parts cancel ho jaate hain, aur us line ke along wale parts bachte hain (Fig. 7.6). Aise saare pairs ko calculus se exactly jodne par centre ki taraf $G M m / r^{2}$ milta hai, jahan $M$ shell ka mass hai aur $r$ particle ki centre se distance.

Andar. Particle par tip wala ek patla double cone banao. Yeh shell se do patches kaatta hai. Door wala patch bada hai, uska area uski distance ke square ki tarah badhta hai, lekin uski kheench usi distance ke inverse square ki tarah kamzor hoti hai. Dono effects exactly cancel hote hain, isliye paas aur door ke patches ulti directions mein barabar zor se kheenchte hain. Har direction aise hi pair banati hai, isliye andar har jagah net force zero hai.

Ek solid ball jiski density sirf uske centre se distance par depend karti hai, aise shells ka ek ghonsla hai, isliye bahar se woh bhi apne centre par rakhe point mass ki tarah kaam karti hai.

2/4Unki sideways kheench cancel ho jaati hai; jo bachta hai woh centre ki taraf point karta hai.

Fig. 7.6Dono shell results, symmetry se.NCERT p. 131

Ek patla spherical shell. Pehle step mein shell ke bahar ek particle m ko do matching patches kheenchte hain, ek centre wali line ke upar aur ek neeche. Doosre step mein un kheenchon ke upar aur neeche wale parts cancel hote dikhte hain, aur ek akela arrow m se centre ki taraf point karta hai. Teesre step mein particle shell ke andar hai; usse guzarne wala ek patla double cone ek chhota paas wala patch aur ek bada door wala patch kaatta hai, jo use ulti directions mein kheenchte hain. Aakhri step mein label kehta hai ki kheench cancel hoti hai aur force zero hai.

7.4 The Gravitational Constant

Kepler ke laws aur Moon test yeh tay kar dete hain ki gravity mass aur distance par kaise depend karti hai, lekin number $G$ nahi de sakte. Astronomy kisi central body ke liye hamesha sirf product $G M$ measure karti hai. Akele $G$ ko nikalne ke liye tumhe lab mein do known masses ke beech ki kheench measure karni hogi, ek force jo itna kamzor hai ki use detect karne mein bhi bahut savdhaani chahiye.

Henry Cavendish ne 1798 mein yeh ek torsion balance se kar dikhaya (Fig. 7.7). Iska dil ek halki horizontal rod hai, jo beech se ek patle wire se latki hai, aur uske dono siron par ek-ek chhoti lead ball lagi hai. Phir chhoti balls ke paas bhaari lead balls rakhi jaati hain: ek left sire ke paas rod ki ek taraf, doosri right sire ke paas doosri taraf. Har bhaari ball apne padosi ko kheenchti hai, dono kheench ulte siron par ulti directions mein point karti hain, aur milkar ek couple banati hain. Rod par koi net force nahi lagta, sirf ek torque jo wire ko ek bahut chhote angle se ghuma deta hai.

2/3Bade spheres rod ke ulte siron ko ulti taraf kheenchte hain: wire twist hota hai aur spot khisakta hai.

Fig. 7.7Upar se dekha gaya torsion balance (twist bada karke dikhaya gaya).NCERT Fig. 7.6 · p. 132

Torsion balance ka top view. Ek rod jiske dono siron par ek-ek chhota sphere hai, beech se ek wire par latki hai, jis par ek chhota mirror laga hai; mirror se light ki ek beam upar ek scale par padti hai. Pehle step mein koi bade spheres nahi hain aur light spot scale ke beech mein hai. Doosre step mein bade spheres chhote spheres ke paas ulti taraf rakhe hain, rod thoda twist hoti hai aur light spot ek taraf khisakta hai. Aakhri step mein bade spheres doosri taraf rakh diye jaate hain, rod ulti taraf twist hoti hai aur spot ulti side par chala jaata hai.

DerivationTwist se

G

padhna

Maan lo $M$ aur $m$ bade aur chhote masses hain, $d$ ek bade sphere aur uske paas wale chhote sphere ke centres ke beech ki distance hai, aur $L$ rod ki length hai. Spheres aise attract karte hain jaise unke masses unke centres par hon, isliye har pair is force se attract karta hai

F = G \frac{M m}{d^{2}}

(7.6)

Ek couple ka torque ek force guna do lines of action ke beech ki distance hota hai, yahan $F L$ . Wire twist hote hue ek torque $κ θ$ se wapas dhakelta hai jo angle ke saath badhta hai; $κ$ , yaani twist ke per radian ka torque, alag se balance ke dheeme free oscillations ka time naap kar nikala jaata hai. Rod wahan ruk jaati hai jahan dono torques barabar hon:

G \frac{M m}{d^{2}} L = κ θ

(7.7)

isliye $θ$ measure karne se $G$ mil jaata hai. Modern versions mein wire par ek chhota mirror lagaya jaata hai aur twist ko door ke scale par light ke spot se padha jaata hai: jab mirror $θ$ ghoomta hai, to reflected beam $2 θ$ ghoomti hai. Bade spheres ko ulti taraf rakhne se torque ulta ho jaata hai aur swing double ho jaata hai, jisse rod ki apni rest position ki koi bhi galti bhi cancel ho jaati hai. Aaj ka accepted value hai

G = 6.67 \times 10^{- 11} N m^{2} k g^{- 2}

(7.8)

(in units mein zyada precisely $6.674 \times 10^{- 11}$ ). Yeh aaj bhi fundamental constants mein sabse kam precisely jaana gaya constant hai.

Example

Twist kitna chhota hai?

