NCERT Plus One (Class 11) Physics • Chapter 7

Gravitation

Kepler's three laws, Newton's universal gravitation & Cavendish G, variation of g with altitude & depth, potential energy, escape speed & satellites.

📄 Official NCERT Textbook PDF

Topic 7.2: Kepler's Three Laws of Planetary Motion

Imagine swinging a coconut-shell toy on a string in an oval loop rather than a perfect circle! Centuries before satellites, Johannes Kepler cracked Tycho Brahe's night-sky charts and discovered that every planet secretly dances around the Sun along a stretched oval (Ellipse), racing faster when closer to the Sun!
1. LAW OF ORBITSPlanets orbit in ellipses with the Sun sitting at one focus
2. LAW OF AREASΔA/Δt = L/(2m) = constant (Consequence of angular momentum conservation!)
3. LAW OF PERIODST² ∝ a³ (Square of year ∝ Cube of semi-major axis a)
🧠 Memory Trick: O-A-P = Orbits (Ellipse), Areas (Equal ΔA/Δt), Periods (T² ∝ a³)! Near Sun (Perihelion) = fast; Far from Sun (Aphelion) = slow!
💡 Simple Kid-Check: Think of an ice-skater effect in space! When Earth swings closer to the Sun (January perihelion), it speeds up so the shorter radius vector sweeps the same slice of area per month as the longer slow sweep in July (aphelion)!
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    First Law (Law of Orbits): All planets move in elliptical orbits with the Sun situated at one of the two foci (semi-major axis a = (rmin + rmax)/2).
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    Second Law (Law of Areas): The line joining any planet to the Sun sweeps equal areas in equal intervals of time: ΔAΔt = L2m = constant. Why? Gravity acts straight toward the Sun, so torque τ = r × F = 0 and angular momentum L is conserved!
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    Third Law (Law of Periods): Square of revolution time period is proportional to cube of semi-major axis: T² = (4π²GMs) a³ ⇒ (T₁/T₂)² = (a₁/a₂)³!
🧒 Super-Easy Example: If a planet orbits 4 times farther from the Sun than Earth (a₂ = 4 a₁), its year takes T₂ = 43/2 = 8 Earth years!
Classic 2-mark Kerala HSE question: Kepler's Second Law (equal areas in equal times) proves conservation of which quantity? Answer: Angular Momentum L — because central gravitational force exerts ZERO torque about the Sun!

Topic 7.3 – 7.4: Newton's Universal Law & Cavendish Experiment

Seeing an apple drop straight onto English grass, Isaac Newton wondered: if Earth pulls that apple, could that EXACT same pull stretch all the way up to keep the Moon curving around Earth instead of flying off into space? That leap united backyard gravity and astronomy in one neat formula!
UNIVERSAL GRAVITATIONF = G m₁ m₂r² (Always attractive, central & inverse-square!)
CAVENDISH CONSTANT GG = 6.67 × 10⁻¹¹ N·m²/kg² • Dim: [M⁻¹L³T⁻²]
SHELL THEOREMOutside hollow shell: acts as mass at centre. Inside cavity: Force = ZERO!
🧠 Memory Trick: G is Universal (NEVER changes anywhere!), whereas g is Local (changes by planet/height)!
💡 Simple Kid-Check: Two schoolfriends sitting beside each other DO pull each other with gravity — yet Cavendish proved G = 0.0000000000667 is so unimaginably tiny that you need planet-sized mass to notice the tug!
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    Inverse-Square Vector Law: Every pair of point masses attract along the line joining them: F₁₂ = −G m₁m₂r² r̂₂₁. Force forms an equal & opposite Newton's 3rd Law pair and obeys superposition for multiple masses!
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    Shell Theorem: A uniform spherical shell pulls any outside particle as if all mass sat at the centre, yet pulls ZERO on any particle anywhere inside the hollow shell!
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    Cavendish Torsion Balance (1798): Using tiny lead spheres twisted on a thin quartz fibre, Henry Cavendish measured G = 6.67 × 10⁻¹¹ N·m²/kg² — which instantly let scientists "weigh" the Earth: ME = gRE² / G ≈ 6 × 10²⁴ kg!
🧒 Super-Easy Example: Two satellites attract with force 160 N at separation r. If moved to 2r, force drops by 2² = 4 times → 160 / 4 = 40 N!
Watch three classic traps: (1) Does G change on the Moon or underwater? NO! G is universal and medium-independent (unlike electrostatic Coulomb's constant!). (2) Gravitational force inside a uniform hollow spherical shell is ZERO everywhere inside! (3) Doubling distance cuts force to one-fourth, NOT half.

