Class 11 Physics · Chapter 7
Short answer:
Class 11 Physics Chapter 7 Gravitation NCERT Solutions — Newton ke universal law se lekar escape velocity aur satellite energy tak, is chapter me saara motion "force ki wajah se" explain hota hai. Ye chapter gravitation class 11 numericals with solutions dhundhne walon ke liye especially important hai kyunki yahan formula-based numericals CBSE me regular aate hain. Neeche saare in-text examples aur exercises step-by-step, poori working ke saath solved hain.
Chapter 6 (System of Particles and Rotational Motion) me humne dekha ki ek rigid body kaise ghoomti hai. Chapter 7 Gravitation ab ek level upar jaata hai — ye batata hai ki Moon Earth ke around kyun ghoomta hai, satellite space me kyun girta nahi, aur astronaut weightless kyun feel karta hai. Chapter 4-5 (Laws of Motion) ka force concept yahan ek specific force — gravitational force — pe apply hota hai, isliye is chapter ko theek se samajhna aage Mechanical Properties of Solids/Fluids aur baad me Oscillations jaise chapters ke liye bhi foundation banata hai.
Agar tum poore class 11 physics ncert solutions series follow kar rahe ho, to is chapter ka sequence important hai: Units and Measurements (Ch.1) → Motion in a Straight Line (Ch.2) → Motion in a Plane (Ch.3) → Laws of Motion (Ch.4) → Work, Energy and Power (Ch.5) → System of Particles and Rotational Motion (Ch.6) → Gravitation (Ch.7). Rationalised 2026-27 syllabus me is chapter se geostationary satellite ka detailed derivation part trim ho gaya hai — concept ka basic idea aur definition syllabus me hai, lekin heavy derivation nahi. Poore class 11 physics all chapters pdf 2026-27 session ke liye official ncert.nic.in se hi latest version download karo, kyunki trimmed portions edition ke saath change ho sakte hain.
Neeche har in-text example aur end-of-chapter exercise full working ke saath solved hai — direct copy karne ke bajaye pehle khud try karo, phir match karo.
Chapter 7 Summary — 5 Minute Revision
Is chapter me humne seekha ki Newton's law of gravitation ke mutabik do masses ke beech force unke masses ke product ke directly proportional aur beech ki distance ke square ke inversely proportional hoti hai (F = Gm₁m₂/r²). Kepler ke teen laws planetary motion ko describe karte hain — elliptical orbits, equal areas in equal times, aur T² ∝ r³. Acceleration due to gravity (g) height ke saath ghatta hai (g' = g(1−2h/R) jab h≪R) aur depth ke saath bhi ghatta hai (g' = g(1−d/R)), aur center of earth pe zero ho jaata hai. Gravitational potential energy hamesha negative hoti hai (U = −GMm/r) kyunki reference infinity pe zero liya jaata hai. Escape speed (v_e = √(2GM/R)) woh minimum speed hai jisse koi object gravity se permanently bach sakta hai — ye v_e = √2 × orbital velocity hota hai. Satellite ki total energy (E = −GMm/2r) negative hoti hai jab tak woh bound orbit me hai, aur astronaut weightless feel karta hai kyunki woh continuously "free fall" me hota hai, gravity absent hone ki wajah se nahi.
In-Text Questions — Solutions
Example 7.1: Do spheres jinka mass 40 kg aur 60 kg hai, 5 m door rakhe hain. Unke beech gravitational force of attraction nikaalo.
Newton's law of gravitation se:
F = Gm₁m₂/r²
Given: m₁ = 40 kg, m₂ = 60 kg, r = 5 m, G = 6.67×10⁻¹¹ N m² kg⁻²
F = (6.67×10⁻¹¹ × 40 × 60) / (5)²
F = (6.67×10⁻¹¹ × 2400) / 25 = 6.4×10⁻⁹ N
Ye force bahut chhota hai kyunki G ki value bahut chhoti hai — everyday objects ke beech gravitational force practically negligible hoti hai.
Example 7.2: Agar Earth ka radius R = 6.4×10⁶ m aur surface pe g = 9.8 m/s² hai, to Earth ka mass nikaalo.
g = GM/R² se, M = gR²/G
M = (9.8 × (6.4×10⁶)²) / (6.67×10⁻¹¹)
M = (9.8 × 4.096×10¹³) / (6.67×10⁻¹¹) = (4.014×10¹⁴)/(6.67×10⁻¹¹)
M ≈ 6.02×10²⁴ kg
Ye Earth ke actual mass (~6×10²⁴ kg) ke bahut close hai — isi tarah g aur R janke kisi bhi planet ka mass nikala ja sakta hai.
