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#gravitational constant

24 public questions tagged with this topic.

Universal Law of Gravitation

What is the physical significance of the gravitational constant G in Newton’s law?

G is the universal gravitational constant, determining the strength of the gravitational force between two masses. It is a fundamental constant that scales the force in F = Gm1m2r2. As per NCERT, applying relevant law/formula with correct units and sign convention leads to It determines the force strength. 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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Universal Law of Gravitation

A satellite orbits a planet at 8×107m from its center with a period of 20 hours. What is the planet’s mass? (G\=6.67×10−

M = 4π2r3GT2. T = 20×3600 = 72000s, T2 = 5.184×109s2. r3 = (8×107)3 = 5.12×1023m3. M = 4×(3.14)2×5.12×10236.67×10−11×5.184×109. M = 2.019×10253.458×10−1≈5.84×1025kg. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 5.8 × 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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Universal Law of Gravitation

A satellite orbits a planet at 3×107m from its center with a period of 6 hours. What is the planet’s mass? (G\=6.67×10−1

M = 4π2r3GT2. T = 6×3600 = 21600s, T2 = 4.6656×108s2. r3 = (3×107)3 = 2.7×1022m3. M = 4×(3.14)2×2.7×10226.67×10−11×4.6656×108. M = 1.065×10243.112×10−2≈3.42×1025kg. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 3.4 × 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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Universal Law of Gravitation

What is the kinetic energy of a 900kg satellite at 11RE from Earth’s center? (ME\=6×1024kg,RE\=6.4×106m,G\=6.67×10−11N m

K = GMEm2r. r = 11RE = 7.04×107m. K = 6.67×10−11×6×1024×9002×7.04×107. K = 3.602×10171.408×108≈2.56×109J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.6 × 10⁹ J. 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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Universal Law of Gravitation

What is the escape speed from a planet with mass 9×1023kg and radius 2.5×106m? (G\=6.67×10−11N m2/kg2)

ve = 2GMR. ve = 2×6.67×10−11×9×10232.5×106. ve = 4.801×107≈6.93×103m/s≈6.9km/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 6.9 km/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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Gravitational Effects and Applications

A planet moves in an elliptical orbit around the Sun with a semi-major axis of 2.25×1011m. If its orbital period is 2 ye

Using Kepler’s third law: T2 = 4π2GMsa3. Rearrange for Ms: Ms = 4π2a3GT2. T = 2×3.156×107 = 6.312×107s. a = 2.25×1011m. T2 = (6.312×107)2 = 3.984×1015s2. a3 = (2.25×1011)3 = 1.139×1033m3. Ms = 4×(3.14)2×1.139×10336.67×10−11×3.984×1015. Ms = 4.49×10342.657×105≈1.69×1030kg.

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

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Gravitational Effects and Applications

A planet has a moon with an orbital radius of 5×108m and a period of 10 days. What is the planet’s mass? (G\=6.67×10−11N

T2 = 4π2GMR3. M = 4π2R3GT2. T = 10×86400 = 8.64×105s. T2 = (8.64×105)2 = 7.465×1011s2. R3 = (5×108)3 = 1.25×1025m3. M = 4×(3.14)2×1.25×10256.67×10−11×7.465×1011. M = 4.93×10264.979×101≈9.9×1024kg. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 9.9 × 10²⁴ kg. 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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Gravitational Effects and Applications

What is the minimum speed to escape from 10RE from Earth’s center? (g\=9.8m/s2,RE\=6.4×106m)

ve = 2gRE210RE = 2×9.8×6.4×10610. ve = 1.254×107≈3.54×103m/s≈3.5km/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 3.5 km/s. 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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Energy of Satellites and Binding Energy

What is the escape speed from a planet of mass 4.8×1024kg and radius 5×106m? (G\=6.67×10−11N m2/kg2)

ve = 2GMR. ve = 2×6.67×10−11×4.8×10245×106. ve = 6.403×107≈8.0×103m/s = 8.0km/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 8.0 km/s. 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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Escape Velocity and Bound Systems

A moon orbits a planet with a period of 11 days and radius 8×108m. What is the planet’s mass? (G\=6.67×10−11N m2/kg2,1da

M = 4π2r3GT2. T = 11×86400 = 9.504×105s. T2 = 9.028×1011s2. r3 = (8×108)3 = 5.12×1026m3. M = 4×(3.14)2×5.12×10266.67×10−11×9.028×1011. M = 2.019×10276.022×101≈3.35×1025kg. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 3.4 × 10²⁵ kg. 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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Escape Velocity and Bound Systems

A moon orbits a planet with a period of 12 days and radius 9×108m. What is the planet’s mass? (G\=6.67×10−11N m2/kg2,1da

M = 4π2r3GT2. T = 12×86400 = 1.0368×106s. T2 = 1.075×1012s2. r3 = (9×108)3 = 7.29×1026m3. M = 4×(3.14)2×7.29×10266.67×10−11×1.075×1012. M = 2.875×10277.171×101≈4.01×1025kg. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 4.0 × 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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