#planet mass

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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Gravitational Effects and Applications

A moon orbits a planet with a period of 10 days and radius 7×108m. What is the planet’s mass? (G\=6.67×10−11N m2/kg2,1da

M = 4π2r3GT2. T = 10×86400 = 8.64×105s. T2 = 7.465×1011s2. r3 = (7×108)3 = 3.43×1026m3. M = 4×(3.14)2×3.43×10266.67×10−11×7.465×1011. M = 1.353×10274.979×101≈2.72×1025kg. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.7 × 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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Gravitational Effects and Applications

What is the escape speed from a planet with mass 1.8×1024kg and radius 3.5×106m? (G\=6.67×10−11N m2/kg2)

ve = 2GMR. ve = 2×6.67×10−11×1.8×10243.5×106. ve = 6.861×107≈8.28×103m/s≈8.3km/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 8.3 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 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

A moon orbits a planet with a period of 7 days and radius 5×108m. What is the planet’s mass? (G\=6.67×10−11N m2/kg2,1day

M = 4π2r3GT2. T = 7×86400 = 6.048×105s. T2 = 3.658×1011s2. r3 = (5×108)3 = 1.25×1025m3. M = 4×(3.14)2×1.25×10256.67×10−11×3.658×1011. M = 4.93×10252.44×101≈2.02×1024kg. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.0 × 10²⁴ 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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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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Orbital Velocity and Time Period

What is the escape speed from a planet with mass 2.4×1024kg and radius 4×106m? (G\=6.67×10−11N m2/kg2)

ve = 2GMR. ve = 2×6.67×10−11×2.4×10244×106. ve = 8.002×107≈8.95×103m/s≈8.95km/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 8.9 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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Orbital Velocity and Time Period

A moon orbits a planet with period 8 days and radius 3×108m. What is the planet’s mass? (G\=6.67×10−11N m2/kg2,1day\=864

M = 4π2r3GT2. T = 8×86400 = 6.912×105s. T2 = 4.776×1011s2. r3 = (3×108)3 = 2.7×1025m3. M = 4×(3.14)2×2.7×10256.67×10−11×4.776×1011. M = 1.065×10263.185×101≈3.34×1024kg. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 3.4 × 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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Orbital Velocity and Time Period

What is the escape speed from a planet with mass 4.5×1024kg and radius 6×106m? (G\=6.67×10−11N m2/kg2)

ve = 2GMR. ve = 2×6.67×10−11×4.5×10246×106. ve = 1.0005×108≈1.00×104m/s = 10.0km/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 10.0 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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Kepler's Laws and Planetary Motion

A moon orbits a planet with a period of 13 days and radius 1.0×109m. What is the planet’s mass? (G\=6.67×10−11N m2/kg2,1

M = 4π2r3GT2. T = 13×86400 = 1.1232×106s. T2 = 1.262×1012s2. r3 = (1.0×109)3 = 1.0×1027m3. M = 4×(3.14)2×10276.67×10−11×1.262×1012. M = 3.947×10278.418×101≈4.69×1025kg. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 4.7 × 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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