Topic 7.5 – 7.6: Acceleration due to Gravity g with Altitude & Depth

Climb Mount Everest and you get a tiny bit lighter; dive down a deep Kolar gold mine and you ALSO get lighter! Keep tunnelling all the way down to Earth's exact centre, and every ton of rock pulls you equally from all sides until g = 0 and you float weightless!
SURFACE g & DENSITYg = GMR² = 43 π G R ρ ≈ 9.8 m/s²
ALTITUDE h ABOVEg(h) = g(1 + h/R)²  |  For h ≪ R: g(h) ≈ g(1 − 2hR)
DEPTH d BELOWg(d) = g(1 − dR) (Exact linear drop; at centre d=R ⇒ g=0!)
🧠 Memory Trick: Surface is King! Gravity g peaks right on Earth's surface and drops BOTH when you climb up (−2h/R, 2× faster!) AND dig down (−d/R)!
💡 Simple Kid-Check: Inside Earth at depth d, the outer shell of thickness d above your head pulls ZERO net force (by Shell Theorem!), so only the smaller inner core of radius (R − d) pulls you!
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    Why Heavy & Light Stones Fall Together: Setting mg = GMm/R² cancels the falling object's mass m! So g = GM / R² depends ONLY on the planet's mass & radius.
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    Going UP (Altitude h): Distance from Earth's centre becomes R + h, so g(h) = g R² / (R + h)². For small heights (h ≪ R), binomial expansion gives g(h) ≈ g(1 − 2h/R)! (Note: at h = R, g drops to g/4!).
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    Going DOWN (Depth d): Only inner sphere of radius r = R − d pulls, whose mass shrinks as r³ while distance squared is r² → net pull scales linearly with r! Thus g(d) = g(1 − d/R)! At Earth's centre (d = R), g = 0!
🧒 Super-Easy Comparison: Climbing 1% of Earth's radius up (h = R/100) cuts g by 2%, while digging 1% down (d = R/100) cuts g by only 1%! To match a drop at small altitude h, you must dig twice as deep (d = 2h)!
Top DHSE graph & numerical trap: (1) Students think g → ∞ as you approach Earth's centre (r → 0)! Wrong — inside a solid sphere, g ∝ r rises linearly from 0 at centre to maximum 9.8 m/s² at surface, then falls as 1/r² outside! (2) Remember for small heights, g decreases twice as fast with height h as with depth d (d = 2h gives same g)!

Topic 7.7: Gravitational Potential Energy

Imagine standing at the bottom of a deep dry well in your Palakkad backyard! We call the open ground far outside "zero level", so sitting down inside the well puts you in energy debt — a negative number! To climb back to ground zero, someone has to pull you up by supplying positive energy. Gravity traps planets inside the exact same negative Potential Well!
POTENTIAL ENERGY U(r)U(r) = −GMmr (Reference U(∞) = 0; always negative!)
LARGE HEIGHT WORKΔU = U(R+h) − U(R) = mgh1 + h/R
GRAVITATIONAL POTENTIALV(r) = U/m = −GMr (Scalar; SI unit: J/kg)
🧠 Memory Trick: U = −GMm/r has ONE power of r (unlike Force 1/r²), and Minus means Bound in Earth's Gravity Well!
💡 Simple Kid-Check: Why don't we use mgh for rockets? Because mgh assumes gravity stays 9.8 all the way up! For high rockets, gravity weakens with distance, so lifting a satellite by a whole Earth radius (h = R) costs only half as much energy as mgR!
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    Why U(r) is Negative: Choosing U = 0 at infinite separation (where gravitational pull fades to zero), since gravity is attractive, bringing mass m from ∞ to distance r releases energy, leaving U(r) = −GMm / r!
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    Work Done Lifting from Surface (R) to Height (R + h):
    ΔU = −GMmR + h − (−GMmR) = GMmhR(R+h) = mgh1 + h/R!
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    Connecting Back to mgh: When h ≪ R (like tossing a ball or climbing stairs), h/R ≈ 0 so ΔU smoothly simplifies to mgh!
🧒 Super-Easy Exam Classic: Energy needed to lift mass m from Earth's surface to height h = R:
Plug h = R into mgh / (1 + h/R) → mgR / (1 + 1) = ½ mgR (NOT mgR)!
Guaranteed 2-mark board trap: Lift a body of mass m from Earth's surface to height h = R (or 2R). Students blindly plug mgh = mgR and get zero marks! You MUST use ΔU = mgh / (1 + h/R) → giving ½ mgR for h = R (and ⅔ mgR for h = 2R)!

