#speed at infinity

Gravitational Effects and Applications

A body is launched from Earth at 15.5km/s. What is its speed at infinity? (Escape speed = 11.2km/s)

vf2 = vi2−ve2. vf2 = (15.5)2−(11.2)2 = 240.25−125.44 = 114.81. vf = 114.81≈10.71km/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 10.7 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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Gravitational Effects and Applications

A body is launched from Earth at 12km/s. What is its speed at infinity? (Escape speed = 11.2km/s)

vf2 = vi2−ve2. vf2 = (12)2−(11.2)2 = 144−125.44 = 18.56. vf = 18.56≈4.31km/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 4.3 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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Escape Velocity and Bound Systems

A body is launched from Earth at 12.5km/s. What is its speed at infinity? (Escape speed = 11.2km/s)

vf2 = vi2−ve2. vf2 = (12.5)2−(11.2)2 = 156.25−125.44 = 30.81. vf = 30.81≈5.55km/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 5.5 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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Work, Energy and Orbital Mechanics

A body is launched from Earth at 13.5km/s. What is its speed at infinity? (Escape speed = 11.2km/s)

vf2 = vi2−ve2. vf2 = (13.5)2−(11.2)2 = 182.25−125.44 = 56.81. vf = 56.81≈7.54km/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 7.5 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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Potential Energy of Mass Systems

A body is launched from Earth at 11.8km/s. What is its speed at infinity? (Escape speed = 11.2km/s)

vf2 = vi2−ve2. vf2 = (11.8)2−(11.2)2 = 139.24−125.44 = 13.8. vf = 13.8≈3.71km/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 3.7 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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