Orbital Velocity and Time Period

Orbital Velocity and Time Period

A 5kg mass is moved from RE to 2RE from Earth’s center. What is the work done? (ME\=6×1024kg,RE\=6.4×106m,G\=6.67×10−11N

W = ΔV = −GMEm(1r2−1r1). r1 = 6.4×106m, r2 = 1.28×107m. W = −6.67×10−11×6×1024×5(11.28×107−16.4×106). W = −2.001×1015(−7.8125×10−8)≈1.56×108J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.6 × 10⁸ J. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

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

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Orbital Velocity and Time Period

At what height above Earth’s surface is g reduced to 6.12m/s2? (g0\=9.8m/s2,RE\=6.4×106m)

g(h) = g0(1+h/RE)2. 6.12 = 9.8(1+h/RE)2. (1+h/RE)2 = 1.601. 1+h/RE = 1.601≈1.265. h/RE = 0.265. h = 0.265×6.4×106≈1.7×106m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.7 × 10⁶ m. 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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Orbital Velocity and Time Period

A 8kg mass is moved from 8RE to 16RE from Earth’s center. What is the change in potential energy? (ME\=6×1024kg,RE\=6.4×

ΔV = −GMEm(1r2−1r1). r1 = 5.12×107m, r2 = 1.024×108m. ΔV = −6.67×10−11×6×1024×8(11.024×108−15.12×107). ΔV = −3.202×1015(−9.766×10−9)≈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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Orbital Velocity and Time Period

A satellite orbits at 3RE from Earth’s center. What is its speed? (g\=9.8m/s2,RE\=6.4×106m)

v = gRE2r. r = 3RE, v = 9.8×6.4×1063. v = 2.09×107≈4.57×103m/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 4.6 × 10³ m/s. 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 600kg satellite orbits Earth at 8RE from the center. What is its total energy? (ME\=6×1024kg,RE\=6.4×106m,G\=6.67×10−1

E = −GMEm2r. r = 8RE = 5.12×107m. E = −6.67×10−11×6×1024×6002×5.12×107. E = −2.401×10171.024×108≈−2.34×109J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to -2.4 × 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

What is the gravitational force on a 7kg mass 6m from the center of a spherical shell of mass 300kg and radius 4m? (G\=6

Outside shell: F = GMmr2. F = 6.67×10−11×300×762. F = 1.401×10−836≈3.89×10−10N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 3.9 × 10⁻¹⁰ N. 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

Three masses of 8kg each form an equilateral triangle with side 6m. What is the net force on one mass? (G\=6.67×10−11N m

Force between two masses: F = Gm2r2 = 6.67×10−118×862 = 1.185×10−10N. Two forces at 60°: FR = F2+F2+2F2cos⁡60∘. FR = (1.185×10−10)2(1+1+1) = 1.185×10−103. FR≈2.05×10−10N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.1 × 10⁻¹⁰ N. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

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

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Orbital Velocity and Time Period

What is the escape speed from a planet with mass 2.4×1024kg and radius 4×106m? (G\=6.67×10−11N m2/kg2)

ve = 2GMR. ve = 2×6.67×10−11×2.4×10244×106. ve = 8.002×107≈8.95×103m/s≈8.95km/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 8.9 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

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

T2∝a3, meaning the period squared is proportional to the semi-major axis cubed, making option 2 correct. As per NCERT, applying relevant law/formula with correct units and sign convention leads to Period squared is proportional to distance cubed. 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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Orbital Velocity and Time Period

A projectile is launched at 5km/s from Earth’s surface. What is its maximum distance from the center? (Escape speed = 11

12vi2−ve22 = −ve22REr. 12.5−62.72 = −62.72REr. rRE = 62.7250.22≈1.25. r = 1.25×6.4×106 = 8.0×106m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 8.0 × 10⁶ 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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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

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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