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#gravitational acceleration

21 public questions tagged with this topic.

At what height above Earth’s surface is g reduced to 5.88m/s2? (g0\=9.8m/s2,RE\=6.4×106m)

g(h) = g0(1+h/RE)2. 5.88 = 9.8(1+h/RE)2. (1+h/RE)2 = 1.667. 1+h/RE = 1.667≈1.291. h/RE = 0.291. h = 0.291×6.4×106≈1.86×106m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.9 × 10⁶ m. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and 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.

At what height above Earth’s surface is g reduced to 8.82m/s2? (g0\=9.8m/s2,RE\=6.4×106m)

g(h) = g0(1+h/RE)2. 8.82 = 9.8(1+h/RE)2. (1+h/RE)2 = 1.111. 1+h/RE = 1.111≈1.054. h/RE = 0.054. h = 0.054×6.4×106≈3.46×105m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 3.5 × 10⁵ m. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

What is the key assumption about Earth’s density in calculating g below its surface?

Considering the Earth has uniform density to simplify the mass contributing to gravity inside (Mr = MEr3RE3), leading to g(d) = g(1−d/RE). Non-uniform density would complicate this. As per NCERT, applying relevant law/formula with correct units and sign convention leads to It is uniform. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

At what height above Earth’s surface is g reduced to 7.84m/s2? (g0\=9.8m/s2,RE\=6.4×106m)

g(h) = g0(1+h/RE)2. 7.84 = 9.8(1+h/RE)2. (1+h/RE)2 = 1.25. 1+h/RE = 1.25≈1.118. h/RE = 0.118. h = 0.118×6.4×106≈7.55×105m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 7.6 × 10⁵ m. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and 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.

Why is the value of g maximum at Earth’s surface?

At the surface. Above, g decreases with 1/(RE+h)2; below, only inner mass contributes, reducing g linearly as g(d) = g(1−d/RE). Thus, g peaks at r = RE. As per NCERT, applying relevant law/formula with correct units and sign convention leads to All mass contributes at the surface. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

What is the minimum speed to escape from 8RE from Earth’s center? (g\=9.8m/s2,RE\=6.4×106m)

ve = 2gRE28RE = 2×9.8×6.4×1068. ve = 1.568×107≈3.96×103m/s≈4.0km/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 4.0 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.

At what depth below Earth’s surface is g reduced to 8.33m/s2? (g0\=9.8m/s2,RE\=6.4×106m)

g(d) = g0(1−d/RE). 8.33 = 9.8(1−d/RE). 1−d/RE = 0.85. d/RE = 0.15. d = 0.15×6.4×106 = 9.6×105m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 9.6 × 10⁵ m. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

A satellite orbits Earth at 15RE from the center. What is its orbital speed? (g\=9.8m/s2,RE\=6.4×106m)

v = gRE2r. r = 15RE, v = 9.8×6.4×10615. v = 4.181×106≈2.04×103m/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.0 × 10³ m/s. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.

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