Chapter 7: Gravitation Long Questions | physics Class 11 CBSE

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Chapter 7: Gravitation Long Questions | physics Class 11 CBSE | ByteNotes.in

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Question 1

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Describe the historical development of models for planetary motion from Ptolemy to Kepler, and explain how Kepler's laws corrected the earlier models.

Ptolemy's geocentric model (2000 years ago) placed the Earth at the centre with all celestial bodies revolving around it in circles; planets moved in circles whose centres themselves moved in larger circles.
The heliocentric model with the Sun at the centre was mentioned by Aryabhatta (5th century AD) and later formalised by Copernicus (1473–1543), who proposed planets moving in circular orbits around the Sun.
Tycho Brahe (1546–1601) spent his lifetime recording precise naked-eye planetary observations, which were later analysed by his assistant Johannes Kepler (1571–1640).
Kepler's First Law corrected circular orbits to elliptical orbits with the Sun at one focus; his Second Law described the varying speed of planets; his Third Law gave the quantitative T² ∝ a³ relation.
Galileo supported the Copernican heliocentric model and faced prosecution; his observations of celestial bodies provided additional empirical support against the geocentric view.
Kepler's three laws, known to Newton, enabled him to derive the universal law of gravitation with its inverse-square dependence, unifying terrestrial and celestial mechanics.

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Question 2

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Derive the expression for the acceleration due to gravity at a height h above and at a depth d below the Earth's surface, and compare how g varies in the two cases.

Above the surface: A mass m at height h is at distance (RE + h) from the Earth's centre, so g(h) = GME/(RE + h)². For h << RE, binomial expansion gives g(h) ≈ g(1 − 2h/RE).
Below the surface: At depth d, only the inner sphere of radius (RE − d) and mass ME(RE − d)³/RE³ contributes to gravity; the outer shell exerts zero net force.
This gives g(d) = GME(RE − d)/RE³ = g(1 − d/RE), a linear decrease with depth.
Comparison: Above the surface, g decreases as an inverse-square law; below the surface, g decreases linearly. The rate of decrease is faster above (factor 2h/RE) than just below (factor d/RE) for small equal distances.
At the centre (d = RE), g = 0; there is no corresponding finite height above the surface where g = 0.
In both cases g is maximum at the surface, confirming that Earth's surface is the unique point of maximum gravitational acceleration.

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Question 3

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Derive an expression for the escape speed from Earth's surface using energy conservation and discuss factors that affect it.

Let an object of mass m be projected with speed Vi from Earth's surface. For the object to escape, it must reach infinity with at least zero kinetic energy.
By energy conservation: ½mVi² − GMEm/RE = 0 (total energy ≥ 0 for escape), giving the minimum escape speed Ve = √(2GME/RE).
Using g = GME/RE², we get Ve = √(2gRE). Substituting g = 9.8 m/s² and RE = 6400 km, Ve ≈ 11.2 km/s.
Escape speed is independent of the mass of the projected body and the direction of projection; it depends only on the mass and radius of the planet.
For the Moon: both g and radius are smaller, giving an escape speed of only ≈ 2.3 km/s, about five times less than Earth's escape speed.
The smaller escape speed explains why the Moon has no atmosphere — gas molecules can easily exceed 2.3 km/s through thermal motion and escape the Moon's gravity over time.

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Question 4

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Derive the total mechanical energy of a satellite in a circular orbit and explain why it is always negative.

For a satellite of mass m in a circular orbit at height h, balancing centripetal and gravitational forces: mV²/(RE + h) = GMEm/(RE + h)², giving V² = GME/(RE + h).
Kinetic energy: KE = ½mV² = GMEm / [2(RE + h)], which is positive.
Potential energy (with V = 0 at infinity): PE = −GMEm/(RE + h), which is negative and equals −2 × KE.
Total energy: E = KE + PE = GMEm/[2(RE+h)] − GMEm/(RE+h) = −GMEm/[2(RE+h)], which is negative.
A negative total energy means the satellite is bound — it cannot escape to infinity without an external energy supply equal to |E|.
If the orbit becomes elliptic, both KE and PE vary point to point, but total energy remains constant and negative; a positive or zero total energy would mean the object escapes.

