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#orbital radius

9 public questions tagged with this topic.

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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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 has a moon with an orbital radius of 5×108m and a period of 10 days. What is the planet’s mass? (G\=6.67×10−11N

T2 = 4π2GMR3. M = 4π2R3GT2. T = 10×86400 = 8.64×105s. T2 = (8.64×105)2 = 7.465×1011s2. R3 = (5×108)3 = 1.25×1025m3. M = 4×(3.14)2×1.25×10256.67×10−11×7.465×1011. M = 4.93×10264.979×101≈9.9×1024kg. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 9.9 × 10²⁴ 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.

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Orbital Velocity and Time Period

A 1000kg satellite orbits Earth at 16RE from the center. What is its total energy? (ME\=6×1024kg,RE\=6.4×106m,G\=6.67×10

E = −GMEm2r. r = 16RE = 1.024×108m. E = −6.67×10−11×6×1024×10002×1.024×108. E = −4.002×10172.048×108≈−1.95×109J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to -1.9 × 10⁹ J. 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.

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Orbital Velocity and Time Period

How much energy is required to move a 300kg satellite from 7RE to 14RE from Earth’s center? (ME\=6×1024kg,RE\=6.4×106m,G

ΔE = −GMEm(1r2−1r1). r1 = 4.48×107m, r2 = 8.96×107m. ΔE = −6.67×10−11×6×1024×300(18.96×107−14.48×107). ΔE = −1.201×1017(−1.116×10−8)≈1.34×109J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.3 × 10⁹ J. 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.

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Orbital Velocity and Time Period

What is the kinetic energy of a 200kg satellite at 4RE from Earth’s center? (ME\=6×1024kg,RE\=6.4×106m,G\=6.67×10−11N m2

K = GMEm2r. r = 4RE = 2.56×107m. K = 6.67×10−11×6×1024×2002×2.56×107. K = 8.004×10165.12×107≈1.56×109J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.6 × 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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Orbital Velocity and Time Period

A moon orbits a planet with period 8 days and radius 3×108m. What is the planet’s mass? (G\=6.67×10−11N m2/kg2,1day\=864

M = 4π2r3GT2. T = 8×86400 = 6.912×105s. T2 = 4.776×1011s2. r3 = (3×108)3 = 2.7×1025m3. M = 4×(3.14)2×2.7×10256.67×10−11×4.776×1011. M = 1.065×10263.185×101≈3.34×1024kg. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 3.4 × 10²⁴ 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.

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