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#Earth gravity

2 public questions tagged with this topic.

At what height above Earth’s surface is g reduced to 7.84m/s2? (g0\=9.8m/s2,RE\=6.4×106m)

g(h) = g0(1+h/RE)2. 7.84 = 9.8(1+h/RE)2. (1+h/RE)2 = 1.25. 1+h/RE = 1.25≈1.118. h/RE = 0.118. h = 0.118×6.4×106≈7.55×105m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 7.6 × 10⁵ m. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

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

What is the gravitational potential due to Earth at 3.2×107m from its center? (ME\=6×1024kg,G\=6.67×10−11N m2/kg2)

U = −GMEr. U = −6.67×10−11×6×10243.2×107. U = −1.25×107J/kg. As per NCERT, applying relevant law/formula with correct units and sign convention leads to -1.2 × 10⁷ 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.