Example 7.3: Earth ki surface se h = R (yaani Earth ke radius jitni) height pe g ki value nikaalo.
Height pe g ka exact formula:
g' = g [R/(R+h)]²
Given h = R, isliye R + h = 2R:
g' = g[R/2R]² = g × (1/4) = 9.8/4 = 2.45 m/s²
Yaani Earth ke radius jitni height pe jaake g apni surface value ka sirf ¼ reh jaata hai.
Example 7.4: Earth ki surface se d = R/4 depth pe g ki value nikaalo.
Depth pe g ka formula:
g' = g(1 − d/R)
Given d = R/4:
g' = g(1 − 1/4) = (3/4) g = 0.75 × 9.8 = 7.35 m/s²
Depth ke saath g linearly ghatti hai (height wale non-linear formula ke opposite), aur center of earth (d = R) pe g' = 0 ho jaati hai.
Example 7.5: Earth ki surface se escape speed nikaalo (R = 6.4×10⁶ m, g = 9.8 m/s²).
Escape speed ka formula:
ve = √(2gR)
ve = √(2 × 9.8 × 6.4×10⁶) = √(1.2544×10⁸)
ve ≈ 11200 m/s = 11.2 km/s
Iska matlab hai koi bhi object jo 11.2 km/s se upar ki speed se launch hota hai, woh Earth ki gravity se permanently escape kar sakta hai — chahe uska mass kuch bhi ho.
Example 7.6: Earth ki surface se h = 400 km height pe orbit karte satellite ki orbital velocity aur time period nikaalo.
Orbit radius: r = R + h = 6.4×10⁶ + 4×10⁵ = 6.8×10⁶ m
GM = gR² = 9.8 × (6.4×10⁶)² = 4.014×10¹⁴
vo = √(GM/r) = √(4.014×10¹⁴ / 6.8×10⁶) = √(5.90×10⁷)
vo ≈ 7683 m/s ≈ 7.68 km/s
Time period:
T = 2πr/vo = (2 × 3.1416 × 6.8×10⁶) / 7683 ≈ 5561 s ≈ 92.7 min
Ye numbers real Low Earth Orbit satellites (jaise ISS) ke actual values ke bahut close hain.
Poore Class 11 Physics ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q18)
7.1 Gravitational force sirf attractive hoti hai, electrostatic force attractive ya repulsive dono ho sakti hai. Iska reason kya hai — briefly explain.
Gravitational force sirf mass ke beech lagti hai, aur mass hamesha positive quantity hoti hai (negative mass exist nahi karta), isliye Gm₁m₂ hamesha positive hi aayega aur force sirf attractive ho sakti hai. Electrostatic force charge ke beech lagti hai, aur charge positive ya negative dono ho sakta hai — same charges repel karte hain, opposite charges attract karte hain, isliye electrostatic force dono directions me possible hai.
7.2 Kepler ke teeno laws of planetary motion state karo.
Law of Orbits: Har planet Sun ke around ek elliptical orbit me ghoomta hai, jisme Sun ek focus pe hota hai.
Law of Areas: Planet aur Sun ko jodne wali line (radius vector) equal time intervals me equal areas sweep karti hai — iska matlab planet Sun ke paas tez aur door dheere chalta hai.
Law of Periods: Kisi planet ke orbital period (T) ka square, uske orbit ke semi-major axis (r) ke cube ke directly proportional hota hai: T² ∝ r³.
7.3 Earth (M = 6×10²⁴ kg) aur Moon (m = 7.4×10²² kg) ke beech gravitational force nikaalo, jab unke centres ke beech average distance 3.84×10⁵ km hai.
r = 3.84×10⁵ km = 3.84×10⁸ m
F = GMm/r² = (6.67×10⁻¹¹ × 6×10²⁴ × 7.4×10²²) / (3.84×10⁸)²
Numerator = 6.67×10⁻¹¹ × 4.44×10⁴⁷ = 2.96×10³⁷
Denominator = 1.475×10¹⁷
F = 2.96×10³⁷ / 1.475×10¹⁷ ≈ 2.01×10²⁰ N
7.4 Moon ki surface pe g, Earth ki surface pe g ka 1/6 hai, aur Moon ka radius Earth ke radius ka 1/4 hai. Moon aur Earth ke mean densities ka ratio nikaalo.
g = GM/R² aur M = (4/3)πR³ρ, isliye g = (4/3)πGRρ, yaani g ∝ Rρ.
g_m/g_e = (R_m ρ_m)/(R_e ρ_e)
Given g_m/g_e = 1/6 aur R_m/R_e = 1/4:
1/6 = (1/4) × (ρ_m/ρ_e)
ρ_m/ρ_e = 4/6 = 2/3
Moon ki mean density Earth ki mean density ki 2/3 hai.
