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State Board · Class 11 · Physics

Lesson 5: Gravitation

7 concepts · 56 practice questions with answers · Free to read

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Kepler's laws of planetary motion

From daily life

Earth is slightly closer to the Sun in January and moves faster; in July it is farther and slower. Link to earlier classes : gravitation, g and orbits of classes 9-10 now follow precise laws (Kepler, Newton).

Simple definition

Kepler's laws give three rules about planetary orbits, speeds and periods.

Kepler's lawsST² ∝ a³ ; dA/dt = L ÷ 2m = constantSun at one focus

In the figure : The green parts are equal areas; the Sun is at a focus.

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Kepler's laws of planetary motion

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Rule 1: First law : orbits — Planets move in ellipses with the Sun at one focus.

  • Easy : Earth's orbit is nearly circular.
  • Medium : Nearest point (perihelion) and farthest (aphelion).
  • Exam (CET) : Shape of the orbit? (1) ellipse (2) circle only (3) parabola (4) straight → answer 1)

Rule 2: Second law : areas — The line from Sun to planet sweeps equal areas in equal times : dA/dt = L ÷ 2m = constant ⇒ angular momentum is conserved, v₁ r₁ = v₂ r₂ (perihelion/aphelion).

  • Easy : r_p = 1, r_a = 4 (units), v_p = 20 ⇒ v_a = 5.
  • Medium : Speed is highest at perihelion.
  • Exam (CET) : The second law reflects conservation of? (1) angular momentum (2) energy (3) mass (4) momentum → answer 1)

Rule 3: Third law : periods — T² ∝ a³ (a = semi-major axis); T² = 4π² a³ ÷ GM. T₁² ÷ T₂² = a₁³ ÷ a₂³.

  • Easy : Mars a = 1.52 AU ⇒ T = 1.52^1.5 = 1.87 years.
  • Medium : r four times ⇒ T = 4^1.5 = 8 times.
  • Exam (CET) : T² ∝ ? (1) a³ (2) a² (3) a (4) 1/a → answer 1) Trick : 'T squared, a cubed'.

Derivation : T² = 4π² r³ ÷ GM (circular orbit)

  • Step 1 : Gravity supplies the centripetal force : GMm/r² = m ω² r.
  • Step 2 : ω = 2π/T ⇒ GM/r² = 4π² r ÷ T².
  • Step 3 : T² = 4π² r³ ÷ GM.
  • Step 4 : T² ∝ r³; the constant 4π²/GM depends only on the Sun.
Remember

T² ∝ a³; v₁r₁ = v₂r₂; dA/dt = L/2m.

Common mistake / CET trap

In T² ∝ a³, a is the semi-major axis; speed is lowest at aphelion, highest at perihelion.

Shortcut

Memory clue : "T squared, a cubed".

Spot the difference

LawStatementRelated conservation
Firstellipse, Sun at focus—
Secondequal areas in equal timesangular momentum
ThirdT² ∝ a³—

Practice questions

Easy
  1. 1State Kepler's three laws.
  2. 2What is perihelion?
  3. 3For which bodies is T²/a³ constant?
Medium
  1. 4a₁ = 1, a₂ = 4 : T₂/T₁?
  2. 5r_p = 2, r_a = 6, v_p = 30 : v_a?
  3. 6a = 9 AU : T in years?
Challenging
  1. 7T = 8 years : a in AU?
  2. 8How does the second law relate to angular momentum?
Show answers (try them yourself first)
  1. 1Ellipse; equal areas; T² ∝ a³
  2. 2Point nearest the Sun
  3. 3Planets of the same Sun
  4. 48
  5. 5v_a = 10
  6. 627 years
  7. 7a = 4 AU
  8. 8dA/dt = L/2m; constant L ⇒ constant areal velocity

Newton's law of gravitation

From daily life

An apple falls and the Moon circles the Earth — one force explains both.

Simple definition

Gravitation is the mutual attractive force between all bodies.

Law of gravitationm₁m₂rF = G m₁ m₂ ÷ r²G = 6.67 × 10⁻¹¹ N m² kg⁻²

In the figure : F = G m₁m₂/r²; the two forces are equal.

