#satellite energy

Gravitational Effects and Applications

How much energy is required to move a 700kg satellite from 11RE to 22RE from Earth’s center? (ME\=6×1024kg,RE\=6.4×106m,

ΔE = −GMEm(1r2−1r1). r1 = 7.04×107m, r2 = 1.408×108m. ΔE = −6.67×10−11×6×1024×700(11.408×108−17.04×107). ΔE = −2.801×1017(−7.102×10−9)≈1.99×109J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.0 × 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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Energy of Satellites and Binding Energy

How much energy is required to move a 500kg satellite from 5RE to 10RE from Earth’s center? (ME\=6×1024kg,RE\=6.4×106m,G

ΔE = −GMEm(1r2−1r1). r1 = 3.2×107m, r2 = 6.4×107m. ΔE = −6.67×10−11×6×1024×500(16.4×107−13.2×107). ΔE = −2.001×1017(−1.5625×10−8)≈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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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

A satellite of mass 500kg orbits Earth at 2RE from the center. What is its kinetic energy? (ME\=6×1024kg,RE\=6.4×106m,G\

K = GMEm2r. r = 2RE = 2×6.4×106 = 1.28×107m. K = 6.67×10−11×6×1024×5002×1.28×107. K = 2.001×10172.56×107≈7.82×109J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 7.8 × 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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Orbital Velocity and Time Period

A 600kg satellite orbits Earth at 8RE from the center. What is its total energy? (ME\=6×1024kg,RE\=6.4×106m,G\=6.67×10−1

E = −GMEm2r. r = 8RE = 5.12×107m. E = −6.67×10−11×6×1024×6002×5.12×107. E = −2.401×10171.024×108≈−2.34×109J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to -2.4 × 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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Kepler's Laws and Planetary Motion

What is the kinetic energy of a 300kg satellite at 5RE from Earth’s center? (ME\=6×1024kg,RE\=6.4×106m,G\=6.67×10−11N m2

K = GMEm2r. r = 5RE = 3.2×107m. K = 6.67×10−11×6×1024×3002×3.2×107. K = 1.201×10176.4×107≈1.88×109J. 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 B is scientifically correct.

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

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Kepler's Laws and Planetary Motion

How much energy is required to move a 600kg satellite from 10RE to 20RE from Earth’s center? (ME\=6×1024kg,RE\=6.4×106m,

ΔE = −GMEm(1r2−1r1). r1 = 6.4×107m, r2 = 1.28×108m. ΔE = −6.67×10−11×6×1024×600(11.28×108−16.4×107). ΔE = −2.401×1017(−7.8125×10−9)≈1.88×109J. 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 B is scientifically correct.

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

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Universal Law of Gravitation

A 400kg satellite orbits Earth at 5RE from the center. What is its total energy? (ME\=6×1024kg,RE\=6.4×106m,G\=6.67×10−1

E = −GMEm2r. r = 5RE = 3.2×107m. E = −6.67×10−11×6×1024×4002×3.2×107. E = −1.601×10176.4×107≈−2.5×109J. 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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