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Question 5

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Explain Newton's reasoning that led him to the universal law of gravitation, starting from the observation that the moon's centripetal acceleration is much smaller than g on Earth's surface.

Newton calculated the centripetal acceleration of the Moon: am = V²/Rm = 4π²Rm/T², where Rm = 3.84 × 10⁸ m and T = 27.3 days, giving am much smaller than g = 9.8 m/s².
He hypothesised that gravitational force, and hence g, decreases with distance as an inverse-square law: am ∝ Rm⁻² and g ∝ RE⁻².
The ratio g/am = Rm²/RE² ≈ 3600, which matches (Rm/RE)² since Rm ≈ 60RE, providing strong evidence for the inverse-square law.
Newton extended this to a universal law: every pair of masses attracts each other with F = Gm₁m₂/r², applicable to all bodies in the universe.
This single law explained both terrestrial gravity and the orbital motion of the Moon, unifying what seemed like separate phenomena.
The universal law also provided a physical basis for Kepler's empirical laws, showing they are natural consequences of inverse-square gravitational attraction.

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Question 6

Long Form

Describe Cavendish's experiment to determine G. Include the principle, key components, the torque balance condition, and the formula used to calculate G.

Principle: A torque produced by gravitational attraction is balanced by the restoring torque of a twisted suspension wire, allowing G to be measured.
Apparatus: A light rod AB with two small lead spheres at its ends is suspended by a fine wire. Two large lead spheres are brought close to the small ones on opposite sides.
The large spheres attract the small ones with equal and opposite forces, creating a net torque F × L (where L is the length of rod AB) that twists the wire.
At equilibrium, gravitational torque equals restoring torque τθ, where τ is the restoring couple per unit angle and θ is the angle of twist: G(Mm/d²)L = τθ.
Solving for G: G = τθd²/(MmL); τ can be independently measured by applying a known torque and recording the resulting angle.
The currently accepted value from refined measurements is G = 6.67 × 10⁻¹¹ N m²/kg²; knowing G enables calculation of the Earth's mass from g = GME/RE².

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Question 7

Long Form

State and prove the shell theorem for a uniform spherical shell, explaining separately the cases of a point mass outside and inside the shell.

For a point mass outside the shell: The gravitational forces from all elements of the shell are resolved into components along and perpendicular to the line joining the point mass to the shell's centre.
The perpendicular components cancel by symmetry, leaving only components along the central line; summing these (using calculus) yields a net force equivalent to all the shell's mass concentrated at its centre.
Mathematical result: F = GMsm/r² (directed towards centre), where Ms is the shell's mass and r is the distance from the centre to the external point.
For a point mass inside the shell: Gravitational forces from all parts of the shell act in various directions and cancel each other completely by symmetry.
This can be understood by noting that for any internal point, larger but more distant shell segments balance smaller but closer segments, resulting in zero net force.
Practical consequence: Inside the Earth at depth d, only the mass of the inner sphere of radius (RE − d) exerts gravity; the outer spherical shell exerts no force.

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Question 8

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Compare the gravitational potential energy of a body at a small height near Earth's surface with the general expression for arbitrary distances, and explain when each formula is appropriate.

Near-surface approximation: For heights h much smaller than RE, gravitational force is essentially constant at mg. The potential energy is W(h) = mgh + W₀, where W₀ is the surface potential energy.
General expression: For arbitrary distances r from Earth's centre, W(r) = −GMEm/r + W₁, where W₁ = 0 by convention, giving W(r) = −GMEm/r.
The mgh formula is an approximation to the difference [W(RE + h) − W(RE)] = −GMEm/(RE + h) + GMEm/RE = GMEm × h / [RE(RE + h)] ≈ mgh for h << RE.
The near-surface formula is convenient for problems involving buildings, projectiles, and everyday heights, but introduces negligible error only when h/RE << 1.
The general formula must be used for satellites, rockets reaching significant fractions of RE in altitude, escape speed calculations, and any scenario involving large distances.
Only differences in potential energy are physically meaningful; the absolute value depends on the choice of reference (zero at surface vs. zero at infinity).