7.5 Earth ki surface pe ek body ka weight 63 N hai. Iski weight Earth ki radius ki half height (h = R/2) pe kya hogi?
g' = g[R/(R+h)]², h = R/2 to R+h = 3R/2:
g'/g = [R/(3R/2)]² = (2/3)² = 4/9
W' = W × (4/9) = 63 × 4/9 = 28 N
7.6 Earth ki surface pe ek body ka weight 250 N hai. Iski weight Earth ki radius ki half depth (d = R/2) pe kya hogi?
g' = g(1 − d/R), d = R/2:
g' = g(1 − 1/2) = g/2
W' = W × (1/2) = 250/2 = 125 N
7.7 Ek kalpanik planet ka mass Earth ke mass ka double hai, lekin radius Earth ke radius jitna hi hai. Us planet se escape speed nikaalo (Earth ki escape speed 11.2 km/s leke).
ve = √(2GM/R). Agar M double ho jaata hai aur R same rehta hai, to v_e, √M ke proportion me badhti hai — yaani √2 factor se:
ve' = 11.2 × √2 ≈ 11.2 × 1.414 ≈ 15.84 km/s
7.8 Ek satellite Earth ki surface se h = R height pe orbit kar raha hai. Iski orbital velocity aur time period nikaalo (R = 6.4×10⁶ m, GM = 4.014×10¹⁴).
r = R + h = 2R = 1.28×10⁷ m
vo = √(GM/r) = √(4.014×10¹⁴/1.28×10⁷) = √(3.14×10⁷) ≈ 5600 m/s = 5.6 km/s
T = 2πr/vo = (2×3.1416×1.28×10⁷)/5600 ≈ 14361 s ≈ 3.99 hr
7.9 Geostationary satellite kya hota hai? Isse stationary appear karne ke liye kya conditions honi chahiye? (Detailed derivation ki jagah sirf concept explain karo.)
Geostationary satellite woh satellite hota hai jo Earth ke equatorial plane me, Earth ke rotation ke exact same direction aur same angular speed (yaani 24-ghante ka period) se ghoomta hai — isliye Earth ki surface se dekhne pe woh stationary dikhta hai. Iske liye zaroori conditions: (i) satellite ka orbit Earth ke equatorial plane me ho, (ii) uska time period Earth ke rotation period (~24 hours) ke barabar ho, aur (iii) orbit circular ho, ek fixed height pe (jisse T² ∝ r³ satisfy ho).
7.10 Earth ki surface se h = R height pe, 10 kg mass ki body ki gravitational potential energy nikaalo (GM = 4.014×10¹⁴, R = 6.4×10⁶ m).
r = R + h = 2R = 1.28×10⁷ m
U = −GMm/r = −(4.014×10¹⁴ × 10)/(1.28×10⁷)
U = −4.014×10¹⁵/1.28×10⁷ ≈ −3.14×10⁸ J
7.11 h = R height pe orbit karte 200 kg mass ke satellite ki total energy aur binding energy nikaalo.
r = 2R = 1.28×10⁷ m
E = −GMm/2r = −(4.014×10¹⁴ × 200)/(2×1.28×10⁷)
E = −8.028×10¹⁶/2.56×10⁷ ≈ −3.14×10⁹ J
Binding energy = −E = +3.14×10⁹ J (yaani itni energy satellite ko orbit se infinity tak escape karne ke liye chahiye).
7.12 Earth ke bilkul center pe kisi body ka weight zero kyun hota hai?
Depth d pe g' = g(1 − d/R). Earth ke center pe d = R, isliye g' = g(1 − R/R) = g(1 − 1) = 0. Physically, center pe body ke chaaron taraf equal Earth-mass symmetrically distributed hota hai, isliye saari directions ki gravitational pulls cancel ho jaati hain aur net force (aur isliye weight) zero ho jaata hai.
7.13 Height h (jahan h ≪ R) pe g ki variation ka approximate expression derive karo, exact formula se shuru karke.
Exact formula:
g' = g[R/(R+h)]² = g[1 + h/R]⁻²
Binomial expansion (h ≪ R ke liye, higher order terms negligible):
g' ≈ g[1 − 2h/R]
Yaani chhoti heights ke liye g linearly, approximately 2h/R fraction se ghatta hai.
