Earth Surface

Question

The weight of an object varies inversely as the second power of the distance from the center of the earth. The radius of earth is approximately 4000 miles. If a person weights 160 pounds on Earth's surface, what would this individual weigh 8000 miles above the surface of Earth?

Answer 1

When we are told that something varies as a function of another variable, then we use some kind of constant to put the equation in a general form.  Here, let's just use "k".

w = the weight of the object.

 

We are told that w varies as 1 divided by the square of the distance, so we can write:

w = k / d^2  (Without "k", we would be saying that the actual weight was equal to 1/ d^2, which would not make sense.

 

Now, we are told the weight at 4000 miles, so we can find k:

 

160 = k / 4000^2 = k / 16* 10^6

Thus k = 2560 * 10^6 = 2.56 * 10^9

 

Now we can find w at 8000 miles

w = 2.56 * 10^9 / 8000^2 = 2.56*10^9 / 64*10^6 = 40

 

Shortcut:

We doubled the distance, therefor the weight must go down by 4X