Rule 1: Newton's law — F = G m₁ m₂ ÷ r², attractive, equal and opposite on both bodies. G = 6.67 × 10⁻¹¹ N m² kg⁻²; dimension [M⁻¹ L³ T⁻²].

  • Easy : 1 kg and 1 kg, 1 m apart ⇒ F = 6.67 × 10⁻¹¹ N.
  • Medium : 100 kg and 50 kg, 2 m apart ⇒ F = 8.34 × 10⁻⁸ N.
  • Exam (CET) : r doubles ⇒ F? (1) ¼ (2) half (3) double (4) four times → answer 1) Trick : inverse square.

Rule 2: Properties — A central, conservative force; acts between all bodies; independent of the medium; the weakest fundamental force. Superposition : total F is the vector sum.

  • Easy : Forces along one line : add/subtract.
  • Medium : Three equal m at the corners of an equilateral triangle : each feels F = √3 Gm²/a².
  • Exam (CET) : The value of G depends on? (1) none, constant (2) medium (3) temperature (4) m → answer 1)

Rule 3: Comparison — For an electron-proton pair gravity is about 10⁻⁴⁰ of the electric force. It dominates for astronomical bodies.

  • Easy : Earth-Moon : F = G M_E M_m / r² = 2.0 × 10²⁰ N.
  • Medium : The force on both bodies is equal.
  • Exam (CET) : Dominant force at the atomic level? (1) electric (2) gravity (3) equal (4) none → answer 1)

Derivation : Inverse square law from Kepler

  • Step 1 : Circular orbit : F = m v² ÷ r, v = 2πr/T.
  • Step 2 : F = 4π² m r ÷ T².
  • Step 3 : Kepler 3 : T² = k r³ ⇒ F = 4π² m ÷ (k r²).
  • Step 4 : F ∝ m ÷ r² ⇒ F = G m M ÷ r².
Remember

F = G m₁m₂/r²; G = 6.67 × 10⁻¹¹; F ∝ 1/r².

Common mistake / CET trap

G (universal) and g (local) differ; r is the distance between centres.

Shortcut

Memory clue : "G m m over r squared".

Spot the difference

ForceAttractive/repulsiveRelative strength
Gravityattractiveweakest
Electricbothstrong
Nuclear—strongest

Practice questions

Easy
  1. 1Write Newton's law of gravitation.
  2. 2Unit of G?
  3. 3r doubled ⇒ F?
Medium
  1. 4m₁ = m₂ = 10 kg, r = 0.5 m : F?
  2. 5F = 8 N at r = 2 m; F at r = 4 m?
  3. 6Both m doubled, r doubled : F?
Challenging
  1. 7Prove the dimension of G.
  2. 8Three masses m at the corners of an equilateral triangle of side a : F on each?
Show answers (try them yourself first)
  1. 1F = Gm₁m₂/r²
  2. 2N m² kg⁻²
  3. 3¼
  4. 42.67 × 10⁻⁸ N
  5. 52 N
  6. 6Same F
  7. 7[F][L²]/[M²] = M⁻¹L³T⁻²
  8. 8√3 Gm²/a²

Gravitational acceleration and its variation

From daily life

Your weight is slightly less on a high mountain; also deep in a mine.

Simple definition

Acceleration due to gravity g is a freely falling body's acceleration; g = GM/R².

Variation of g with distanceRgrg ∝ rg ∝ 1/r²g = G M ÷ R²

In the figure : Inside g ∝ r; outside g ∝ 1/r².

Rule 1: Formula for g — g = G M ÷ R² ≈ 9.8 m/s². Independent of the body's mass. Dimension [L T⁻²].

  • Easy : g = 6.67 × 10⁻¹¹ × 6 × 10²⁴ ÷ (6.4 × 10⁶)² = 9.77 m/s².
  • Medium : On the Moon g ≈ 1.6 m/s² (≈ 1/6 of Earth's).
  • Exam (CET) : g depends on? (1) M and R (2) falling body's mass (3) shape (4) colour → answer 1)

Rule 2: Altitude and depth — g_h = g R² ÷ (R + h)² ≈ g (1 − 2h/R); g_d = g (1 − d/R). At the centre g = 0.

