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#escape speed

22 public questions tagged with this topic.

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.

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.

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.

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.

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.

A projectile is launched from Earth with a speed of 15km/s. What is its speed at infinity? (Escape speed = 11.2km/s)

Total energy: E = 12mvi2−GMEmRE. At infinity: E = 12mvf2, ve = 2GMERE. 12vi2−12ve2 = 12vf2. vf2 = (15)2−(11.2)2 = 225−125.44 = 99.56. vf = 99.56≈10km/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 10 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.

A projectile is launched at 1km/s from Earth’s surface. What is its maximum distance from the center? (Escape speed = 11

12vi2−ve22 = −ve22REr. 0.5−62.72 = −62.72REr. rRE = 62.7262.22≈1.008. r = 1.008×6.4×106≈6.45×106m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 6.5 × 10⁶ 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.

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.

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.

What is the minimum speed to escape from 7RE from Earth’s center? (g\=9.8m/s2,RE\=6.4×106m)

ve = 2gRE27RE = 2×9.8×6.4×1067. ve = 1.79×107≈4.23×103m/s≈4.2km/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 4.2 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.