Ek classroom torsion balance mein $M = 1.50 k g$ , $m = 20.0 g$ , $d = 5.00 c m$ , rod ki length $L = 15.0 c m$ aur torsion constant $κ = 2.00 \times 10^{- 8} N m / r a d .$ hai. Bade spheres apni jagah rakhne par mirror se $2.00 m$ door scale par light spot $2.40 c m$ khisakta hai. $G$ nikalo, aur spheres ke ek pair ke beech ka force bhi.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

2 mein se 1

Dono bade spheres rod ki ulti taraf kyun rakhe jaate hain?

har chhote sphere par kheench double karne ke liyeek couple banane ke liye: rod par torque, bina net force kerod par Earth ki kheench cancel karne ke liyewire ko khinchne se rokne ke liye

Answer dekho

Answer (B) ek couple banane ke liye: rod par torque, bina net force ke

Dono siron par ulte forces se net force zero hota hai lekin torque

F L

milta hai, jo wire ko twist karta hai. Dono spheres ek hi taraf rakhne se rod twist hone ke bajaye side mein dhakel di jaati.

7.5 Earth ka Acceleration due to Gravity

Shell result (1) pyaaz jaisi Earth ki har layer par ek saath lagta hai. Zameen se upar kahin bhi khade ho, to core se crust tak har layer aise kheenchti hai jaise uska mass centre par ho. Saari layers jodo to poori Earth centre par rakhe $M_{E}$ mass ke ek particle ki tarah behave karti hai.

Zameen ke neeche hisaab badal jaata hai. Ek mine mein neeche centre se $r$ distance wale point tak jao (Fig. 7.8). Ab tumhare upar ki layers tumhe gher leti hain, aur shell result (2) se woh kheench mein kuch nahi jodti. Sirf tumhare pairon ke neeche $r$ radius wali ball count hoti hai, jo centre par rakhe $M_{r}$ mass ke particle ki tarah kaam karti hai.

3/3Sirf $r$ radius wali ball kheenchti hai, jaise uska mass $M_{r}$ centre par ho.

Fig. 7.8Sirf woh mass tumhe kheenchta hai jo tumse zyada centre ke paas hai.NCERT Fig. 7.7 · p. 132

Earth ke liye ek kata hua circle, jismein surface se ek mine shaft andar centre se r distance par rakhi body m tak jaata hai. Pehle step mein poori Earth ek jaisi shaded hai. Doosre step mein radius r ke bahar ka shell grey kar diya gaya hai, note ke saath ki yeh koi net kheench nahi deta. Aakhri step mein r radius wali andar ki ball highlight hai aur ek arrow body se centre ki taraf point karta hai, note ke saath ki andar ki ball aise kaam karti hai jaise uska saara mass centre par ho.

DerivationUniform Earth ke andar force

Body ke neeche ki ball point mass ki tarah kheenchti hai:

F = \frac{G m M_{r}}{r^{2}}

(7.9)

Agar Earth ki density $ρ$ har jagah same hai, to mass volume ke proportional hai:

M_{r} = \frac{4}{3} π r^{3} ρ = M_{E} \frac{r^{3}}{R_{E}^{3}}

F = \frac{G m M_{E}}{R_{E}^{3}} r

(7.10)

Uniform Earth ke andar force centre se distance ke proportional badhta hai aur centre par khud zero ho jaata hai. Surface par, $r = R_{E}$ :

F = \frac{G M_{E} m}{R_{E}^{2}}

(7.11)

Newton ke second law se yeh force $m g$ ke barabar hai, isliye surface par acceleration due to gravity hai

g = \frac{F}{m} = \frac{G M_{E}}{R_{E}^{2}}

(7.12)

Ismein $m$ ka koi nishaan nahi: har body same $g$ se girti hai, bilkul jaisa Galileo ne paaya tha.

Kisi point par per unit mass force, $𝐅 / m$ , ko wahan ka gravitational field (ya gravitational intensity) kehte hain. Earth ke paas yeh centre ki taraf point karta hai, aur surface par iska size $g$ hai. Iska unit $N / k g$ hai, jo $m / s^{2}$ ke barabar hai.

$G$ pata ho to Eq. (7.12) do aasaan measurements, $g = 9.8 m / s^{2}$ aur $R_{E} = 6.37 \times 10^{6} m,$ ko planet ke mass mein badal deta hai: $M_{E} = g R_{E}^{2} / G \approx 6.0 \times 10^{24} k g .$ Isiliye kaha jaata hai ki Cavendish ne 'Earth ko tola'. Earth ke volume se divide karne par uski mean density milti hai:

\begin{aligned} ρ & = \frac{3 g}{4 π G R_{E}} \\ \approx 5.5 \times 10^{3} k g / m^{3} \end{aligned}

Surface ki chattanon ki density iski sirf lagbhag aadhi hai, lagbhag $2.7 \times 10^{3} k g / m^{3},$ isliye gehra andar ka hissa kaafi zyada dense hona chahiye: Earth ke iron core ka ek shuruaati, mazboot clue.

Quick check

2 mein se 1

Mars ka mass

6.42 \times 10^{23} k g

aur radius

3.39 \times 10^{6} m .

hai. Uske surface par

g

nikalo,

m / s^{2}

mein.

m/s²

Answer dekho

Answer 3.73 m/s²

g = G M / R^{2} = (6.67 \times 10^{- 11}) (6.42 \times 10^{23}) / (3.39 \times 10^{6})^{2} = 3.73 m / s^{2},

yaani Earth ke ek-tihaai se thoda zyada.

7.6 Earth ke Surface ke neeche aur upar Acceleration due to Gravity

Equation (7.12) surface par $g$ deta hai. Pahaad par chadhne, plane mein udne ya mine mein neeche jaane par yeh kaise badalta hai?