7.14 Space station me orbit karta astronaut weightless kyun feel karta hai, jabki uss height pe gravity almost surface jitni hi strong hoti hai?
Astronaut weightless isliye feel karta hai kyunki woh aur poora space station dono Earth ke gravity ke under continuously 'free fall' me hote hain — dono ka acceleration same hota hai, isliye astronaut ka body space station ki floor pe koi normal force nahi lagata. Weight ka feeling actually normal reaction force ki wajah se hoti hai (jaise lift me), gravity ki absence ki wajah se nahi — gravity to us height pe bhi kaafi strong hoti hai (surface value ka ~90%), lekin free-fall condition me contact force zero ho jaata hai.
7.15 Free space me 1 kg mass ke do point objects 1 m door rakhe hain. Unke system ki gravitational potential energy nikaalo.
U = −Gm₁m₂/r = −(6.67×10⁻¹¹ × 1 × 1)/1
U = −6.67×10⁻¹¹ J
7.16 Ek planet ka mass Earth ke mass ka 1/10 hai aur radius Earth ke radius ka 1/2 hai. Us planet ki surface pe g ki value nikaalo.
gp = GMp/Rp² = G(Me/10)/(Re/2)²
gp = (GMe/10) / (Re²/4) = (4/10) × (GMe/Re²) = 0.4 ge
gp = 0.4 × 9.8 = 3.92 m/s²
7.17 Ek rocket Earth ki surface se exactly escape speed se vertically fire kiya jaata hai. Air resistance aur Earth ki rotation ignore karte hue, uski speed nikaalo jab woh Earth ke center se 2R distance pe ho.
Energy conservation lagate hue (total mechanical energy launch pe zero hoti hai, kyunki escape speed pe body infinity tak bilkul ruk ke pahunchti hai):
½ve² − GM/R = ½v² − GM/(2R)
Left side zero hai kyunki ve² = 2GM/R, to ½(2GM/R) − GM/R = 0:
0 = ½v² − GM/(2R) ⟹ v² = GM/R = gR
v = √(gR) = √(9.8 × 6.4×10⁶) ≈ 7920 m/s ≈ 7.92 km/s
7.18 Acceleration due to gravity (g) aur Universal gravitational constant (G) me difference batao.
G ek universal constant hai (6.67×10⁻¹¹ N m² kg⁻²), jo poore universe me sabhi jagah same rehta hai, aur do masses ke beech force define karta hai. g ek variable quantity hai (Earth ki surface pe ~9.8 m/s²), jo location (height, depth, latitude) aur planet ke mass/radius ke saath change hoti hai — g = GM/R² formula se G aur g connected hote hain, lekin dono alag physical quantities hain jinki units bhi different hain (N m² kg⁻² vs m/s²).
Important Equations — Ek Nazar Me
| Concept | Formula | Notes |
|---|---|---|
| Newton's Law of Gravitation | F = Gm1m2/r² | G = 6.67×10-11 N m² kg-2, hamesha attractive |
| Kepler's Third Law | T² ∝ r³ | Elliptical orbit ke semi-major axis pe apply hota hai |
| g on Earth's surface | g = GM/R² | M, R = Earth ka mass aur radius |
| g at height h (h ≪ R) | g' = g(1 − 2h/R) | Exact form: g' = g[R/(R+h)]² |
| g at depth d | g' = g(1 − d/R) | d = R pe g' = 0 (Earth ka center) |
| Gravitational Potential Energy | U = −GMm/r | Reference: U = 0 at r = ∞ |
| Escape Speed | ve = √(2GM/R) = √(2gR) | Earth ke liye ≈ 11.2 km/s; mass-independent |
| Orbital Velocity | vo = √(GM/r) | r = R + h (satellite ka orbit radius) |
| Relation between ve and vo | ve = √2 · vo | Surface pe orbit maan ke |
| Time Period of Satellite | T = 2π√(r³/GM) | Kepler's third law ka direct application |
| Total Energy of Satellite | E = −GMm/2r | Negative = bound orbit; Binding energy = +GMm/2r |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- G (universal gravitational constant, 6.67×10⁻¹¹ N m² kg⁻²) aur g (acceleration due to gravity, 9.8 m/s²) ko confuse karna — dono alag-alag physical quantities hain, units bhi alag.
- Gravitational PE ke formula U = −GMm/r me negative sign bhool jaana — students isse mgh wale positive-PE-jaisa treat kar dete hain jo galat hai.
