The weight (in pounds) w=f(d) of an object varies inversely as the square of its distance (in thousands of miles) d from the center of earth. an astronaut weighs 160 pounds at sea level (about 4,000 miles from earth center) find an equation for f. how much would the astronaut weigh at 4 thousand miles above earths surface?

The joint variation that is given in the problem above can be expressed as follows:
                                    w = k/d²
where w is the weight, k is the variation constant, and d is the distance from the center of the earth. Substituting the first scenario to obtain the value of k,
                                     160 = k/(4000)²
                                        k = 2.56 x 10^9
Using this value for the second scenario,
                                       w = (2.56 x 10^9) / (8,000)² = 40 lbs
Thus, the weight of the man, 8000 miles above the center of the earth is 40 lbs. 

The answer is 4.56. I hope this helps you