  • Easy : h = R ⇒ g_h = g/4 = 2.45 m/s².
  • Medium : d = R/2 ⇒ g_d = g/2 = 4.9 m/s².
  • Exam (CET) : g at the centre? (1) 0 (2) g (3) 2g (4) infinite → answer 1)

Rule 3: Latitude and rotation — g' = g − ω² R cos²λ; at the equator g − ω² R (minimum), at the poles g (maximum). ω = 7.27 × 10⁻⁵ rad/s.

  • Easy : ω²R = 0.034 m/s²; at the equator g ≈ 9.766 m/s².
  • Medium : No effect at the poles.
  • Exam (CET) : g is greatest at? (1) the poles (2) the equator (3) 45° (4) same everywhere → answer 1)

Derivation : g at a depth

  • Step 1 : Uniform density ρ ⇒ g = G M ÷ R² = (4/3) π G ρ R.
  • Step 2 : Mass inside radius r (= R − d) is M' = ρ (4/3) π r³.
  • Step 3 : g_d = G M' ÷ r² = (4/3) π G ρ r.
  • Step 4 : g_d = g × r ÷ R = g (1 − d/R).
Remember

g = GM/R²; g_h = gR²/(R+h)²; g_d = g(1 − d/R).

Common mistake / CET trap

At h = R, g is g/4, not g/2; g falls linearly with depth but as 1/r² with height.

Shortcut

Memory clue : "Height = inverse square; depth = linear".

Spot the difference

Placeg
SurfaceGM/R²
Height hg R²/(R+h)²
Depth dg(1 − d/R)
Centre0

Practice questions

Easy
  1. 1Formula for g?
  2. 2g at the centre?
  3. 3Where is g maximum?
Medium
  1. 4h = R/2 : g_h?
  2. 5d = R/4 : g_d?
  3. 6Equator : value of ω²R?
Challenging
  1. 7h = 3R : g_h?
  2. 8Planet M = 2M_E, R = R_E : g?
Show answers (try them yourself first)
  1. 1GM/R²
  2. 20
  3. 3At the poles
  4. 44.36 m/s²
  5. 57.35 m/s²
  6. 60.034 m/s²
  7. 70.61 m/s²
  8. 819.6 m/s²

Gravitational field, potential and potential energy

From daily life

A stone rolling down a hill changes its energy; the exact energy is −GMm/r.

Simple definition

Gravitational potential = work done in bringing a unit mass from infinity (negative).

Gravitational potential : V = −GM/r0r →V−GM/RU = −GMm ÷ rE = GM ÷ r²W = m (V₂ − V₁)

In the figure : V is 0 at infinity; negative nearer.

Rule 1: Gravitational field — Intensity E = F ÷ m = GM ÷ r² (toward the centre, N/kg); at the surface E = g.

  • Easy : r = 2R ⇒ E = g/4 = 2.45 N/kg.
  • Medium : At the surface E = 9.8 N/kg.
  • Exam (CET) : Unit of E? (1) N/kg (2) N (3) J (4) kg → answer 1)

Rule 2: Potential and potential energy — V = −GM ÷ r (J/kg); U = −GMm ÷ r (J). Zero at infinity; becomes more negative nearer.

  • Easy : At the surface V = −gR = −6.27 × 10⁷ J/kg.
  • Medium : 1 kg from R to 2R : W = GMm(1/R − 1/2R) = mgR/2 = 3.14 × 10⁷ J.
  • Exam (CET) : U at infinity? (1) 0 (2) most negative (3) +GMm/r (4) mgh → answer 1)

Rule 3: Relation to mgh — For small h, ΔU = U(R + h) − U(R) ≈ mgh.

