#orbital period

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.

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Universal Law of Gravitation

A satellite orbits a planet at 8×107m from its center with a period of 20 hours. What is the planet’s mass? (G\=6.67×10−

M = 4π2r3GT2. T = 20×3600 = 72000s, T2 = 5.184×109s2. r3 = (8×107)3 = 5.12×1023m3. M = 4×(3.14)2×5.12×10236.67×10−11×5.184×109. M = 2.019×10253.458×10−1≈5.84×1025kg. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 5.8 × 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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Universal Law of Gravitation

A satellite orbits a planet at 3×107m from its center with a period of 6 hours. What is the planet’s mass? (G\=6.67×10−1

M = 4π2r3GT2. T = 6×3600 = 21600s, T2 = 4.6656×108s2. r3 = (3×107)3 = 2.7×1022m3. M = 4×(3.14)2×2.7×10226.67×10−11×4.6656×108. M = 1.065×10243.112×10−2≈3.42×1025kg. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 3.4 × 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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Universal Law of Gravitation

A satellite near Earth has a period of 89 minutes. What is its period at h\=7RE? (RE\=6.4×106m)

T2∝(RE+h)3. T02 = kRE3, h = 7RE, r = 8RE. T2 = k(8RE)3 = 512kRE3. T = T0512 = 89×22.63≈2014min. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2010 min. 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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Gravitational Effects and Applications

A moon orbits a planet with a period of 10 days and radius 7×108m. What is the planet’s mass? (G\=6.67×10−11N m2/kg2,1da

M = 4π2r3GT2. T = 10×86400 = 8.64×105s. T2 = 7.465×1011s2. r3 = (7×108)3 = 3.43×1026m3. M = 4×(3.14)2×3.43×10266.67×10−11×7.465×1011. M = 1.353×10274.979×101≈2.72×1025kg. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.7 × 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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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.

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Gravitational Effects and Applications

A planet moves in an elliptical orbit around the Sun with a semi-major axis of 2.25×1011m. If its orbital period is 2 ye

Using Kepler’s third law: T2 = 4π2GMsa3. Rearrange for Ms: Ms = 4π2a3GT2. T = 2×3.156×107 = 6.312×107s. a = 2.25×1011m. T2 = (6.312×107)2 = 3.984×1015s2. a3 = (2.25×1011)3 = 1.139×1033m3. Ms = 4×(3.14)2×1.139×10336.67×10−11×3.984×1015. Ms = 4.49×10342.657×105≈1.69×1030kg.

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

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Gravitational Effects and Applications

A satellite near Earth has a period of 86 minutes. What is its period at h\=5RE? (RE\=6.4×106m)

T2∝(RE+h)3. T02 = kRE3, h = 5RE, r = 6RE. T2 = k(6RE)3 = 216kRE3. T = T0216 = 86×14.7≈1264min. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1270 min. This satisfies dimensional consistency and physical conditions given, so option D is scientifically correct.

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

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Gravitational Effects and Applications

A moon orbits a planet with a period of 7 days and radius 5×108m. What is the planet’s mass? (G\=6.67×10−11N m2/kg2,1day

M = 4π2r3GT2. T = 7×86400 = 6.048×105s. T2 = 3.658×1011s2. r3 = (5×108)3 = 1.25×1025m3. M = 4×(3.14)2×1.25×10256.67×10−11×3.658×1011. M = 4.93×10252.44×101≈2.02×1024kg. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.0 × 10²⁴ 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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Escape Velocity and Bound Systems

Why does a satellite in a higher orbit have a longer orbital period?

T2∝r3 for satellites. A higher orbit (larger r) increases T2 by r3, so the period T increases as the radius grows, reflecting weaker gravitational pull at greater distances. As per NCERT, applying relevant law/formula with correct units and sign convention leads to Gravitational force decreases. 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

The time period of a satellite near Earth’s surface is 84 minutes. What is its time period at a height h\=RE? (RE\=6.4×1

T2 = k(RE+h)3, where k = 4π2GME. For h = 0, T0 = 84min, T02 = kRE3. For h = RE, r = 2RE, T2 = k(2RE)3 = 8kRE3. T = 8T0 = 2.828×84≈237min.

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

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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.

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