SOLUTION: I need help with a Precalculus problem: 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 t

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Question 650148: I need help with a Precalculus problem:
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 220 lb. Find their weight when they are orbiting 400 mi above sea level.

Found 2 solutions by Alan3354, josmiceli:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
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 220 lb. Find their weight when they are orbiting 400 mi above sea level.
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W = k/r^2
220 = k/3978^2
You can solve for k, but it's not necessary.
220%2A3978%5E2+=+W%2A%283978+%2B+400%29%5E2
W = 181.64 pounds

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+W+=+k%2A%281%2Fd%29+
+k+ is called the constant of proportionality
given:
+W+=+220+ pounds
+d+=+3978+ mi
+220+=+k%2A%28+1%2F3978+%29+
+k+=+220%2A3978+
+k+=+875160+
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+W+=+875160%2A%281%2F%28+3978+%2B+400+%29%29+
+W+=+875160+%2F+4378+
+W+=+199.8995+
The astronaut weighs about 200 pounds