  • Easy : m = 2 kg, h = 10 m ⇒ ΔU ≈ 196 J.
  • Medium : mgh holds only when h ≪ R.
  • Exam (CET) : When is mgh accurate? (1) h ≪ R (2) h = R (3) h ≫ R (4) always → answer 1)

Derivation : ΔU ≈ mgh

  • Step 1 : ΔU = −GMm/(R + h) + GMm/R = GMm h ÷ [R (R + h)].
  • Step 2 : h ≪ R ⇒ R + h ≈ R.
  • Step 3 : ΔU ≈ GMm h ÷ R².
  • Step 4 : GM/R² = g ⇒ ΔU = m g h.
Remember

E = GM/r²; V = −GM/r; U = −GMm/r; ΔU ≈ mgh.

Common mistake / CET trap

U is negative; mgh only for small h; V is energy per kg.

Shortcut

Memory clue : "V = −GM/r; U = −GMm/r".

Spot the difference

QuantityFormulaUnit
EGM/r²N/kg
V−GM/rJ/kg
U−GMm/rJ
ΔU (small h)mghJ

Practice questions

Easy
  1. 1Formula for V?
  2. 2What is E?
  3. 3U at infinity?
Medium
  1. 4r = 3R : E?
  2. 5r = 2R : V (R = 6.4 × 10⁶)?
  3. 6m = 5, h = 20 : ΔU?
Challenging
  1. 71 kg from R to 3R : W?
  2. 8Why is U negative?
Show answers (try them yourself first)
  1. 1−GM/r
  2. 2F/m, gravitational field intensity
  3. 30
  4. 41.09 N/kg
  5. 5−3.14 × 10⁷ J/kg
  6. 6980 J
  7. 74.18 × 10⁷ J
  8. 8Work done on bringing it from infinity in an attractive field is negative

Escape velocity

From daily life

A rocket must break free of Earth's pull — what minimum speed?

Simple definition

Escape velocity = minimum speed to reach infinity.

Escape velocityM, Rv_e∞½ m v² − GMm ÷ R = 0v_e = √(2GM ÷ R) = √(2gR)Earth : 11.2 km/s

In the figure : On Earth v_e = 11.2 km/s.

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Escape velocity

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Rule 1: Escape velocity — Minimum launch speed to reach infinity : v_e = √(2GM ÷ R) = √(2gR). Independent of m.

  • Easy : Earth : √(2 × 9.8 × 6.4 × 10⁶) = 11.2 km/s.
  • Medium : Moon (g = 1.6, R = 1.74 × 10⁶) : 2.4 km/s.
  • Exam (CET) : v_e depends on? (1) M, R (2) the projectile's m (3) direction (4) colour → answer 1) Trick : 'two g R'.

Rule 2: By planet — M four times, same R ⇒ v_e doubles. R doubled with same density ⇒ v_e doubles.

  • Easy : M = 4M_E, R = R_E ⇒ v_e = 22.4 km/s.
  • Medium : M = M_E, R = 4R_E ⇒ v_e = 5.6 km/s.
  • Exam (CET) : v_e ∝ ? (1) √(M/R) (2) M/R (3) MR (4) R/M → answer 1)

Rule 3: Atmosphere link — Gases escape from the Moon because molecular speeds are close to its small v_e. Earth's v_e is large.

  • Easy : v_e = 2.4 km/s ⇒ light gases escape.
  • Medium : Speeds far below 11.2 ⇒ atmosphere stays.
  • Exam (CET) : The Moon has no atmosphere because? (1) small v_e (2) large g (3) large M (4) cold → answer 1)

Derivation : v_e = √(2GM ÷ R)

  • Step 1 : Energy conservation : KE + U at the surface = value at infinity (minimum KE 0, U 0).
  • Step 2 : ½ m v_e² − GMm ÷ R = 0.
  • Step 3 : v_e² = 2GM ÷ R.
  • Step 4 : GM = gR² ⇒ v_e = √(2gR).
Remember

v_e = √(2GM/R) = √(2gR) = 11.2 km/s.

Common mistake / CET trap

v_e is independent of direction and m; R is from the centre.

Shortcut

Memory clue : "Two g R".