DerivationSurface se

h

height par

Height $h$ par body Earth ke bahar hai, centre se $R_{E} + h$ distance par:

F (h) = \frac{G M_{E} m}{(R_{E} + h)^{2}}

(7.13)

\begin{aligned} g (h) & = \frac{G M_{E}}{(R_{E} + h)^{2}} \\ = \frac{g}{(1 + h / R_{E})^{2}} \end{aligned}

(7.14)

Earth ki radius ke comparison mein chhoti heights ( $h ≪ R_{E}$ ) ke liye, $∣ x ∣ ≪ 1$ ke liye binomial approximation $(1 + x)^{- 2} \approx 1 - 2 x$ se milta hai

g (h) \approx g (1 - \frac{2 h}{R_{E}})

(7.15)

DerivationSurface se

d

depth par

Depth $d$ par body centre se $R_{E} - d$ par hai. Uske upar $d$ motai wala shell koi net force nahi lagata, aur $R_{E} - d$ radius wali andar ki ball aise kheenchti hai jaise uska mass $M_{s}$ centre par ho. Uniform density ke saath,

\frac{M_{s}}{M_{E}} = \frac{(R_{E} - d)^{3}}{R_{E}^{3}}

(7.16)

F (d) = \frac{G M_{s} m}{(R_{E} - d)^{2}}

(7.17)

F (d) = \frac{G M_{E} m}{R_{E}^{3}} (R_{E} - d)

(7.18)

\begin{aligned} g (d) & = \frac{F (d)}{m} \\ = g (1 - \frac{d}{R_{E}}) \end{aligned}

(7.19)

Is baar koi approximation nahi chahiye tha: uniform Earth ke liye result exact hai.

Dono results ko saath dekho (Fig. 7.9), to surface ek peak ban jaata hai: kisi bhi direction mein, upar aasmaan ki taraf ya neeche zameen mein, jao to $g$ chhota hota jaata hai. Surface ke paas upar jaane par girawat do guna tez hai, isliye chhoti height $h$ par utna hi $g$ ghatta hai jitna $2 h$ depth par.

3/3Surface ke theek upar curve, andar wali line ke chadhne se do guna tezi se girta hai.

Fig. 7.9Earth ke centre se distance

r

ke saath

g

kaise badalta hai (uniform Earth).NCERT Fig. 7.8 · p. 134

g ka graph, uski surface value ke fraction ke roop mein, centre se distance ke against, Earth ki radius ki units mein 0 se 4 tak. Pehle step mein ek seedhi line centre par zero se surface par 1 tak chadhti hai. Doosre step mein ek curve surface par 1 se chaar radius par ek-solahvaan tak girta hai, label ke saath ki yeh one over r squared ki tarah girta hai. Aakhri step mein surface par ek dashed tangent dikhata hai ki bahar ka curve andar ki line ke chadhne se do guna tezi se girta hai.

Example

g 1 % kahan ghatta hai?

$R_{E} = 6370 k m .$ lo. (a) Surface se kitni height par $g$ surface se 1 % kam hai? (b) Kitni depth par? (c) Sea level se $8.85 k m$ upar Mount Everest ki choti par $g$ kitne percent kam hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Surface se kitni height par

g

apni surface value ka ek-chauthaai hai?

R_{E} = 6400 k m

lo aur height km mein do.

Answer dekho

Answer 6400 km

g (h) / g = 1 / (1 + h / R_{E})^{2} = 1 / 4,

isliye

1 + h / R_{E} = 2

aur

h = R_{E} = 6400 k m .

Approximate formula (7.15) use nahi ho sakta:

h

chhota nahi hai.

7.7 Gravitational Potential Energy

Ek kitaab ko shelf par uthao, to tumhara kiya work bekaar nahi jaata: kitaab ko girne do to gravity use kinetic energy ke roop mein lauta deti hai. Position ki wajah se is tarah stored energy potential energy hai. Yeh sirf conservative forces ke liye define ho sakti hai, jinka do points ke beech work har raaste par same hota hai. Gravity conservative hai, isliye uski ek potential energy hai.

Kuch metre uthane mein kheench mushkil se badalti hai, isliye use constant $m g$ maano. Tab $m$ mass ki body ko height $h_{1}$ se $h_{2}$ tak uthane mein work lagta hai

W_{12} = m g (h_{2} - h_{1})

(7.20)

Koi bhi function jiska difference yeh work de, potential energy ka kaam kar sakta hai. Sabse simple neeche wala hai, aur yeh work ko difference ke roop mein deta hai:

U (h) = m g h + U_{0}

(7.21)

W_{12} = U (h_{2}) - U (h_{1})

(7.22)

Additive constant $U_{0}$ free hai: yeh bas ground level ke liye tumhari chuni hui value hai, aur zyada tar problems mein ise zero rakha jaata hai.

DerivationEarth se door potential energy

Badi distances par poora inverse-square force use karna padta hai. $r \geq R_{E}$ ke liye yeh hai

F = \frac{G M_{E} m}{r^{2}}

(7.23)

Body ko $r_{1}$ se badi distance $r_{2}$ tak dheere-dheere uthao. Agent ka force raaste ke har point par gravity ko balance karta hai, isliye uska kiya work hai

W_{12} = \int_{r_{1}}^{r_{2}} \frac{G M_{E} m}{r^{2}} d r

W_{12} = G M_{E} m (\frac{1}{r_{1}} - \frac{1}{r_{2}})

(7.24)

Ek function jiska difference $U (r_{2}) - U (r_{1})$ theek yahi de, woh hai

U (r) = - \frac{G M_{E} m}{r} + U_{\infty}

(7.25)

jismein ek free constant $U_{\infty}$ hai, yaani infinite distance par value. Hum hamesha sirf differences measure karte hain, isliye ise fix kar sakte hain; universal choice $U_{\infty} = 0$ se milta hai

U (r) = - \frac{G M_{E} m}{r}

$r \geq R_{E}$ ke liye. Ise work ke baare mein ek statement ki tarah padho: body ko infinity se dheere-dheere andar laate waqt agent use ek aisi kheench ke against rokta hai jo use tez kar deti, isliye agent ka work, $U (r) - 0$ , negative hai. Body ko phir se door le jaane ke liye tumhe $G M_{E} m / r$ dena padega. Zero ko khiskane se $U$ ki har value same constant se shift ho jaati, aur koi force ya energy difference nahi badalta.

