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#escape velocity

38 public questions tagged with this topic.

PHYSICS

What is the minimum speed to escape from 6 R_E from Earth’s nter? ( g = 9.8 m/s², R_E = 6.4 × 10⁶m )

Given: What is the minimum speed to escape from 6 R_E from Earth’s nter? ( g = 9.8 m/s², R_E = 6.4 × 10⁶m ) Formula: v_e = √2 g R_E²/6 R_E = √2 × 9.8 × 6.4 × 10⁶/6. Substitution & Calculation: v_e = √2.09 × 10⁷approx 4.57 × 10³m/s approx 4.6 km/s . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27),Topic: Fundamental laws, definitions and applications as per latest NCERT. The section explains governing laws, formulas like μ₀ = 4π.

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PHYSICS

Why does an object need a minimum speed to escape Earth’s gravity?

Escape speed ( v_e = √2 G M/R ) is the minimum speed where kinetic energy equals the gravitational potential energy’s magnitude, ensuring total energy is zero, allowing escape to infinity.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27),Topic: Fundamental laws, definitions and applications as per latest NCERT. The section explains governing laws, formulas like μ₀ = 4π.

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Universal Law of Gravitation

What is the minimum speed to escape from 5RE from Earth’s center? (g\=9.8m/s2,RE\=6.4×106m)

ve = 2gRE25RE = 2×9.8×6.4×1065. ve = 2.509×107≈5.01×103m/s = 5.0km/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 5.0 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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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 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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Gravitational Effects and Applications

What condition must an object’s energy satisfy to remain bound to Earth?

For an object to remain bound, its total mechanical energy (E = KE+PE) must be negative, indicating it lacks sufficient energy to reach infinity where E = 0. As per NCERT, applying relevant law/formula with correct units and sign convention leads to Total energy must be negative. 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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Gravitational Effects and Applications

What is the minimum speed to escape from 3RE from Earth’s center? (g\=9.8m/s2,RE\=6.4×106m)

ve = 2gRE23RE = 2×9.8×6.4×1063. ve = 4.181×107≈6.47×103m/s≈6.5km/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 6.5 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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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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Energy of Satellites and Binding Energy

A projectile is launched at 9km/s from Earth’s surface. What is its maximum distance from the center? (Escape speed = 11

12vi2−ve22 = −ve22REr. 40.5−62.72 = −62.72REr. rRE = 62.7222.22≈2.82. r = 2.82×6.4×106≈1.80×107m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.8 × 10⁷ m. 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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