#planetary physics

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

What is the escape speed from a planet with mass 3.6×1024kg and radius 5×106m? (G\=6.67×10−11N m2/kg2)

ve = 2GMR. ve = 2×6.67×10−11×3.6×10245×106. ve = 9.607×107≈9.80×103m/s≈9.8km/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 9.8 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

What is the escape speed from a planet with g\=4.9m/s2 and radius 3.2×106m?

ve = 2gR. ve = 2×4.9×3.2×106 = 3.136×107. ve≈5.6×103m/s = 5.6km/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 5.6 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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Variation of g with Height and Depth

What is the escape speed from a planet with mass 1.2×1024kg and radius 3×106m? (G\=6.67×10−11N m2/kg2)

ve = 2GMR. ve = 2×6.67×10−11×1.2×10243×106. ve = 5.336×107≈7.3×103m/s = 7.3km/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 7.3 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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Universal Law of Gravitation

What is the escape speed from a planet with mass 2.7×1024kg and radius 4.5×106m? (G\=6.67×10−11N m2/kg2)

ve = 2GMR. ve = 2×6.67×10−11×2.7×10244.5×106. ve = 8.001×107≈8.94×103m/s≈8.9km/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 C is scientifically correct.

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

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