To phir $m g h$ kahan se aata hai? Ek body ko surface se chhoti height $h$ tak uthao aur $r_{1} = R_{E}$ aur $r_{2} = R_{E} + h$ ke saath Eq. (7.24) use karo:

\begin{aligned} Δ U & = \frac{G M_{E} m h}{R_{E} (R_{E} + h)} \\ \approx \frac{G M_{E} m}{R_{E}^{2}} h = m g h \end{aligned}

$h ≪ R_{E}$ ke liye. Yaani $m g h$ exact potential energy ke difference ka ek approximation hai. Shelf ya pahaad ke liye yeh bahut badhiya hai aur satellite heights par bilkul galat (Fig. 7.10).

3/3Surface ke paas curve lagbhag seedha hai, slope $m g$ ke saath: yahi woh jagah hai jahan $m g h$ kaam karta hai.

Fig. 7.10

U = - G M_{E} m / r

: infinity par zero, surface par sabse zyada negative.NCERT p. 135

Earth ke centre se distance r ke against potential energy ka graph, zero line upar hai aur left mein Earth ka andar wala hissa shaded hai. Pehle step mein curve surface par gehra aur negative shuru hota hai aur r ke saath badhta hai. Doosre step mein ek note dikhata hai ki door jaakar curve zero ki taraf flat hota hai. Aakhri step mein surface par ek chhota seedha tangent dikhata hai ki surface ke paas curve lagbhag seedhi line hai.

Definition

Gravitational potential

Kisi point par gravitational potential wahan rakhi body ki per unit mass gravitational potential energy hai: kisi body ko infinity se us point tak dheere-dheere laane mein per kilogram kiya gaya work.

ExampleMass $M$ ke paas $V = - G M / r$ ; iska SI unit $J / k g$ hai.

Gravitational potential energy bodies ke ek pair ki hoti hai. $r$ doori par rakhe do particles $m_{1}$ aur $m_{2}$ ke liye, infinite separation par zero lekar,

U = - \frac{G m_{1} m_{2}}{r}

Kai particles ke liye superposition se har pair ka ek aisa term jodo: teen particles ke teen pairs hote hain, chaar ke chheh.

Example 7.3NCERT Example 7.3 · p. 135

Triangle par teen masses

Teen particles, har ek $2.0 k g$ mass ka, $0.50 m$ side wale equilateral triangle ke corners par rakhe hain. (a) System ki gravitational potential energy nikalo. (b) Triangle ke centroid par gravitational potential nikalo. (c) Jab $1.0 k g$ mass ka chautha particle door se centroid tak laaya jaata hai, to kitni energy release hoti hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Infinity par

U = 0

lekar, orbit mein satellite ki gravitational potential energy hai

positivezeronegativepositive ya negative, uski speed par depend karta hai

Answer dekho

Answer (C) negative

U = - G M m / r

har finite distance par negative hai. Speed kinetic energy ko affect karti hai,

U

ko nahi.

7.8 Escape Speed

Upar uchhali cricket ball kuch seconds mein neeche aa jaati hai; Mars bheja gaya probe kabhi wapas nahi aata. In dono ke beech kahin ek threshold launch speed hai, aur conservation of energy use kuch hi lines mein nikal deta hai.

DerivationSabse kam speed jo kabhi wapas nahi aati

$m$ mass ki body ko Earth ke centre se $R_{E} + h$ distance wale point se $v_{i}$ speed se launch karo, air resistance aur baaki saari bodies ko ignore karke. Socho ki agar woh kabhi infinitely door pahunche, jahan uski potential energy zero hai aur speed koi $v_{f}$ hai, to kya bachta hai. Mechanical energy conserve hoti hai, isliye

\begin{aligned} \frac{1}{2} m v_{i}^{2} - \frac{G M_{E} m}{R_{E} + h} \\ = \frac{1}{2} m v_{f}^{2} \end{aligned}

(7.28)

Right-hand side kabhi negative nahi ho sakti, isliye body infinity tak tabhi pahunchti hai jab

\frac{1}{2} m v_{i}^{2} - \frac{G M_{E} m}{R_{E} + h} \geq 0

(7.29)

Sabse chhoti launch speed left side ko theek zero kar deti hai, taaki body dheere-dheere infinity tak pahunche aur uski speed zero ki taraf jaaye:

\frac{1}{2} m (v_{i})_{\min}^{} = \frac{G M_{E} m}{R_{E} + h}

(7.30)

Surface se launch ( $h = 0$ ) ke liye yeh minimum escape speed (पलायन चाल) hai:

v_{e} = \sqrt{\frac{2 G M_{E}}{R_{E}}}

(7.31)

aur kyunki Eq. (7.12) se $G M_{E} = g R_{E}^{2}$ ,

v_{e} = \sqrt{2 g R_{E}}

(7.32)

$g = 9.8 m / s^{2}$ aur $R_{E} = 6.37 \times 10^{6} m,$ ke saath $v_{e} = 11.2 k m / s,$ yaani lagbhag $40 000$ km/h. Mass $m$ cancel ho gaya, isliye ek kankad aur ek spacecraft ko same speed chahiye. Launch direction bhi kahin nahi aayi: energy ek scalar hai, isliye koi bhi direction chalegi jab tak path zameen se na takraye (aur hum hawa ko ignore karein). Isiliye dhyaan rakhne wali books escape velocity ke bajaye escape speed kehti hain.

2/3 $E = 0$ par woh chadhti rehti hai; uski speed sirf infinity par zero hoti hai. Yahi escape speed hai.

