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#orbital energy

14 public questions tagged with this topic.

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 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.

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Gravitational Effects and Applications

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.

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Energy of Satellites and Binding Energy

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.

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Energy of Satellites and Binding Energy

A 250kg satellite orbits Earth at 6RE from the center. What is its total energy? (ME\=6×1024kg,RE\=6.4×106m,G\=6.67×10−1

E = −GMEm2r. r = 6RE = 3.84×107m. E = −6.67×10−11×6×1024×2502×3.84×107. E = −1.001×10177.68×107≈−1.30×109J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to -1.3 × 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.

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Escape Velocity and Bound Systems

What is the total energy of a 100kg satellite orbiting Earth at 3RE from the center? (ME\=6×1024kg,RE\=6.4×106m,G\=6.67×

E = −GMEm2r. r = 3RE = 3×6.4×106 = 1.92×107m. E = −6.67×10−11×6×1024×1002×1.92×107. E = −4.002×10163.84×107≈−1.04×109J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to -1.0 × 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.

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Escape Velocity and Bound Systems

How much energy is required to move a 100kg satellite from 4RE to 8RE from Earth’s center? (ME\=6×1024kg,RE\=6.4×106m,G\

ΔE = −GMEm(1r2−1r1). r1 = 2.56×107m, r2 = 5.12×107m. ΔE = −6.67×10−11×6×1024×100(15.12×107−12.56×107). ΔE = −4.002×1016(−1.953×10−8)≈7.82×108J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 7.8 × 10⁸ 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.

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Escape Velocity and Bound Systems

Energy required to move a 300kg satellite from 3RE to 6RE is? ( ME\=6×1024kg,RE\=6.4×106m,G\=6.67×10−11Nm2/kg2)

ΔE = −GMEm(1r2−1r1). r1 = 3RE = 1.92×107m, r2 = 6RE = 3.84×107m. ΔE = −6.67×10−11×6×1024×300(13.84×107−11.92×107). ΔE = −1.201×1017(−2.604×10−8)≈3.13×109J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 3.1 × 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.

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Work, Energy and Orbital Mechanics

What is the kinetic energy of a 800kg satellite at 10RE from Earth’s center? (ME\=6×1024kg,RE\=6.4×106m,G\=6.67×10−11N m

K = GMEm2r. r = 10RE = 6.4×107m. K = 6.67×10−11×6×1024×8002×6.4×107. K = 3.202×10171.28×108≈2.50×109J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.5 × 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.

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Work, Energy and Orbital Mechanics

How much energy is required to move a 500kg satellite from 9RE to 18RE from Earth’s center? (ME\=6×1024kg,RE\=6.4×106m,G

ΔE = −GMEm(1r2−1r1). r1 = 5.76×107m, r2 = 1.152×108m. ΔE = −6.67×10−11×6×1024×500(11.152×108−15.76×107). ΔE = −2.001×1017(−8.681×10−9)≈1.74×109J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.7 × 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.

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Work, Energy and Orbital Mechanics

A 800kg satellite orbits Earth at 12RE from the center. What is its total energy? (ME\=6×1024kg,RE\=6.4×106m,G\=6.67×10−

E = −GMEm2r. r = 12RE = 7.68×107m. E = −6.67×10−11×6×1024×8002×7.68×107. E = −3.202×10171.536×108≈−2.08×109J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to -2.1 × 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.

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