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#Kepler's laws

16 public questions tagged with this topic.

A planet orbits the Sun with a period of 10 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 = 10years, aE = 1.5×1011m. 10212 = ap3(1.5×1011)3. 100 = ap33.375×1033. ap3 = 100×3.375×1033 = 3.375×1035. ap = (3.375×1035)1/3≈6.96×1011m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 7.0 × 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.

What role does the Sun play in planetary motion according to Kepler’s laws?

The Sun is at one focus of the elliptical orbits (first law) and provides the central gravitational force driving the equal-area sweep (second law) and period-distance relation (third law). As per NCERT, applying relevant law/formula with correct units and sign convention leads to It provides the central 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.

Which of Kepler’s laws implies that the gravitational force acting on a planet is a central force?

Kepler’s second law (law of areas) states that the line joining a planet to the Sun sweeps out equal areas in equal times. This is a consequence of the conservation of angular momentum, which holds true for a central force directed along the line joining the two bodies (e.g., the Sun and planet). A central force ensures no torque, preserving angular momentum.

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

A satellite’s period is 100 minutes at h\=0. What is its period at h\=2RE? (RE\=6.4×106m)

T2∝(RE+h)3. T02 = kRE3, T2 = k(3RE)3 = 27kRE3. T = T027 = 100×5.196≈520min. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 520 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.

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.

Which of the following statements is correct about Kepler’s laws?

Kepler’s laws apply to any central force following an inverse-square law (e.g., gravitation), making option 4 correct as they describe planetary motion around the Sun. They require circular orbits 1 2 3 4 Correct ans: Option (4) They apply only to Earth They depend on planet mass They apply to inverse-square forces 4

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

A planet orbits the Sun with a period of 3 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 = 3years, aE = 1.5×1011m. 3212 = ap3(1.5×1011)3. 9 = ap33.375×1033. ap3 = 9×3.375×1033 = 3.0375×1034. ap = (3.0375×1034)1/3≈3.12×1011m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 3.2 × 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.