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#Earth radius

20 public questions tagged with this topic.

A satellite orbits Earth at 2.5RE from the center. What is its orbital speed? (g\=9.8m/s2,RE\=6.4×106m)

v = gRE2r. r = 2.5RE, v = 9.8×6.4×1062.5. v = 2.509×107≈5.01×103m/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 5.0 × 10³ m/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 body weighs 147N on Earth’s surface. What is its weight at a height h\=RE/3? (g\=9.8m/s2)

g(h) = g(1+h/RE)2. h = RE/3, 1+h/RE = 4/3. g(h) = 9.8(4/3)2 = 9.8×916 = 5.51m/s2. Mass: m = 147/9.8 = 15kg. Weight: W = 15×5.51≈82.7N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 83 N. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

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

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.

What is the kinetic energy of a 900kg satellite at 11RE from Earth’s center? (ME\=6×1024kg,RE\=6.4×106m,G\=6.67×10−11N m

K = GMEm2r. r = 11RE = 7.04×107m. K = 6.67×10−11×6×1024×9002×7.04×107. K = 3.602×10171.408×108≈2.56×109J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.6 × 10⁹ J. 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 kinetic energy of a 150kg satellite at 3RE from Earth’s center? (ME\=6×1024kg,RE\=6.4×106m,G\=6.67×10−11N m2

K = GMEm2r. r = 3RE = 1.92×107m. K = 6.67×10−11×6×1024×1502×1.92×107. K = 6.003×10163.84×107≈1.56×109J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.6 × 10⁹ J. 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 kinetic energy of a 500kg satellite at 7RE from Earth’s center? (ME\=6×1024kg,RE\=6.4×106m,G\=6.67×10−11N m2

K = GMEm2r. r = 7RE = 4.48×107m. K = 6.67×10−11×6×1024×5002×4.48×107. K = 2.001×10178.96×107≈2.23×109J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.3 × 10⁹ J. 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.

A 1100kg satellite orbits Earth at 18RE from the center. What is its total energy? (ME\=6×1024kg,RE\=6.4×106m,G\=6.67×10

E = −GMEm2r. r = 18RE = 1.152×108m. E = −6.67×10−11×6×1024×11002×1.152×108. E = −4.402×10172.304×108≈−1.91×109J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to -1.9 × 10⁹ J. 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.

A satellite orbits Earth at 19RE from the center. What is its orbital speed? (g\=9.8m/s2,RE\=6.4×106m)

v = gRE2r. r = 19RE, v = 9.8×6.4×10619. v = 3.301×106≈1.82×103m/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.8 × 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.

A satellite orbits Earth at 11RE from the center. What is its orbital speed? (g\=9.8m/s2,RE\=6.4×106m)

v = gRE2r. r = 11RE, v = 9.8×6.4×10611. v = 5.698×106≈2.39×103m/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.4 × 10³ m/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 700kg satellite orbits Earth at 10RE from the center. What is its total energy? (ME\=6×1024kg,RE\=6.4×106m,G\=6.67×10−

E = −GMEm2r. r = 10RE = 6.4×107m. E = −6.67×10−11×6×1024×7002×6.4×107. E = −2.801×10171.28×108≈−2.19×109J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to -2.2 × 10⁹ J. 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.