27 public questions tagged with this topic.
Escape speed depends on planet parameters (option 1 correct), is independent of direction (option 3 correct), and mass (option 4 correct). Option 2 is incorrect as it decreases with altitude.
Ref: Halliday & Resnick, Fundamentals of Physics, 11th ed., Chapter 13: Gravitation.
Negative total energy (E = KE+PE
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
ΔE = −GMEm(1r2−1r1). r1 = 7.04×107m, r2 = 1.408×108m. ΔE = −6.67×10−11×6×1024×700(11.408×108−17.04×107). ΔE = −2.801×1017(−7.102×10−9)≈1.99×109J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.0 × 10⁹ J. 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.
In an elliptical orbit, total energy is constant, but kinetic energy (KE = 12mv2) varies as speed changes with distance from the center (faster closer, slower farther), balancing the changing potential energy.
Ref: Halliday & Resnick, Fundamentals of Physics, 11th ed., Chapter 13: Gravitation.
ΔE = −GMEm(1r2−1r1). r1 = 3.2×107m, r2 = 6.4×107m. ΔE = −6.67×10−11×6×1024×500(16.4×107−13.2×107). ΔE = −2.001×1017(−1.5625×10−8)≈3.13×109J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 3.1 × 10⁹ J. 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.
Total energy: E = 12mvi2−GMEmRE. At infinity: E = 12mvf2, ve = 2GMERE. 12vi2−12ve2 = 12vf2. vf2 = (15)2−(11.2)2 = 225−125.44 = 99.56. vf = 99.56≈10km/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 10 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.
Gravity is a conservative force, so the total mechanical energy (E = KE+PE) is conserved in an elliptical orbit, with kinetic and potential energies trading off but summing to a constant.
Ref: Halliday & Resnick, Fundamentals of Physics, 11th ed., Chapter 13: Gravitation.
12vi2−ve22 = −ve22REr. 0.5−62.72 = −62.72REr. rRE = 62.7262.22≈1.008. r = 1.008×6.4×106≈6.45×106m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 6.5 × 10⁶ 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.
M = 4π2r3GT2. T = 6×86400 = 5.184×105s. T2 = 2.689×1011s2. r3 = (4×108)3 = 6.4×1025m3. M = 4×(3.14)2×6.4×10256.67×10−11×2.689×1011. M = 2.523×10261.794×101≈1.41×1025kg. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.4 × 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.
M = 4π2r3GT2. T = 11×86400 = 9.504×105s. T2 = 9.028×1011s2. r3 = (8×108)3 = 5.12×1026m3. M = 4×(3.14)2×5.12×10266.67×10−11×9.028×1011. M = 2.019×10276.022×101≈3.35×1025kg. 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 C is scientifically correct.
Ref: Halliday & Resnick, Fundamentals of Physics, 11th ed., Chapter 13: Gravitation.
M = 4π2r3GT2. T = 12×86400 = 1.0368×106s. T2 = 1.075×1012s2. r3 = (9×108)3 = 7.29×1026m3. M = 4×(3.14)2×7.29×10266.67×10−11×1.075×1012. M = 2.875×10277.171×101≈4.01×1025kg. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 4.0 × 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.
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.