- Approximation g' = g(1 − 2h/R) ko bade heights (jaise h comparable to R) pe bhi use kar dena — ye formula sirf h ≪ R ke liye valid hai, warna exact formula g[R/(R+h)]² use karo.
- Escape velocity object ke mass pe depend karti hai — ye galat dhaarna hai. v_e sirf planet ke M aur R pe depend karti hai, escaping body ke mass ya launch direction pe nahi.
- Escape velocity aur orbital velocity ke formulas ko mix kar dena — yaad rakho v_e = √2 × v_o, factor of √2 bhoolna common mistake hai.
- Satellite ka orbit radius 'r' likhte waqt sirf height 'h' le lena, jabki r = R (earth radius) + h hota hai — ye unit conversion se bhi zyada common calculation error hai.
Board-Style Important Questions
- Kepler ke teenon laws of planetary motion state karo.
- Earth ki surface se kisi body ki escape speed ke liye expression derive karo.
- Height ke saath acceleration due to gravity (g) ki variation ka expression derive karo, jab h ≪ R ho.
- Ek two-body system ki gravitational potential energy ka expression derive karo, aur isse ek orbiting satellite ki total energy ka expression bhi obtain karo.
- Orbit karte spacecraft me astronaut weightless kyun feel karta hai? Free fall ke concept se explain karo.
- Ek satellite kisi planet ki surface ke paas orbit kar raha hai. Dikhao ki uski orbital velocity satellite ke mass pe depend nahi karti, aur iska expression g aur R ke terms me derive karo.
Aksar Poochhe Jaane Wale Sawaal
Class 11 Physics Chapter 7 Gravitation me kitne numericals aate hain aur weightage kitni hai?
Exact marks-weightage saal-dar-saal aur board ke exam pattern ke hisaab se change hoti hai, isliye koi fixed number claim karna sahi nahi hoga — CBSE apna official sample paper/marking scheme release karta hai jisme har chapter ki weightage batayi jaati hai, wahi check karo. Jo fix hai woh ye ki is chapter se numericals (escape velocity, orbital velocity, g ki variation) bahut consistently poochhe jaate hain, isliye formula table achhe se yaad rakhna zaroori hai.
Class 11 Physics ke saare chapters ka PDF 2026-27 session ke liye kahan milega?
Sabse reliable source ncert.nic.in ka official textbook portal hai — wahan se rationalised 14-chapter Physics syllabus (Units and Measurements se shuru hoke) ka latest PDF download karo. Third-party sites purana pre-2023 syllabus bhi carry karte hain, isliye chapter count (14 vs purane 15-16) match karke hi confirm karo ki tumhara PDF current session ka hai.
2026-27 ke deleted syllabus me Gravitation chapter se kya hata hai?
Poora chapter syllabus me hai, hataya nahi gaya — sirf geostationary satellite wala detailed derivation portion trim kiya gaya hai. Basic definition aur concept (ki geostationary satellite Earth ke rotation ke saath sync me ghoomta hai) syllabus me reh sakta hai, lekin full mathematical derivation ab expected nahi hai. Exact trimmed lines ke liye apni current NCERT copy ka table of contents official PDF se cross-check karo.
Gravitation ke numericals solve karne ka best approach kya hai?
Pehle identify karo kaunsa formula apply hoga (force, g-variation, escape/orbital velocity, ya energy), phir saari given values ko SI units me convert karo (especially km ko m me), aur satellite wale questions me hamesha r = R + h use karo, sirf h nahi. Step-by-step working dikhana marking scheme ke liye zaroori hai, sirf final answer likhna kaafi nahi.
Gravitation chapter aage kaunse chapters ke liye foundation banata hai?
Escape velocity aur PE ka energy-conservation approach Work-Energy chapter se seedha connect hota hai, aur circular-orbit ka concept Oscillations (SHM) chapter me pendulum-type problems samajhne me help karta hai. Isliye Mechanical Properties of Fluids aur Thermodynamics jaise agle chapters shuru karne se pehle is chapter ke formulas solid honi chahiye.
Class 11 Physics ke aur chapters ke NCERT solutions kahan practice karein?
Isi tarah ka structured practice tum Motion in a Straight Line (Chapter 2), Laws of Motion, System of Particles and Rotational Motion jaise chapters ke liye bhi kar sakte ho — har chapter ke important questions aur exemplar problems alag se practice karo, especially Units and Measurements ke important questions jo foundational hain poore mechanics section ke liye.
Class 11 Physics — Saare Chapters
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