A body of mass 1 kg starts from rest at infinity and falls freely toward the Earth. Neglect air resistance and Earth's rotation. Taking the Earth's radius as 6400 km and g = 9.8 m·s⁻² at the surface, what are the speed and kinetic energy of the body just before it reaches the Earth's surface?
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Question

A body of mass 1 kg starts from rest at infinity and falls freely toward the Earth. Neglect air resistance and Earth's rotation. Taking the Earth's radius as 6400 km and g = 9.8 m·s⁻² at the surface, what are the speed and kinetic energy of the body just before it reaches the Earth's surface?

Options

G1 Physics · Accepted Answer

Take gravitational potential energy to be zero at infinity. Initially, Kᵢ = 0, Uᵢ = 0. At the Earth's surface, U(f) = -GMm ÷ R. Conservation of mechanical energy gives 0 = (1 ÷ 2)mv² - GMm ÷ R. Therefore, v² = 2GM ÷ R. Since g = GM ÷ R², we have GM ÷ R = gR. Hence, v = √(2gR). Substitute g = 9.8 m·s⁻², R = 6400 km = 6.4×10⁶ m. Thus, v = √[2×9.8×6.4×10⁶] ≈ 1.12×10⁴ m·s⁻¹ = 11.2 km·s⁻¹. The kinetic energy is K = (1 ÷ 2)mv². For m = 1 kg, K = (1 ÷ 2)(1)(1.12×10⁴)² ≈ 6.27×10⁷ J. Correct option: A.