Chapter 7: Gravitation Case Study Questions | physics Class 11 CBSE

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

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A student is studying planetary motion using data from Table 7.1 of the NCERT textbook. The table lists eight planets with their semi-major axes (a) and time periods (T). For Mercury, a = 5.79 × 10¹⁰ m and T = 0.24 years. For Earth, a = 15.0 × 10¹⁰ m and T = 1 year. For Jupiter, a = 77.8 × 10¹⁰ m and T = 11.9 years. The student notices that for each planet, the ratio Q = T²/a³ is nearly constant at about 2.97 × 10⁻³⁴ y² m⁻³. The student wants to use this data to understand Kepler's third law and verify that it applies universally to all planets orbiting the Sun.

Question 1: State Kepler's law of periods and identify what the nearly constant value Q = T²/a³ represents physically.

Kepler's law of periods states that the square of the time period of revolution of a planet is proportional to the cube of the semi-major axis of its elliptical orbit: T² ∝ a³.
The constant Q = T²/a³ = 4π²/(GMs), where Ms is the mass of the Sun; its near-constant value for all planets confirms that the proportionality constant depends only on the Sun's mass.
The constancy of Q across all eight planets (Mercury to Neptune) is the empirical verification of Kepler's third law derived from Tycho Brahe's observational data.

Question 2: Using Kepler's third law and the Earth's data (a = 15.0 × 10¹⁰ m, T = 1 year), predict the time period of a hypothetical planet with semi-major axis 60.0 × 10¹⁰ m.

By Kepler's third law: (T_new/T_Earth)² = (a_new/a_Earth)³ = (60.0/15.0)³ = (4)³ = 64.
Therefore T_new = T_Earth × √64 = 1 × 8 = 8 Earth years.
This shows that doubling the semi-major axis increases the period by a factor of 2^(3/2) ≈ 2.83, confirming the non-linear (power-law) relationship between orbital size and period.

Question 3: Derive the expression T² = (4π²/GMs)R³ from Newton's law of gravitation for a circular orbit, and explain how this derivation validates Kepler's empirical third law.

For a planet of mass m in a circular orbit of radius R around the Sun (mass Ms), the centripetal force equals the gravitational force: mV²/R = GmMs/R², giving V = √(GMs/R).
The orbital period T = 2πR/V = 2πR/√(GMs/R) = 2π√(R³/GMs); squaring gives T² = (4π²/GMs)R³.
The constant k = 4π²/GMs depends only on the Sun's mass, explaining why Q is the same for all planets — it is a property of the Sun, not the individual planets.

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Case Set 2

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ISRO engineers are designing a new Earth observation satellite to be placed in a circular orbit at a height of 500 km above Earth's surface. The satellite has a mass of 500 kg. Given data: mass of Earth ME = 6 × 10²⁴ kg, radius of Earth RE = 6.4 × 10⁶ m, G = 6.67 × 10⁻¹¹ N m²/kg², and g = 9.8 m/s² at the surface. The engineers need to calculate the orbital speed, time period, and total mechanical energy of the satellite to ensure correct fuel budgeting and mission planning. They also want to compare these values with a satellite at 1000 km altitude.

Question 1: Calculate the orbital speed of the satellite at 500 km altitude.

At height h = 500 km = 5 × 10⁵ m, the orbital radius is r = RE + h = 6.4 × 10⁶ + 5 × 10⁵ = 6.9 × 10⁶ m.
Orbital speed V = √(GME/r) = √(6.67 × 10⁻¹¹ × 6 × 10²⁴ / 6.9 × 10⁶) = √(5.80 × 10⁷) ≈ 7615 m/s ≈ 7.6 km/s.
This speed is slightly less than the surface orbital speed of √(gRE) = √(9.8 × 6.4 × 10⁶) ≈ 7920 m/s, consistent with V decreasing as altitude increases.

Question 2: Calculate the total mechanical energy of the 500 kg satellite at 500 km altitude and state what its sign implies.

