The weight of an object above the surface of the Earth varies inversely with the square of the
distance from the center of the Earth. If a body weighs 50 pounds when it is 3,960 miles from
Earth's center, what would it weigh if it were 4,015 miles from Earth's center?

[tex]w = \frac{k}{ {d}^{2} } \\ 50 = \frac{k}{ {3960}^{2} } \\ \\ k = 50 \times {3960}^{2} \\ k = 50 \times 15681600 \\ k = 784080000 \\ \\ w = \frac{784080000}{ {d}^{2} } \\ w = \frac{784080000}{16120225} \\ \\ w = 48.71228786 \\ w = 48.7pounds[/tex]

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The weight of an object above the surface of the Earth varies inversely with the square of the distance from the center of the Earth. If a body weighs 50 pounds when it is 3,960 miles from Earth's center, what would it weigh if it were 4,015 miles from Earth's center?

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The weight of an object above the surface of the Earth varies inversely with the square of the
distance from the center of the Earth. If a body weighs 50 pounds when it is 3,960 miles from
Earth's center, what would it weigh if it were 4,015 miles from Earth's center?