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#orbital motion

8 public questions tagged with this topic.

Gravitational Effects and Applications

What is the primary source of centripetal force for an Earth satellite?

For a satellite in circular orbit, the centripetal force (F = mv2r) is provided by Earth’s gravitational force (F = GMEmr2), which keeps it in orbit. As per NCERT, applying relevant law/formula with correct units and sign convention leads to Gravitational force. 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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Escape Velocity and Bound Systems

A 7kg mass is moved from 7RE to 14RE from Earth’s center. What is the change in potential energy? (ME\=6×1024kg,RE\=6.4×

ΔV = −GMEm(1r2−1r1). r1 = 4.48×107m, r2 = 8.96×107m. ΔV = −6.67×10−11×6×1024×7(18.96×107−14.48×107). ΔV = −2.801×1015(−1.116×10−8)≈3.13×107J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 3.1 × 10⁷ J. 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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Escape Velocity and Bound Systems

A body is launched from Earth at 11.5km/s. What is its speed at infinity? (Escape speed = 11.2km/s)

vf2 = vi2−ve2. vf2 = (11.5)2−(11.2)2 = 132.25−125.44 = 6.81. vf = 6.81≈2.61km/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.6 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.

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Orbital Velocity and Time Period

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.

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Kepler's Laws and Planetary Motion

A moon orbits a planet with a period of 9 days and radius 6×108m. What is the planet’s mass? (G\=6.67×10−11N m2/kg2,1day

M = 4π2r3GT2. T = 9×86400 = 7.776×105s. T2 = 6.046×1011s2. r3 = (6×108)3 = 2.16×1026m3. M = 4×(3.14)2×2.16×10266.67×10−11×6.046×1011. M = 8.51×10264.033×101≈2.11×1025kg. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.1 × 10²⁵ kg. 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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Variation of g with Height and Depth

A planet’s orbital period around the Sun is 8 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 = 8years, aE = 1.5×1011m. 8212 = ap3(1.5×1011)3. 64 = ap33.375×1033. ap3 = 64×3.375×1033 = 2.16×1035. ap = (2.16×1035)1/3≈6.0×1011m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 6.0 × 10¹¹ m. 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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