Total energy E = −GMEm/[2(RE + h)] = −(6.67 × 10⁻¹¹ × 6 × 10²⁴ × 500)/(2 × 6.9 × 10⁶).
E = −(2.001 × 10¹⁷)/(1.38 × 10⁷) ≈ −1.45 × 10¹⁰ J (approximately −14.5 GJ).
The negative sign indicates the satellite is gravitationally bound to the Earth; energy equal to 14.5 GJ must be supplied to move the satellite to infinity and free it from Earth's gravity.

Question 3: If the satellite at 500 km is to be transferred to a circular orbit at 1000 km, calculate the change in total energy and explain what happens to its kinetic energy during the transfer.

E at 500 km: E₁ = −GMEm/[2(RE+h₁)] = −GME×500/[2×6.9×10⁶] ≈ −1.45×10¹⁰ J; E at 1000 km: E₂ = −GME×500/[2×7.4×10⁶] ≈ −1.35×10¹⁰ J.
Change ΔE = E₂ − E₁ ≈ +1.0×10⁹ J; energy must be supplied to the satellite to move it to the higher orbit.
Kinetic energy at 500 km: KE₁ ≈ 1.45×10¹⁰ J; at 1000 km: KE₂ ≈ 1.35×10¹⁰ J; so KE decreases by ≈10⁹ J — despite gaining total energy, the satellite moves slower in the higher orbit, because the decrease in PE (−2×10⁹ J) exceeds the decrease in KE.

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Henry Cavendish performed his celebrated experiment in 1798 to measure the universal gravitational constant G. He used a torsion balance consisting of a light rod AB of length L = 2 m with small lead spheres of mass m = 0.5 kg at each end, suspended by a fine wire. Large lead spheres of mass M = 50 kg were placed at a separation d = 0.1 m from the small spheres. The wire twisted by angle θ = 0.01 rad and the restoring couple per unit angle was τ = 3.0 × 10⁻⁸ N m/rad. The experiment enabled the first laboratory measurement of G and subsequently the mass of the Earth. The accepted value obtained from modern refinements is G = 6.67 × 10⁻¹¹ N m²/kg².

Question 1: Describe the principle behind Cavendish's experiment and state the equilibrium condition used to calculate G.

The principle is the balance between gravitational torque (produced by attraction between large and small spheres) and the restoring torque of the twisted suspension wire.
At equilibrium: gravitational torque = restoring torque, i.e., G(Mm/d²) × L = τθ, from which G = τθd²/(MmL).
The restoring couple per unit angle τ is measured independently by applying a known torque and observing the angle of twist, making the experiment self-calibrating.

Question 2: Using the given values (L = 2 m, m = 0.5 kg, M = 50 kg, d = 0.1 m, θ = 0.01 rad, τ = 3.0 × 10⁻⁸ N m/rad), calculate G from Cavendish's formula.

From the equilibrium condition: G = τθd²/(MmL).
Substituting: G = (3.0 × 10⁻⁸ × 0.01 × (0.1)²)/(50 × 0.5 × 2) = (3.0 × 10⁻¹²)/(50) = 6.0 × 10⁻¹¹ N m²/kg².
This is close to the accepted value of 6.67 × 10⁻¹¹ N m²/kg²; small discrepancies arise from simplifying assumptions such as treating spheres as point masses.

Question 3: Once G was known from Cavendish's experiment, explain in detail how the mass of the Earth ME could be calculated, and why this led to the statement 'Cavendish weighed the Earth'.

The acceleration due to gravity at Earth's surface is g = GME/RE²; rearranging gives ME = gRE²/G.
With g = 9.8 m/s², RE = 6.4 × 10⁶ m (known from geodetic measurements), and G from Cavendish: ME = 9.8 × (6.4 × 10⁶)²/(6.67 × 10⁻¹¹) ≈ 6.0 × 10²⁴ kg.
Before Cavendish, G was unknown, so ME could not be computed despite knowing g and RE; his measurement of G was the missing link that allowed the Earth's mass to be determined for the first time — hence 'Cavendish weighed the Earth'.