Fig. 7.11Total energy tay karti hai ki launch ki gayi body wapas aayegi ya nahi.NCERT p. 136

Earth ke centre se distance ke against potential energy ka curve, jo surface par gehra shuru hota hai aur zero ki taraf badhta hai. Pehle step mein negative total energy ki ek horizontal line curve ko ek turning point par milti hai, aur surface par ek bar launch par kinetic energy dikhata hai: body wapas mud jaati hai. Doosre step mein total-energy line zero axis par hai: body bas escape kar paati hai, aur uski speed sirf infinity par zero hoti hai. Aakhri step mein line zero se upar hai: body escape karti hai aur door jaakar bhi kuch kinetic energy rakhti hai.

Yahi formula kisi bhi planet ya moon ke liye kaam karta hai. Moon ka $g = 1.6 m / s^{2}$ aur radius $1740 k m$ hai, isliye uski escape speed sirf lagbhag $2.4 k m / s$ hai.

Example 7.4NCERT Example 7.4 · p. 136

Do spheres ke beech launch

Do uniform spheres fixed rakhe hain, unke centres $10 R$ door hain: A ka mass $M$ aur radius $R$ hai; B ka mass $16 M$ aur radius $2 R$ hai. Ek chhota probe A ke surface se seedha B ke centre ki taraf fire kiya jaata hai. (a) Unke beech kahan probe par net force zero hai? (b) Woh sabse kam launch speed nikalo jisse probe B tak pahunche. (c) Us launch speed ke saath woh B se kitni speed se takrata hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Ek planet ki mean density Earth jitni hai lekin radius do guna. Uski escape speed Earth ki kitne guna hai?

Answer dekho

Answer 2

M = \frac{4}{3} π R^{3} ρ

ke saath

v_{e} = \sqrt{2 G M / R}

v_{e} \propto R \sqrt{ρ},

milta hai, isliye

R

double karne se

v_{e}

double hota hai.

\sqrt{2}

jawab dene ka matlab bhool gaye ki mass

R^{3}

ki tarah badhta hai.

7.9 Earth Satellites

Newton ne ek bahut oonche pahaad par horizontal fire karti ek cannon ki kalpana ki (Fig. 7.12). Dheemi ball neeche mudti hai aur paas hi gir jaati hai. Tez ball aur door girti hai, kyunki zameen uske neeche mud kar door hoti jaati hai. Ek khaas speed par ball utni hi tezi se girti hai jitni tezi se zameen mudti hai, isliye woh kabhi zameen par nahi girti: woh orbit mein hai, hamesha ke liye Earth ke chaaron taraf girti hui. Aur tez karo to orbit khinch kar ellipse ban jaata hai, aur escape speed par ball kabhi wapas nahi aati.

Jo bhi cheez is tarah kisi planet ke chaaron taraf girti hai woh ek satellite (उपग्रह) hai. Moon arabon saalon se yahi kar raha hai, har chakkar mein lagbhag $27.3$ din leta hai; woh har chakkar mein apni axis par bhi ek baar ghoomta hai, isliye uska ek hi chehra hamesha humein dikhta hai. Pehla artificial satellite, Sputnik 1, 1957 mein upar gaya, aur aaj hazaaron satellites phone calls aur television relay karte hain, navigation guide karte hain, aur mausam aur faslon par nazar rakhte hain. Kepler ke laws bina badle lagte hain, bas Sun ki jagah Earth: satellite orbits circles ya ellipses hote hain jinke ek focus par Earth ka centre hota hai.

4/5 $v_{o}$ se thoda tez, aur orbit ek ellipse ban jaata hai jo bahut door tak jaata hai.

Fig. 7.12Newton ki cannon (pahaad aur speeds bahut bada karke dikhaye gaye).NCERT p. 137

Earth jiske upar ek ooncha pahaad hai aur ek cannon right ki taraf horizontal fire karti hai. Har step ek zyada launch speed wala path jodta hai aur pichhle paths halke ho jaate hain. Pehle step mein ball neeche mudti hai aur pahaad ke paas girti hai. Doosre mein woh lagbhag ek-chauthaai chakkar door girti hai. Teesre mein woh pahaad ki choti ke theek upar ek circle par poora chakkar lagaati hai. Chauthe mein woh ek ellipse par chalti hai jo Earth ke bahut neeche tak jaakar wapas aata hai. Aakhri step mein path khul kar picture se bahar chala jaata hai: ball escape kar jaati hai.

DerivationCircular orbit mein speed aur period

$m$ mass ka ek satellite $R_{E} + h$ radius ke circle par $v$ speed se chalta hai, jahan $h$ surface se uski height hai. Circle mein chalne ke liye centripetal force chahiye

F_{c} = \frac{m v^{2}}{R_{E} + h}

(7.33)

jo centre ki taraf ho, aur gravity theek yahi deti hai:

F_{g} = \frac{G M_{E} m}{(R_{E} + h)^{2}}

(7.34)

$F_{c} = F_{g}$ rakh kar aur $m$ cancel karke orbital speed milti hai:

v = \sqrt{\frac{G M_{E}}{R_{E} + h}}

(7.35)

Oonche orbits dheeme hote hain. Surface ko chhoote orbit ke liye, hawa ignore karke aur $h$ ko zero rakh kar, yeh ban jaata hai

\begin{aligned} v_{o} & = \sqrt{\frac{G M_{E}}{R_{E}}} \\ = \sqrt{g R_{E}} \end{aligned}

(7.36)

jo lagbhag $7.9 k m / s$ hai. Eq. (7.31) se compare karo: $v_{e} = \sqrt{2} v_{o}$ , isliye escape karne mein Earth ke chaaron taraf chhoote hue ghoomne se sirf 41 % zyada speed chahiye. Ek chakkar $2 π (R_{E} + h)$ hai, isliye period hai

\begin{aligned} T & = \frac{2 π (R_{E} + h)}{v} \\ = \frac{2 π (R_{E} + h)^{3 / 2}}{\sqrt{G M_{E}}} \end{aligned}

(7.37)

Dono sides ka square karne par,

\begin{aligned} T^{2} & = k (R_{E} + h)^{3}, \\ k & = \frac{4 π^{2}}{G M_{E}} \end{aligned}