Spot the difference

Bodyv_e
Earth11.2 km/s
Moon2.4 km/s
M = 4M_E, R = R_E22.4 km/s

Practice questions

Easy
  1. 1Formula for v_e?
  2. 2Earth's v_e?
  3. 3Does v_e depend on m?
Medium
  1. 4g = 4.9, R = 3.2 × 10⁶ : v_e?
  2. 5M doubled, same R : v_e?
  3. 6R four times, same M : v_e?
Challenging
  1. 7Derive v_e.
  2. 8Escape speed = orbital speed × ?
Show answers (try them yourself first)
  1. 1√(2gR)
  2. 211.2 km/s
  3. 3No
  4. 45.6 km/s
  5. 5√2 times = 15.8 km/s
  6. 6half = 5.6 km/s
  7. 7Use KE + U = 0
  8. 8√2

Satellite orbital speed, period and energy

From daily life

The harder a stone is thrown, the farther it lands; hard enough and it circles the Earth.

Simple definition

A satellite orbits a planet; gravity supplies the centripetal force.

Satellite orbitMvFrv₀ = √(GM ÷ r)T = 2π√(r³ ÷ GM)E = −GMm ÷ 2r

In the figure : F points to the centre, v is tangential.

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Satellite orbital speed, period and energy

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Rule 1: Orbital speed — v₀ = √(GM ÷ r) = √(g R² ÷ r); independent of satellite's m. Near the surface v₀ = √(gR) ≈ 7.9 km/s.

  • Easy : v₀ = √(9.8 × 6.4 × 10⁶) = 7.9 km/s.
  • Medium : r = 4R ⇒ v₀ = 7.9 ÷ 2 = 3.95 km/s.
  • Exam (CET) : If r grows, v₀? (1) decreases (2) increases (3) same (4) zero → answer 1) Trick : higher = slower.

Rule 2: Period — T = 2π √(r³ ÷ GM); near the surface T = 2π√(R/g) ≈ 84 min. T² ∝ r³.

  • Easy : T = 2π√(6.4 × 10⁶ ÷ 9.8) = 85 min.
  • Medium : r four times ⇒ T = 8 times.
  • Exam (CET) : T² ∝ ? (1) r³ (2) r² (3) r (4) 1/r → answer 1)

Rule 3: Energy — KE = GMm ÷ 2r, U = −GMm ÷ r, E = −GMm ÷ 2r = −KE. U = −2 KE.

  • Easy : KE = 100 J ⇒ E = −100 J, U = −200 J.
  • Medium : Negative E ⇒ the satellite is bound.
  • Exam (CET) : E = ? (1) −KE (2) +KE (3) 0 (4) 2KE → answer 1)

Derivation : E = −GMm ÷ 2r

  • Step 1 : GMm/r² = m v² / r ⇒ m v² = GMm / r.
  • Step 2 : KE = ½ m v² = GMm ÷ 2r.
  • Step 3 : U = −GMm ÷ r.
  • Step 4 : E = KE + U = −GMm ÷ 2r.
Remember

v₀ = √(GM/r); T = 2π√(r³/GM); E = −GMm/2r.

Common mistake / CET trap

v₀ and T do not involve m; more energy lifts it outward.

Shortcut

Memory clue : "Higher = slower and longer period".

Spot the difference

QuantityFormulaAs r increases
v₀√(GM/r)decreases
T2π√(r³/GM)increases
KEGMm/2rdecreases
E−GMm/2rincreases (less negative)

Practice questions

Easy
  1. 1Formula for v₀?
  2. 2T near the surface?
  3. 3Formula for E?
Medium
  1. 4r = 9R : v₀?
  2. 5KE = 50 J : E?
  3. 6r₂ = 4 r₁ : T₂/T₁?
Challenging
  1. 7v_e ÷ v₀ = ?
  2. 8If the satellite gains energy, r?
Show answers (try them yourself first)
  1. 1√(GM/r)
  2. 285 min
  3. 3−GMm/2r
  4. 42.64 km/s
  5. 5−50 J
  6. 68
  7. 7√2
  8. 8r increases (E less negative)

Geostationary satellite, weightlessness and binding energy

From daily life

A TV dish always faces the same direction — the satellite seems fixed in the sky.

Simple definition

A geostationary satellite has T = 24 h and seems fixed over a spot.

