Question

The weight of a person on or above the surface of the earth varies inversely as the square of the distance the person is from the center of the earth. If a person weighs 180 pounds on the surface of the earth and the radius of the earth is 3900 miles, what will the person weigh if he or she is 850 miles above the earth's surface? Round your answer to the nearest hundredth of a pound.

Answer 1

If you do this with a complex set of fractions and let
K = G Me m_space be the same in both parts, there should be some cancellation.

W_earth = 180 lbs

W_earth = K /3900 miles^2
W_space =K  /(3900 + 850)^2

180 / W_space = k/3900^2
x = k / (4750)^2

$\frac{180}{Wspace} = \frac{ \frac{k}{3900^{2} } }{ \frac{k}{4750^{2}} /[tex] Now you need to invert and multiply the bottom fraction on the left. [tex] \frac{180}{x} {=} \frac{k}{3900^{2}} {*} \frac{4750^{2}}{k}$ 

The ks cancel out.

You are left with 180/x = 4750^2 / 3900^2 Now cross multiply
180 * 3900^2 = 4750^2  = x
180 * 3900^2 / 4750^2 = x 
180 * 0.67413 = x
x = 121 pounds. Weight is a force, but because all the units on one side are equivalent to the units on the other, the conversions become part of k. Normally you would have to do the conversions, but not in this particular case.

Answer 2

Answer:

What the person said above. They rounded, but they got 121.  

But if you round to the nearest hundredth of a pound, the answer is 121.34.  

SO the answer is A: 121.34  

Step-by-step explanation:

The formula is 180*3900^2=x(3900+850)^2

The answer is 121.34  

I hope this helps ^v^