(7.38)

Yeh Earth ke satellites ke liye Kepler ka third law hai. Constant sirf central mass par depend karta hai; Sun ke mass $M_{S}$ ke saath yahi steps planets ka law $T^{2} = K_{S} a^{3}$ dete hain, jahan $K_{S} = 4 π^{2} / G M_{S}$ . Surface ko chhoote satellite ke liye,

\begin{aligned} T_{0} & = 2 π \sqrt{\frac{R_{E}}{g}} \\ \approx 84 m i n \end{aligned}

(7.39)

Asli low orbits kuch sau kilometre upar hote hain, isliye unhe thoda zyada time lagta hai: International Space Station, lagbhag $420 k m$ oonchaai par, lagbhag 93 minute mein chakkar lagaata hai. Circle ke formulas planets ke liye bhi itne achhe kyun chalte hain? Kyunki unke ellipses mushkil se khinche hue hain. Earth ke orbit ka $e \approx 0.017$ hai; scale par banaya jaaye to bilkul gol dikhega, $b$ aur $a$ mein 0.02 % se bhi kam fark hai. Mercury ( $e \approx 0.21$ ) saaf exception hai, aur Mars ( $e \approx 0.09$ ) kaafi peeche doosre number par.

InteractiveOrbit launcher

Bani hui Earth ke upar chuni gayi height se ek satellite horizontal launch hota hai, us speed par jo tum slider se set karte ho. Predicted path turant dikhta hai: woh Earth se takra sakta hai, circle ban sakta hai, ellipse ban sakta hai ya khul kar escape kar sakta hai. Play dabane par satellite path par chalta hai. Readouts us height par circular-orbit aur escape speeds, per kilogram total energy, sabse door pahunchi distance aur, closed orbits ke liye, period dikhate hain. Buttons speed ko circular ya escape value par set kar dete hain.

Fig. 7.13Ek height aur horizontal launch speed chuno, phir Play dabao. Panel orbit ka naam batata hai aur uski energy aur period deta hai.

Example 7.5NCERT Example 7.5 · p. 137

Saturn ko tolna, Ceres ka time naapna

(a) Saturn ka sabse bada moon, Titan, Saturn ke centre se $1.22 \times 10^{9} m$ par $15.9$ din ke period ke saath ghoomta hai. Saturn ke mass ka estimate lagao. (b) Dwarf planet Ceres, Earth ki orbital radius ki 2.77 guna doori par Sun ke chaaron taraf ghoomta hai. Uska saal Earth ke kitne din ka hai? Dono orbits ko circles maano.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 7.6NCERT Example 7.6 · p. 138

Earth ke mass tak pahunchne ke do raaste

Earth ka mass nikalo (a) surface par $g = 9.80 m / s^{2}$ aur $R_{E} = 6.37 \times 10^{6} m,$ se, aur (b) International Space Station ke orbit se, jiski radius $6.79 \times 10^{6} m$ aur period $92.8$ min hai. $G = 6.67 \times 10^{- 11} N m^{2} k g^{- 2} .$ lo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 7.7NCERT Example 7.7 · p. 138

Kaam ki units mein Kepler ka constant

Sun ke chaaron taraf orbits ke liye $T^{2} = K_{S} a^{3}$ , jahan $K_{S} = 4 π^{2} / G M_{S}$ . (a) $M_{S} = 1.99 \times 10^{30} k g .$ lekar $K_{S}$ ko $s^{2} m^{- 3},$ mein nikalo. (b) $1 A U = 1.496 \times 10^{11} m$ aur $1 y r = 3.156 \times 10^{7} s .$ use karke ise years² per AU³ mein likho. (c) Ek asteroid ke orbit ka semi-major axis $4.0 A U$ hai. Uska period nikalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

$T^{2} = (4 π^{2} / G M_{E}) a^{3}$ form mein Kepler ka third law elliptical orbits par bhi lagta hai, jahan $a$ semi-major axis hai aur Earth ek focus par hai. Isliye jis satellite ki Earth ke centre se distance $7000 k m$ aur $13 000 k m$ ke beech badalti hai, uska period $10 000 k m$ radius ke circle wale satellite jitna hi hai: lagbhag 2.8 ghante.

Quick check

3 mein se 1

Satellite A ek planet ke chaaron taraf radius

r

par aur satellite B radius

4 r

par ghoomta hai. Unke periods ka ratio

T_{B} / T_{A}

hai

41682

Answer dekho

Answer (C) 8

T \propto r^{3 / 2}

, isliye

T_{B} / T_{A} = 4^{3 / 2} = 8.

4 jawab dene ka matlab

T \propto r

maan liya; 16 mein

r^{2}

use hua; 2 mein

\sqrt{r}

7.10 Orbit mein ghoomte Satellite ki Energy

DerivationKinetic, potential aur total energy

Radius $r = R_{E} + h$ ke circular orbit ke liye Eq. (7.35) se $v^{2} = G M_{E} / r$ milta hai, isliye kinetic energy hai

\begin{aligned} K & = \frac{1}{2} m v^{2} \\ = \frac{G M_{E} m}{2 r} \end{aligned}

(7.40)

Infinity par zero lekar potential energy hai

U = - \frac{G M_{E} m}{r}

(7.41)

aur total mechanical energy hai

\begin{aligned} E & = K + U \\ = - \frac{G M_{E} m}{2 r} \end{aligned}

(7.42)

Isse teen saaf relations milte hain: $K = - E$ , $U = 2 E$ aur $∣ U ∣ = 2 K$ . Total energy negative hai: satellite Earth se bound hai.

Negative total energy ka matlab saaf hai. Satellite ko poori tarah azaad karne ke liye, yaani use zero speed ke saath infinity tak le jaane ke liye jahan $E = 0$ , tumhe $G M_{E} m / 2 r$ dena padega. Yeh uski binding energy hai. $E \geq 0$ wali body satellite hoti hi nahi: woh escape kar jaati, jaisa tumne Section 7.8 mein dekha.

