A planet has a mass of 4.0 × 10^24 kg and a radius of 6.0 × 10^6 m. What is the acceleration due to gravity at the surface of this planet? (Use G = 6.67 × 10^-11 N·m²/kg²)
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Correct answer
C. 7.4 m/s²
Principle or equation
Newton's law of universal gravitation gives the gravitational field strength at a planet's surface: g = GM / R².
Why this answer is correct
g = (6.67 × 10^-11 × 4.0 × 10^24) / (6.0 × 10^6)² = (2.668 × 10^14) / (3.6 × 10^13) = 7.41 m/s², approximately 7.4 m/s².
Example
For Earth, M = 6.0 × 10^24 kg, R = 6.4 × 10^6 m, g = 6.67 × 10^-11 × 6.0 × 10^24 / (6.4 × 10^6)² ≈ 9.8 m/s².
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