#semi-major axis

Universal Law of Gravitation

A planet orbits the Sun with a period of 5 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 = 5years, aE = 1.5×1011m. 5212 = ap3(1.5×1011)3. 25 = ap33.375×1033. ap3 = 25×3.375×1033 = 8.4375×1034. ap = (8.4375×1034)1/3≈4.39×1011m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 4.4 × 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.

View Details

Gravitational Effects and Applications

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.

View Details

Orbital Velocity and Time Period

A planet orbits the Sun with a period of 12 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 = 12years, aE = 1.5×1011m. 12212 = ap3(1.5×1011)3. 144 = ap33.375×1033. ap3 = 144×3.375×1033 = 4.86×1035. ap = (4.86×1035)1/3≈7.86×1011m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 7.9 × 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.

View Details

Kepler's Laws and Planetary Motion

A planet orbits the Sun with a period of 7 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 = 7years, aE = 1.5×1011m. 7212 = ap3(1.5×1011)3. 49 = ap33.375×1033. ap3 = 49×3.375×1033 = 1.65375×1035. ap = (1.65375×1035)1/3≈5.49×1011m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 5.5 × 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.

View Details

Work, Energy and Orbital Mechanics

A planet has an orbital period of 4 years around the Sun. If Earth’s orbital radius is 1.5×1011m, what is the planet’s s

Kepler’s third law: T2∝a3, Tp2TE2 = ap3aE3. TE = 1year, Tp = 4years, aE = 1.5×1011m. 4212 = ap3(1.5×1011)3. 16 = ap33.375×1033. ap3 = 16×3.375×1033 = 5.4×1034. ap = (5.4×1034)1/3≈3.78×1011m.

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

View Details

Work, Energy and Orbital Mechanics

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.

View Details