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Question 9

Long Form

Derive Kepler's third law from Newton's law of gravitation for circular planetary orbits and verify it using the given data for any two planets.

For a planet of mass m in a circular orbit of radius R around the Sun (mass Ms), the centripetal force equals gravitational force: mV²/R = GmMs/R².
Solving for V: V = √(GMs/R). The orbital period T = 2πR/V = 2π√(R³/GMs).
Squaring: T² = (4π²/GMs)R³, i.e., T² ∝ R³ with constant k = 4π²/GMs, confirming Kepler's third law.
For Earth: a = 15.0 × 10¹⁰ m, T = 1 year; Q = T²/a³ = 2.96 × 10⁻³⁴ y² m⁻³.
For Jupiter: a = 77.8 × 10¹⁰ m, T = 11.9 years; Q = (11.9)²/(77.8)³ × 10⁻³⁰ ≈ 3.01 × 10⁻³⁴ y² m⁻³.
The near-constant value of Q for all planets confirms that T²/a³ = 4π²/GMs is indeed the same for every planet orbiting the Sun.

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Question 10

Long Form

Analyse the energy changes when a satellite is transferred from a lower circular orbit of radius 2RE to a higher circular orbit of radius 4RE around the Earth.

Total energy in orbit of radius r is E = −GMEm/(2r). Initial total energy: Ei = −GMEm/(2×2RE) = −GMEm/(4RE).
Final total energy: Ef = −GMEm/(2×4RE) = −GMEm/(8RE). Change in total energy: ΔE = Ef − Ei = −GMEm/(8RE) + GMEm/(4RE) = GMEm/(8RE) > 0.
Since ΔE is positive, energy must be supplied to the satellite to move it to the higher orbit; for a 400 kg satellite: ΔE = gmRE/8 ≈ 3.13 × 10⁹ J.
Kinetic energy change: ΔK = Kf − Ki = −ΔE = −3.13 × 10⁹ J. The satellite moves slower in the higher orbit, so its KE decreases.
Potential energy change: ΔV = Vf − Vi = 2ΔE = +6.25 × 10⁹ J. The increase in PE is twice the total energy increase.
The result |ΔPE| = 2|ΔKE| reflects the virial theorem: for a bound gravitational system, the average KE is half the magnitude of the average PE.

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Chapter 7: Gravitation | Long Questions

Describe the historical development of models for planetary motion from Ptolemy to Kepler, and explain how Kepler's laws corrected the earlier models.

Derive the expression for the acceleration due to gravity at a height h above and at a depth d below the Earth's surface, and compare how g varies in the two cases.

Derive an expression for the escape speed from Earth's surface using energy conservation and discuss factors that affect it.

Derive the total mechanical energy of a satellite in a circular orbit and explain why it is always negative.

Explain Newton's reasoning that led him to the universal law of gravitation, starting from the observation that the moon's centripetal acceleration is much smaller than g on Earth's surface.

Describe Cavendish's experiment to determine G. Include the principle, key components, the torque balance condition, and the formula used to calculate G.

State and prove the shell theorem for a uniform spherical shell, explaining separately the cases of a point mass outside and inside the shell.

Compare the gravitational potential energy of a body at a small height near Earth's surface with the general expression for arbitrary distances, and explain when each formula is appropriate.

Derive Kepler's third law from Newton's law of gravitation for circular planetary orbits and verify it using the given data for any two planets.

Analyse the energy changes when a satellite is transferred from a lower circular orbit of radius 2RE to a higher circular orbit of radius 4RE around the Earth.

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