# Weight of An Astronaut

• Feb 2nd 2010, 10:04 AM
sologuitar
Weight of An Astronaut
The weight of an object varies inversely with the square of its distance from the center of the earth. Assume that the radius of the earth is 3960 miles. If an astronaut weighs 180 pounds on the surface of the earth, what does that astronaut weigh when traveling 300 miles above the surface of the earth?

MY WORK:

I developed the equation W = k/(r^2) from the given information, where W = weight of an object, r = radius distance from the center of the earth. After doing the math, I got nowhere near what the answer should be.

I then tried setting up the following proportion:

180 pounds/x pounds = 3960 miles/300 miles

After solving for x, I got 13.64 pounds, which is wrong.

The correct answer is 1.56 pounds.

How do I get that answer? What is the correct equation?
• Feb 2nd 2010, 11:31 AM
HallsofIvy
Quote:

Originally Posted by sologuitar
The weight of an object varies inversely with the square of its distance from the center of the earth. Assume that the radius of the earth is 3960 miles. If an astronaut weighs 180 pounds on the surface of the earth, what does that astronaut weigh when traveling 300 miles above the surface of the earth?

MY WORK:

I developed the equation W = k/(r^2) from the given information, where W = weight of an object, r = radius distance from the center of the earth. After doing the math, I got nowhere near what the answer should be.

I then tried setting up the following proportion:

180 pounds/x pounds = 3960 miles/300 miles

After solving for x, I got 13.64 pounds, which is wrong.

The correct answer is 1.56 pounds.

How do I get that answer? What is the correct equation?

You have three errors.

1) The 300 miles is above the surface of the earth, not from the center of the earth. If the astronaut is 300 miles above the earth and the radius of the earth is 3960 miles, the astronaut is 3960+ 300= 4260 miles from the center of the earth.

2) Your formula has $r^2$ but you didn't square!

3) Since weight is inversely proportional to the distance, you should have $\frac{weight_1}{weight_2}= \frac{radius_2^2}{radius_1^2}$

Try $\frac{x}{180}= \frac{3960^2}{4260^2}$.

But that is NOT the 1.54 pounds. 300 miles is relatively small compared to 3960 so you really haven't gone a great distance, comparatively. And how is the astronauts weight to be measured? Is there more to the problem you haven't told us?
• Feb 2nd 2010, 02:35 PM
sologuitar
ok...
Quote:

Originally Posted by HallsofIvy
You have three errors.

1) The 300 miles is above the surface of the earth, not from the center of the earth. If the astronaut is 300 miles above the earth and the radius of the earth is 3960 miles, the astronaut is 3960+ 300= 4260 miles from the center of the earth.

I tried setting up a right triangle in a circle to find the distance from the surface of the earth to the location of the astronuat not realizing that it was given to be 300 miles. After reading your reply, I drew a picture to see where my error stemmed from.

2) Your formula has $r^2$ but you didn't square!

Squaring would not have made a difference because I had the wrong information in the proportion.

3) Since weight is inversely proportional to the distance, you should have $\frac{weight_1}{weight_2}= \frac{radius_2^2}{radius_1^2}$

Try $\frac{x}{180}= \frac{3960^2}{4260^2}$.

But that is NOT the 1.54 pounds. 300 miles is relatively small compared to 3960 so you really haven't gone a great distance, comparatively. And how is the astronauts weight to be measured? Is there more to the problem you haven't told us?

There is nothing missing from this word problem. I checked the precalculus textbook to make sure and it's all the same wording.