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The weight \( W \) of an object varies inversely as the square of the distance \( d \) from the center of the Earth. At sea level (3978 mi from the center of the Earth), an astronaut weighs 124 lb.

Find her weight when she is 102 mi above the surface of the Earth and the spacecraft is not in motion.

(Round to the nearest hundredth as needed.)

Answer :

Final answer:

The weight of an object varies inversely as the square of the distance from the center of the earth. An astronaut weighing 124 lb at sea level (3978 miles from the center) would weigh approximately 118.11 lb when 102 miles above the earth's surface, at a distance of 4080 miles from the center.

Explanation:

The weight W of an object varies inversely as the square of the distance d from the center of the earth. This relationship is given by the formula W = k / d^2, where k is a constant of proportionality.

At sea level, the distance from the center of the earth is 3978 miles, and the astronaut weighs 124 lb. When the astronaut is 102 miles above the surface, the total distance from the center of the earth is 3978 + 102 = 4080 miles.

To find the astronaut's weight at this new distance, we set up a proportion based on the inverse square relationship: 124 / 3978^2 = W / 4080^2.

Solving for W, we find that W = 124 * (3978^2 / 4080^2)

= 124 * (15820884 / 16646400)

≠ 118.11 lb.

Therefore, the astronaut would weigh approximately 118.11 lb when 102 miles above the surface of the earth.

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Rewritten by : Barada