Geostationary satelliteEarthT = 24 hh ≈ 36000 kmv ≈ 3.07 km/sabove the equator

In the figure : Over the equator, west to east.

Rule 1: Geostationary satellite — T = 24 h, west to east, over the equator; r = (GM T² ÷ 4π²)^(1/3) ≈ 4.2 × 10⁷ m; h ≈ 3.6 × 10⁴ km.

  • Easy : r = 4.23 × 10⁷ m; h = r − R = 35898 km.
  • Medium : v = 2πr/T = 3.08 km/s.
  • Exam (CET) : Period of a geostationary satellite? (1) 24 h (2) 12 h (3) 84 min (4) 365 d → answer 1)

Rule 2: Weightlessness — Inside an orbiting satellite everything freely falls with the same acceleration ⇒ N = 0, 'weightless'. Gravity is present but not felt.

  • Easy : Cable cut in a lift ⇒ N = 0 inside.
  • Medium : Astronauts float although g ≈ 8.7 m/s² there.
  • Exam (CET) : Weightless because? (1) free fall (2) g = 0 (3) no air (4) cold → answer 1)

Rule 3: Binding energy — Binding energy = energy needed to take it to infinity = −E = GMm ÷ 2r.

  • Easy : m = 100 kg, r = R : GMm/2R = mgR/2 = 3.14 × 10⁹ J.
  • Medium : E = −B.E.
  • Exam (CET) : Binding energy = ? (1) GMm/2r (2) GMm/r (3) 2GMm/r (4) 0 → answer 1)

Derivation : Radius of the geostationary orbit

  • Step 1 : T = 2π√(r³ ÷ GM) ⇒ r³ = GM T² ÷ 4π².
  • Step 2 : T = 86400 s; GM = gR².
  • Step 3 : r = 4.23 × 10⁷ m.
  • Step 4 : h = r − R ≈ 35898 km.
Remember

r = (GMT²/4π²)^(1/3); B.E. = GMm/2r.

Common mistake / CET trap

Weightless ≠ g = 0; geostationary only over the equator.

Shortcut

Memory clue : "24 hours, 36 thousand km".

Spot the difference

OrbitTHeight
Near84 minabout 0
Geostationary24 h≈ 36000 km
Moon27.3 days≈ 3.8 × 10⁵ km

Practice questions

Easy
  1. 1T of a geostationary satellite?
  2. 2What is binding energy?
  3. 3Why weightlessness?
Medium
  1. 4100 kg, r = 2R, B.E.?
  2. 5Height of a geostationary orbit?
  3. 6Speed of a geostationary satellite?
Challenging
  1. 7Can a geostationary satellite be over the pole?
  2. 8Derive binding energy.
Show answers (try them yourself first)
  1. 124 hours
  2. 2Energy needed to reach infinity
  3. 3Free fall ⇒ N = 0
  4. 41.57 × 10⁹ J
  5. 5≈ 36000 km
  6. 6≈ 3.07 km/s
  7. 7No; the orbit plane must contain the centre and equator
  8. 8B.E. = −E = GMm/2r

Quick revision of the whole lesson

ConceptKey formula / factUnit / dimensions
Kepler 3T² ∝ a³—
GravitationF = G m₁m₂/r²N; M L T⁻²
G6.67 × 10⁻¹¹M⁻¹ L³ T⁻²
gGM/R²m/s²; L T⁻²
PotentialV = −GM/rJ/kg; L² T⁻²
Escape speed√(2gR) = 11.2 km/sm/s; L T⁻¹
Orbital speed√(GM/r)m/s
Orbital energyE = −GMm/2rJ; M L² T⁻²

🎯 Most important points for the exam

  • T² ∝ a³; second law = angular momentum.
  • F = Gm₁m₂/r²; inverse square.
  • g falls as 1/r² with height, linearly with depth.
  • V and U are 0 at infinity, negative elsewhere.
  • v_e = √2 × v₀ = 11.2 km/s.
  • Higher orbit = slower and longer T.
  • Geostationary : 24 h, 36000 km, equator.
Study mantra

Kepler → Newton → g → V, U → v_e, v₀, E.