Kinetic Energy of a 400 kg Satellite at 6RE

Practice question

Question

What is the kinetic energy of a 400kg satellite at 6RE from Earth’s center? (ME\=6×1024kg,RE\=6.4×106m,G\=6.67×10−11N m2/kg2)

Options

Choose one · Correct answer highlighted

2.0 × 10⁹ J 2.1 × 10⁹ J Correct 2.2 × 10⁹ J 2.3 × 10⁹ J

Explanation

K = GMEm2r. r = 6RE = 3.84×107m. K = 6.67×10−11×6×1024×4002×3.84×107. K = 1.601×10177.68×107≈2.08×109J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.1 × 10⁹ J. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

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