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Case Set 4

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During a physics class, a teacher explains the variation of acceleration due to gravity using a thought experiment. A borehole is drilled from Earth's surface to the centre, and a mass is dropped through it. Simultaneously, another mass is lifted in steps above the surface. The teacher notes that g = 9.8 m/s² at the surface. The Earth is assumed to be a uniform-density sphere with RE = 6400 km and ME = 6 × 10²⁴ kg. Students are asked to calculate g at a depth of 3200 km (half the Earth's radius) and at a height of 3200 km above the surface, and then compare the two values to understand the asymmetric reduction of g.

Question 1: Calculate g at a depth d = 3200 km (= RE/2) below the Earth's surface.

Using g(d) = g(1 − d/RE) with d = RE/2: g(RE/2) = 9.8 × (1 − 1/2) = 9.8 × 0.5 = 4.9 m/s².
At the Earth's centre (d = RE), g = 0; the linear decrease with depth is a consequence of only the inner sphere of radius (RE − d) contributing to gravity.
This result assumes uniform Earth density; the actual Earth is denser at the core, so real g at this depth is somewhat higher than the uniform-density estimate.

Question 2: Calculate g at a height h = 3200 km (= RE/2) above the Earth's surface and compare it with the value at depth RE/2.

Using the exact formula: g(h) = GME/(RE + h)² = g × RE²/(RE + h)² = 9.8 × (6400)²/(9600)² = 9.8 × (2/3)² = 9.8 × 4/9 ≈ 4.36 m/s².
Comparing: g at depth RE/2 is 4.9 m/s², while g at height RE/2 is ≈ 4.36 m/s²; at the same fractional distance (RE/2), g is lower above the surface than below.
This shows that g decreases faster above the surface (inverse-square) than below (linear); the asymmetry confirms that the surface is the unique maximum of g.

Question 3: Starting from Newton's law of gravitation and the shell theorem, derive the expression g(d) = g(1 − d/RE) for a point inside a uniform-density Earth at depth d.

By the shell theorem, a point at depth d is inside shells of thickness d (which exert zero force) and outside the inner sphere of radius (RE − d).
Mass of inner sphere: Ms = ME × (RE − d)³/RE³ (since mass ∝ volume ∝ r³ for uniform density).
Gravitational acceleration at depth d: g(d) = GMs/(RE−d)² = G × ME(RE−d)³/[RE³ × (RE−d)²] = GME(RE−d)/RE³ = (GME/RE²) × (1 − d/RE) = g(1 − d/RE), completing the derivation.

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Case Set 5

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A space agency is planning a mission to land on the Moon and return to Earth. Mission scientists need to calculate the escape speed from the Moon's surface to determine if a small rocket can launch the return capsule. Given data: mass of Moon MM = 7.4 × 10²² kg, radius of Moon RM = 1.74 × 10⁶ m, G = 6.67 × 10⁻¹¹ N m²/kg², mass of Earth ME = 6 × 10²⁴ kg, RE = 6.4 × 10⁶ m. The Moon has no atmosphere. The mission planners also want to understand why the Moon has no atmosphere and whether a gas molecule at 500 K could escape the Moon.

Question 1: Calculate the escape speed from the Moon's surface and state its formula.

Escape speed from the Moon: ve(Moon) = √(2GMM/RM) = √(2 × 6.67 × 10⁻¹¹ × 7.4 × 10²²/1.74 × 10⁶).
Numerating: 2 × 6.67 × 10⁻¹¹ × 7.4 × 10²² = 9.87 × 10¹²; dividing by 1.74 × 10⁶: 5.67 × 10⁶; ve = √(5.67 × 10⁶) ≈ 2380 m/s ≈ 2.3 km/s.
This is about 5 times smaller than Earth's escape speed of 11.2 km/s, because both the Moon's mass and radius are much smaller than Earth's.