Ellipse par $K$ aur $U$ ka bantwara orbit ke saath badalta rehta hai: satellite perigee, yaani Earth ke sabse paas wale point, ki taraf girte hue speed pakadta hai, aur wapas upar chadhte hue speed khota hai. Total nahi badalta. Woh $E = - G M_{E} m / 2 a$ par rehta hai, $r$ ki jagah semi-major axis $a$ ke saath, aur circle ki tarah negative.

Yahan ek surprise hai (Fig. 7.14). Satellite ko oonche orbit mein le jaane ke liye energy chahiye, kyunki $E$ kam negative ho jaata hai. Phir bhi oonche orbit mein woh zyada dheere chalta hai: $K$ utna hi ghatta hai jitna $E$ badhta hai, jabki $U$ uska do guna badhta hai. Di gayi energy, aur chhodi gayi kinetic energy, sab potential energy mein chali jaati hai.

3/3Upar jaane ke liye $E$ mein badhat jitni energy chahiye: $G M_{E} m / r$ ka ek-chauthaai.

Fig. 7.14Energies

G M_{E} m / r

ki units mein.NCERT p. 138

Ek zero line ke around bar charts. Pehle step mein, radius r ke orbit ke liye, kinetic-energy bar zero se aadha unit upar jaata hai, potential-energy bar ek unit neeche aur total-energy bar aadha unit neeche. Doosre step mein radius 2r ke orbit ke bars dikhte hain, har ek aadha lamba: kinetic energy ek-chauthaai unit, potential energy minus aadha, total energy minus ek-chauthaai. Aakhri step mein ek arrow total energy mein badhat dikhata hai, ek-chauthaai unit, jo di jaane wali energy hai.

Example 7.8NCERT Example 7.8 · p. 139

Satellite ko upar le jaana

$1500 k g$ ka ek satellite $7.00 \times 10^{6} m,$ radius ke circular orbit se, jo lagbhag $630 k m$ upar hai, $4.22 \times 10^{7} m .$ radius wale orbit mein le jaaya jaata hai. Uski total, kinetic aur potential energy mein changes nikalo. $G M_{E} = 3.98 \times 10^{14} m^{3} / s^{2} .$ use karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Circular orbit mein satellite ke liye kaun sa relation sahi hai?

E = K

U = - K

E = U

E = - K

Answer dekho

Answer (D) $E = - K$

K = G M m / 2 r

aur

E = - G M m / 2 r

, isliye

E = - K

U = - K

ek aam galti hai: asal mein

U = - 2 K

Recap

Summary aur key points

Kepler ke laws: orbits ellipses hain jinke ek focus par Sun hai; Sun–planet line barabar time mein barabar areas sweep karti hai; $T^{2} \propto a^{3}$ (Section 7.2).
Law of areas angular momentum conservation hai, $d A / d t = L / 2 m$ , aur yeh kisi bhi central force ke liye sahi hai (Derivation).
Newton ka law: do particles ko jodne wali line ke along $F = G m_{1} m_{2} / r^{2}$ ; kai bodies ke forces vectors ki tarah judte hain (Section 7.3).
Uniform shell bahar ke particle ko aise kheenchta hai jaise uska mass centre par ho aur andar koi force nahi lagata; gravity ko shield nahi kiya ja sakta (Theorem).
Cavendish ke torsion balance se $G = 6.67 \times 10^{- 11} N m^{2} k g^{- 2}$ mila (Section 7.4).
Surface par $g = G M_{E} / R_{E}^{2}$ ; uniform Earth ke andar kheench $r$ ke proportional badhti hai (Section 7.5).
$h ≪ R_{E}$ ke liye $g (h) \approx g (1 - 2 h / R_{E})$ aur $g (d) = g (1 - d / R_{E});$ $g$ surface par sabse zyada hai (Section 7.6).
Infinity par $U = 0$ ke saath $U = - G m_{1} m_{2} / r$ ; $V = - G M / r$ ; $m g h$ surface ke paas ka approximation hai; kai bodies ke liye saare pairs jodo (Section 7.7).
Escape speed $v_{e} = \sqrt{2 G M_{E} / R_{E}} = \sqrt{2 g R_{E}} \approx 11.2 k m / s,$ har mass aur direction ke liye same (Section 7.8).
Circular orbits: $v = \sqrt{G M_{E} / r}$ , $T^{2} = (4 π^{2} / G M_{E}) r^{3};$ surface ke paas $v_{o} \approx 7.9 k m / s$ aur $T_{0} \approx 84 m i n$ (Section 7.9).
Orbit mein astronauts weightless isliye mehsoos karte hain kyunki woh free fall mein hain, isliye nahi ki gravity nahi hai (Tairna hi girna hai).
Satellite energy: $K = G M_{E} m / 2 r$ , $U = - G M_{E} m / r$ , $E = - G M_{E} m / 2 r$ (ya ellipse ke liye $- G M_{E} m / 2 a$ ); bound orbits mein $E < 0$ (Section 7.10).

