44 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.
Escape energy from altitude h is GMmR+h, while energy to reach h from the surface is GMm(1R−1R+h), which is less. Escape requires overcoming all potential energy to infinity. As per NCERT, applying relevant law/formula with correct units and sign convention leads to Escape needs energy to infinity. 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.
ve = 2gRE23RE = 2×9.8×6.4×1063. ve = 4.181×107≈6.47×103m/s≈6.5km/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 6.5 km/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.
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.
Escape speed as ve = 2GMR. The mass m of the object cancels out in the energy conservation equation (12mve2 = GMmR), making ve dependent only on the planet’s mass M and radius R.
Ref: Halliday & Resnick, Fundamentals of Physics, 11th ed., Chapter 13: Gravitation.
Distance to 8kg: 8−3 = 5m. U = −Gm1r1−Gm2r2. U = −6.67×10−11(43+85). U = −6.67×10−11(1.333+1.6) = −1.96×10−10J/kg. As per NCERT, applying relevant law/formula with correct units and sign convention leads to -2.0 × 10⁻¹⁰ J/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.
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.
U = −GMEr. U = −6.67×10−11×6×10241.28×107. U = −3.125×107J/kg. As per NCERT, applying relevant law/formula with correct units and sign convention leads to -3.1 × 10⁷ J/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.
Distance to 20kg: 20−9 = 11m. U = −Gm1r1−Gm2r2. U = −6.67×10−11(109+2011). U = −6.67×10−11(1.111+1.818)≈−1.95×10−10J/kg. As per NCERT, applying relevant law/formula with correct units and sign convention leads to -1.9 × 10⁻¹⁰ J/kg. 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.
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.
E = −GMEm2r. r = 18RE = 1.152×108m. E = −6.67×10−11×6×1024×11002×1.152×108. E = −4.402×10172.304×108≈−1.91×109J. 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.
