At a distance of 14,000 km from the center of Planet Z-99,the free-fall acceleration is 32 m/s².What is the free-fall acceleration at a point 28,000 km from the center of this planet? A)8.0 m/s² B)16 m/s² C)128 m/s² D)4.0 m/s² E)2.0 m/s²

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Quizplus · Accepted Answer

The free-fall acceleration is inversely proportional to the square of the distance from the center of the planet. Doubling the distance (from 14,000 km to 28,000 km) decreases the acceleration by a factor of 4 (since 2^2 = 4), so 32 m/s^2 divided by 4 equals 8 m/s^2.

At a Distance of 14,000 Km from the Center of Planet