Table 7.3Gravitational quantities ek nazar meinNCERT p. 140

Quantity	Symbol	Expression	SI unit	Dimensions
Gravitational constant	$G$	$6.67 \times 10^{- 11}$	$N m^{2} k g^{- 2}$	$[M^{- 1} L^{3} T^{- 2}]$
Gravitational potential energy	$U$	$- G m_{1} m_{2} / r$	$J$	$[M L^{2} T^{- 2}]$
Gravitational potential	$V$	$- G M / r$	$J k g^{- 1}$	$[L^{2} T^{- 2}]$
Gravitational field (intensity)	$𝐠$	$- (G M / r^{2}) \hat{𝐫}$	$N k g^{- 1}$	$[L T^{- 2}]$
Escape speed	$v_{e}$	$\sqrt{2 G M / R}$	$m s^{- 1}$	$[L T^{- 1}]$
Orbital speed	$v$	$\sqrt{G M / r}$	$m s^{- 1}$	$[L T^{- 1}]$

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Exercise 7.1	Practice Q1
NCERT Exercise 7.2	Practice Q2
NCERT Exercise 7.3	Practice Q3
NCERT Exercise 7.4	Practice Q4
NCERT Exercise 7.5	Practice Q5
NCERT Exercise 7.6	Practice Q6
NCERT Exercise 7.7	Practice Q7
NCERT Exercise 7.8	Practice Q8
NCERT Exercise 7.9	Practice Q9
NCERT Exercise 7.10	Practice Q10
NCERT Exercise 7.11	Practice Q11
NCERT Exercise 7.12	Practice Q12
NCERT Exercise 7.13	Practice Q13
NCERT Exercise 7.14	Practice Q14
NCERT Exercise 7.15	Practice Q15
NCERT Exercise 7.16	Practice Q16
NCERT Exercise 7.17	Practice Q17
NCERT Exercise 7.18	Practice Q18
NCERT Exercise 7.19	Practice Q19
NCERT Exercise 7.20	Practice Q20
NCERT Exercise 7.21	Practice Q21

Quick check: MCQs

7.1 Introduction2 questions
7.2 Kepler's Laws3 questions
7.3 Universal Law of Gravitation4 questions
7.4 The Gravitational Constant2 questions
7.5 Earth ka Acceleration due to Gravity2 questions
7.6 Earth ke Surface ke neeche aur upar Acceleration due to Gravity3 questions
7.7 Gravitational Potential Energy3 questions
7.8 Escape Speed3 questions
7.9 Earth Satellites3 questions
7.10 Orbit mein ghoomte Satellite ki Energy3 questions

Glossary · 14

Key terms (English + Hindi)

acceleration due to gravityगुरुत्वीय त्वरणgurutviya tvaran: Free fall karti body ka acceleration $g$ ; Earth ke surface par $g = G M_{E} / R_{E}^{2} \approx 9.8 m / s^{2} .$ Chapter mein pehli baar yahan aaya →

binding energyबंधन ऊर्जाbandhan oorja: Ek bound body ko poori tarah azaad karne ke liye zaroori energy; circular orbit mein satellite ke liye yeh $G M m / 2 r$ hai.Chapter mein pehli baar yahan aaya →

central forceकेंद्रीय बलkendriya bal: Aisa force jo hamesha body ko ek fixed centre se jodne wali line ke along point karta hai.Chapter mein pehli baar yahan aaya →

ellipseदीर्घवृत्तdeerghvritt: Ek band curve jiske saare points ki do fixed points, yaani foci, se distances ka jod same hota hai.Chapter mein pehli baar yahan aaya →
escape speedपलायन चालpalayan chaal: Sabse kam launch speed jisse koi body planet ke surface ko chhod kar kabhi wapas na aaye, $v_{e} = \sqrt{2 G M / R} .$ Chapter mein pehli baar yahan aaya →

gravitationगुरुत्वाकर्षणgurutvakarshan: Woh attraction jo har mass baaki har mass par lagata hai.Chapter mein pehli baar yahan aaya →
gravitational constantगुरुत्वीय नियतांकgurutviya niyatank: Newton ke law of gravitation mein constant $G = 6.67 \times 10^{- 11} N m^{2} k g^{- 2}$ ; yeh har jagah same hai.
gravitational potentialगुरुत्वीय विभवgurutviya vibhav: Kisi point par per unit mass gravitational potential energy, mass $M$ ke paas $V = - G M / r$ , $J / k g$ mein.Chapter mein pehli baar yahan aaya →
gravitational potential energyगुरुत्वीय स्थितिज ऊर्जाgurutviya sthitij oorja: Masses ko infinite separation se dheere-dheere paas laane mein kiya gaya work; do masses ke liye $U = - G m_{1} m_{2} / r$ .Chapter mein pehli baar yahan aaya →

orbital speedकक्षीय चालkakshiya chaal: $r$ radius ke circular orbit mein satellite ki speed, $v = \sqrt{G M / r}$ .Chapter mein pehli baar yahan aaya →

perihelionउपसौरupsaur: Planet ke orbit ka Sun ke sabse paas wala point; sabse door wala point aphelion hai.Chapter mein pehli baar yahan aaya →

satelliteउपग्रहupagrah: Aisi body jo kisi badi body ke chaaron taraf ghoomti hai, jaise Moon ya Earth ke chaaron taraf artificial satellite.Chapter mein pehli baar yahan aaya →
semi-major axisअर्ध-दीर्घ अक्षardh-deergh aksh: Ellipse ke sabse lambe diameter ka aadha, jise $a$ likhte hain; orbit ke liye yeh sabse paas aur sabse door ki distances ka average hai.Chapter mein pehli baar yahan aaya →

weightlessnessभारहीनताbharheenta: Free fall karti body ki state, jo koi support force mehsoos nahi karti, jabki gravity us par abhi bhi lagti hai.Chapter mein pehli baar yahan aaya →

Gravitation: Class 11 Physics Chapter 7 Hinglish mein

Is chapter mein aap kya seekhenge

7.1 Introduction

7.2 Kepler's Laws

7.3 Universal Law of Gravitation

7.4 The Gravitational Constant

7.5 Earth ka Acceleration due to Gravity

7.6 Earth ke Surface ke neeche aur upar Acceleration due to Gravity

7.7 Gravitational Potential Energy

7.8 Escape Speed

7.9 Earth Satellites

7.10 Orbit mein ghoomte Satellite ki Energy

Summary aur key points

Practice questions answers ke saath (NCERT “Exercises”)

Apna NCERT exercise question dhoondo

Quick check: MCQs

Key terms (English + Hindi)

Tum chapter ke end tak pahunch gaye