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Work, Energy and Orbital Mechanics

Questions covering work, energy principles, and orbital mechanics, including applications in physics exams. Suitable for students studying these topics.

25 questions

A planet orbits the Sun with a period of 6 years. If Earth’s orbital radius is 1.5×1011m, what is its semi-major axis?

Kepler’s third law: Tp2TE2 = ap3aE3. TE = 1year, Tp = 6years, aE = 1.5×1011m. 6212 = ap3(1.5×1011)3. 36 = ap33.375×1033. ap3 = 36×3.375×1033 = 1.215×1035. ap = (1.215×1035)1/3≈4.95×1011m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 5.0 × 10¹¹ m. 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 satellite orbits Earth at 17RE from the center. What is its orbital speed? (g\=9.8m/s2,RE\=6.4×106m)

v = gRE2r. r = 17RE, v = 9.8×6.4×10617. v = 3.694×106≈1.92×103m/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.9 × 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 near Earth has a period of 90 minutes. What is its period at h\=3RE? (RE\=6.4×106m)

T2∝(RE+h)3. T02 = kRE3, h = 3RE, r = 4RE. T2 = k(4RE)3 = 64kRE3. T = T064 = 90×8 = 720min. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 720 min. 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 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.

Three masses of 18kg each form an equilateral triangle with side 12m. What is the net force on one mass? (G\=6.67×10−11N

Force between two masses: F = Gm2r2 = 6.67×10−1118×18122 = 1.50×10−10N. Two forces at 60°: FR = F2+F2+2F2cos⁡60∘. FR = (1.50×10−10)2(1+1+1) = 1.50×10−103. FR≈2.60×10−10N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.6 × 10⁻¹⁰ N. 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 a planet at 2.5×107m from its center with a period of 5 hours. What is the planet’s mass? (G\=6.67×10

M = 4π2r3GT2. T = 5×3600 = 18000s, T2 = 3.24×108s2. r3 = (2.5×107)3 = 1.5625×1022m3. M = 4×(3.14)2×1.5625×10226.67×10−11×3.24×108. M = 6.158×10232.161×10−2≈2.85×1025kg. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.9 × 10²⁵ kg. 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 does Kepler’s third law reveal about the motion of planets farther from the Sun?

Kepler’s third law (T2∝a3) shows that planets farther from the Sun (larger semi-major axis a) have longer orbital periods (T), as the period increases with the distance cubed. As per NCERT, applying relevant law/formula with correct units and sign convention leads to They have longer periods. 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.