Topic 7.8: Escape Speed from Earth (and Planets)

Throw a cricket ball straight up and gravity drags it back down. Throw it faster and it flies higher before returning. But fire an ISRO rocket fast enough at Sriharikota so it hits 11.2 km/s, and it climbs completely out of Earth's potential well — never falling back again!
ESCAPE SPEED FORMULAve = √(2GMR) = √(2gR)
EARTH & MOON VALUESEarth: 11.2 km/s • Moon: 2.38 km/s (lost air!)
WHAT ve IGNORESIndependent of projectile mass m & launch angle!
🧠 Memory Trick: Pay Off Your Gravity Debt! Set Kinetic boost ½mve² = +GMm/R (exact debt) ⇒ mass m cancels cleanly out!
💡 Simple Kid-Check: Whether you launch a tiny paperclip or a huge 500-ton spaceship, both need the exact same Escape Speed 11.2 km/s — because projectile mass m cancels from both sides!
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    Energy Conservation Derivation: Launch at minimum speed ve so rocket reaches ∞ with K∞ = 0, U∞ = 0:
    ½ m ve² − GMmR = 0 ⇒ ve = √(2GMR) = √(2gR) (since GM = gR²).
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    Earth & Moon Comparison: Plugging Earth's g = 9.8 m/s², R = 6.4 × 10⁶ m gives ve ≈ 11.2 km/s. Moon's tiny ve ≈ 2.3 km/s is slower than warm gas molecules, so the Moon's air escaped into space!
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    Launching Faster than ve: If thrown with v > ve, remaining speed at infinity comes from energy conservation: ½mv∞² = ½mv² − ½mve² ⇒ v∞ = √(v² − ve²)!
Watch three classic 1-mark traps: (1) Escape speed does NOT depend on mass m of the rocket or the angle of launch (neglecting air drag)! (2) Why is it called escape speed and not velocity? Because direction doesn't matter! (3) If projected at v = 2ve, speed at infinity is √(4ve² − ve²) = √3 ve (NOT ve!).

Topic 7.9 – 7.10: Earth Satellites, Orbital Speed & Total Energy

Look up at the night sky — an ISRO weather satellite gliding above Kerala does NOT stay up because gravity disappeared! At 400 km altitude, gravity still pulls at ~90% of its surface strength! The satellite simply flies sideways so fast (~7.9 km/s) that as it falls toward Earth, Earth curves away beneath it!
ORBITAL SPEED & ESCAPEvo = √(GMR+h) • Near Earth: vo = √(gR) ≈ 7.92 km/s • ve = √2 vo!
TIME PERIOD & KEPLERT = 2π √((R+h)³GM) (Near Earth ≈ 85 min)
GOLDEN ENERGY RATIOK = +GMm2r  |  U = −GMmr  |  E = −GMm2r = −K = U/2
🧠 Memory Trick: E = −K & U = 2E! Check signs: K is positive (+1), E is negative (−1), and U is double negative (−2) — so +1 + (−2) = −1!
💡 Simple Kid-Check: Why do astronauts inside the space station float completely weightless? NOT because gravity is zero — because BOTH the astronaut and the station are falling freely around Earth together at the exact same speed, so the floor exerts zero push (N = 0)!
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    Orbital Speed Derivation: Gravity supplies the exact centripetal force needed for circular orbit of radius r = R + h:
    mvo²r = GMmr² ⇒ vo = √(GMR+h). Right near Earth (h ≪ R), vo = √(gR) ≈ 7.92 km/s, so ve = √2 vo!
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    Orbit Time Period: T = 2πrvo = 2π √(r³GM) (matches Kepler's 3rd law T² ∝ r³!). Close to Earth's surface, T₀ = 2π√(R/g) ≈ 85 minutes.
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    Energy of an Orbiting Satellite: Kinetic K = ½mvo² = +GMm2r; Potential U = −GMmr; Total E = K + U = −GMm2r. Negative total energy means the satellite is bound to Earth!
Two super-sneaky DHSE traps: (1) If an orbiting satellite loses energy from faint air drag, its orbit shrinks (r ↓) — what happens to its orbital speed vo and kinetic energy K? They actually INCREASE (E = −K becomes more negative, so K goes UP while U drops twice as much)! (2) Weightlessness in orbit happens due to free fall (zero normal reaction), NOT zero gravity!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March • 2 Marks): State Kepler's Second and Third Laws of planetary motion. Which physical conservation law underlies the Second Law?

