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#gravitation

93 public questions tagged with this topic.

PHYSICS

A body weighs 196 N on Earth’s surface. What is its weight at a depth d = R_E/3 ? ( g = 9.8 m/s² )

Given: A body weighs 196 N on Earth’s surface. What is its weight at a depth d = R_E/3 ? ( g = 9.8 m/s² ) These values define the system as per NCERT data. Formula: g(d) = g (1 - d/R_E). This is the standard NCERT relation for this phenomenon. Substitution & Calculation: d = R_E/3, g(d) = 9.8 (1 - 1/3) = 9.8 × 2/3 = 6.53 m/s² . Mass: m = 196/9.8 = 20 kg . Weight: W = 20 × 6.53 approx 130.6 N . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.

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Escape Velocity and Bound Systems

A moon orbits a planet with a period of 6 days and radius 4×108m. What is the planet’s mass? (G\=6.67×10−11N m2/kg2,1day

M = 4π2r3GT2. T = 6×86400 = 5.184×105s. T2 = 2.689×1011s2. r3 = (4×108)3 = 6.4×1025m3. M = 4×(3.14)2×6.4×10256.67×10−11×2.689×1011. M = 2.523×10261.794×101≈1.41×1025kg. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.4 × 10²⁵ kg. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: Halliday & Resnick, Fundamentals of Physics, 11th ed., Chapter 13: Gravitation.

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Kepler's Laws and Planetary Motion

Three 2kg masses form an equilateral triangle of side 1m. What is the potential energy? (G\=6.67×10−11N m2/kg2)

3 pairs: V = −3Gm2r. V = −36.67×10−11×2×21. V = −3×2.668×10−10 = −8.004×10−10J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to -8.0 × 10⁻¹⁰ J. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Kepler's Laws and Planetary Motion

What ensures that a satellite remains in a stable circular orbit?

A stable circular orbit requires the gravitational force (GMEmr2) to equal the centripetal force (mv2r), balancing the forces to maintain constant radius and speed. As per NCERT, applying relevant law/formula with correct units and sign convention leads to Balance of gravitational and centripetal forces. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: Halliday & Resnick, Fundamentals of Physics, 11th ed., Chapter 13: Gravitation.

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Kepler's Laws and Planetary Motion

What is the gravitational potential due to Earth at 3.84×107m from its center? (ME\=6×1024kg,G\=6.67×10−11N m2/kg2)

U = −GMEr. U = −6.67×10−11×6×10243.84×107. U = −1.0417×107J/kg. As per NCERT, applying relevant law/formula with correct units and sign convention leads to -1.0 × 10⁷ J/kg. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.

Ref: Halliday & Resnick, Fundamentals of Physics, 11th ed., Chapter 13: Gravitation.

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Kepler's Laws and Planetary Motion

Four 5kg masses form a square of side 6m. What is the potential energy of the system? (G\=6.67×10−11N m2/kg2)

4 sides: r = 6m, 2 diagonals: r = 62m. V = −4Gm26−2Gm262. V = −6.67×10−11×25(46+262). V = −1.6675×10−9(0.667+0.236)≈−1.51×10−9J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to -1.5 × 10⁻⁹ J. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Kepler's Laws and Planetary Motion

A satellite orbits Earth at 7RE from the center. What is its orbital speed? (g\=9.8m/s2,RE\=6.4×106m)

v = gRE2r. r = 7RE, v = 9.8×6.4×1067. v = 8.966×106≈2.99×103m/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 3.0 × 10³ m/s. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: Halliday & Resnick, Fundamentals of Physics, 11th ed., Chapter 13: Gravitation.

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Kepler's Laws and Planetary Motion

A moon orbits a planet with a period of 9 days and radius 6×108m. What is the planet’s mass? (G\=6.67×10−11N m2/kg2,1day

M = 4π2r3GT2. T = 9×86400 = 7.776×105s. T2 = 6.046×1011s2. r3 = (6×108)3 = 2.16×1026m3. M = 4×(3.14)2×2.16×10266.67×10−11×6.046×1011. M = 8.51×10264.033×101≈2.11×1025kg. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.1 × 10²⁵ kg. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: Halliday & Resnick, Fundamentals of Physics, 11th ed., Chapter 13: Gravitation.

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Kepler's Laws and Planetary Motion

Four 4kg masses form a square of side 5m. What is the potential energy of the system? (G\=6.67×10−11N m2/kg2)

4 sides: r = 5m, 2 diagonals: r = 52m. V = −4Gm25−2Gm252. V = −6.67×10−11×16(45+252). V = −1.067×10−9(0.8+0.283)≈−1.15×10−9J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to -1.2 × 10⁻⁹ J. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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