Question 2: Using the escape speed result, explain why the Moon has no atmosphere, with reference to gas molecule speeds.

Gas molecules at moderate temperatures have thermal speeds that can exceed the Moon's escape speed of 2.3 km/s; for example, hydrogen molecules at room temperature have rms speeds of ≈ 1.9 km/s, and lighter molecules or those in the tail of the Maxwell distribution can easily exceed 2.3 km/s.
Over geological time, any gas molecules formed on or outgassed from the Moon with speeds ≥ 2.3 km/s escape the Moon's gravitational field permanently, and since there is no mechanism to retain them, the atmosphere is continually depleted.
Earth retains its atmosphere because its escape speed of 11.2 km/s is much higher than typical molecular thermal speeds at Earth's surface temperatures, so the fraction of molecules that can escape is negligibly small.

Question 3: Compare the escape speed from Earth with that from the Moon using the formula ve = √(2gR), and derive a general expression showing how escape speed depends on the planet's density and radius.

ve = √(2gR) = √(2 × GME/RE² × RE) = √(2GME/RE); for Earth: ve = √(2 × 9.8 × 6.4 × 10⁶) ≈ 11.2 km/s; for Moon: ve ≈ 2.3 km/s (as calculated).
Writing M = (4/3)πR³ρ (uniform density ρ), escape speed becomes ve = √(2G × (4/3)πR³ρ / R) = √(8πGρ/3) × R.
Therefore ve ∝ R√ρ: larger planets with higher density have larger escape speeds; the Moon's small radius and lower mean density (∼3340 kg/m³ vs Earth's ∼5500 kg/m³) both contribute to its low escape speed.

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Chapter 7: Gravitation | Case Study Questions

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Question 1: State Kepler's law of periods and identify what the nearly constant value Q = T²/a³ represents physically.

Question 2: Using Kepler's third law and the Earth's data (a = 15.0 × 10¹⁰ m, T = 1 year), predict the time period of a hypothetical planet with semi-major axis 60.0 × 10¹⁰ m.

Question 3: Derive the expression T² = (4π²/GMs)R³ from Newton's law of gravitation for a circular orbit, and explain how this derivation validates Kepler's empirical third law.

Passage

Question 1: Calculate the orbital speed of the satellite at 500 km altitude.

Question 2: Calculate the total mechanical energy of the 500 kg satellite at 500 km altitude and state what its sign implies.

Question 3: If the satellite at 500 km is to be transferred to a circular orbit at 1000 km, calculate the change in total energy and explain what happens to its kinetic energy during the transfer.

Passage

Question 1: Describe the principle behind Cavendish's experiment and state the equilibrium condition used to calculate G.

Question 2: Using the given values (L = 2 m, m = 0.5 kg, M = 50 kg, d = 0.1 m, θ = 0.01 rad, τ = 3.0 × 10⁻⁸ N m/rad), calculate G from Cavendish's formula.

Question 3: Once G was known from Cavendish's experiment, explain in detail how the mass of the Earth ME could be calculated, and why this led to the statement 'Cavendish weighed the Earth'.

Passage

Question 1: Calculate g at a depth d = 3200 km (= RE/2) below the Earth's surface.

Question 2: Calculate g at a height h = 3200 km (= RE/2) above the Earth's surface and compare it with the value at depth RE/2.

Question 3: Starting from Newton's law of gravitation and the shell theorem, derive the expression g(d) = g(1 − d/RE) for a point inside a uniform-density Earth at depth d.

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Question 1: Calculate the escape speed from the Moon's surface and state its formula.

Question 2: Using the escape speed result, explain why the Moon has no atmosphere, with reference to gas molecule speeds.

Question 3: Compare the escape speed from Earth with that from the Moon using the formula ve = √(2gR), and derive a general expression showing how escape speed depends on the planet's density and radius.

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