• Law of Areas (2nd): The radius vector from Sun to planet sweeps equal areas in equal intervals of time (ΔA/Δt = constant). It is based on Conservation of Angular Momentum (L = 2m ΔA/Δt, zero central torque).
• Law of Periods (3rd): Square of orbital time period is proportional to cube of semi-major axis: T² ∝ a³.

Q2 (Kerala HSE Say • 1 Marks): Does the universal gravitational constant G depend on the nature of the medium between two masses or on location (Earth vs Moon)? Write its dimensional formula.

No! G = 6.67 × 10⁻¹¹ N·m²/kg² is truly universal and independent of medium or planet.
Dimensional formula: [G] = [F r² / m²] = [MLT⁻²][L²]/[M²] = [M⁻¹ L³ T⁻²].

Q3 (Kerala HSE March • 3 Marks): Derive an expression for acceleration due to gravity at a depth d below the Earth's surface. What happens to g at the centre of the Earth?

Assuming uniform density ρ, surface g = (4/3)π G R ρ.
At depth d (distance r = R − d from centre), the outer shell exerts zero force; only inner sphere of radius (R − d) pulls:
g(d) = (4/3)π G (R − d) ρ = g (1 − d/R).
At Earth's centre (d = R), g(R) = g(1 − 1) = 0 (weightless!).

Q4 (Kerala HSE Model • 3 Marks): Derive the variation of acceleration due to gravity g with altitude h above Earth's surface (for both general h and small h ≪ R).

At height h, g(h) = GM / (R + h)². Dividing by g = GM / R²:
g(h) = g / (1 + h/R)² = g(1 + h/R)⁻².
For h ≪ R, binomial expansion gives g(h) ≈ g (1 − 2h/R).

Q5 (Kerala HSE March • 2 Marks): At what height above Earth's surface is the value of g the same as at a depth of 80 km below the surface? (Assume h, d ≪ R)

Equating small-distance formulas: g(1 − 2h/R) = g(1 − d/R) ⇒ 2h = d.
Thus h = d / 2 = 80 / 2 = 40 km.

Q6 (Kerala HSE Say • 2 Marks): Calculate the work done in lifting a body of mass m from the surface of the Earth (radius R) to a height equal to R.

Using exact potential change ΔU = U(2R) − U(R):
ΔU = −GMm/(2R) − (−GMm/R) = +GMm/(2R).
Since GM = gR², ΔU = ½ mgR!

Q7 (Kerala HSE March • 3 Marks): Define Escape Speed. Derive an expression for escape speed from Earth's surface and calculate its value (g = 9.8 m/s², R = 6.4 × 10⁶ m).

Definition: Minimum launch speed needed to escape a planet's gravitational pull to infinity with zero residual velocity.
By energy conservation: ½mve² − GMm/R = 0 ⇒ ve = √(2GM/R) = √(2gR).
Calculation: ve = √(2 × 9.8 × 6.4 × 10⁶) ≈ 11.2 km/s.

Q8 (Kerala HSE Model • 3 Marks): Derive the expressions for Orbital Speed, Kinetic Energy, Potential Energy, and Total Energy of a satellite of mass m orbiting Earth at radius r = R + h. What does the negative sign in Total Energy signify?

Centripetal force = Gravity: mvo²/r = GMm/r² ⇒ vo = √(GM/r).
• Kinetic: K = ½mvo² = +GMm / (2r)
• Potential: U = −GMm / r
• Total Energy: E = K + U = −GMm / (2r) = −K.
The negative sign signifies that the satellite is gravitationally bound to Earth.