If an imaginary line is drawn from the centers of the earth and the then the net
ID: 2880099 • Letter: I
Question
If an imaginary line is drawn from the centers of the earth and the then the net gravitational force F acting on an object situated on this line segment is F = - K/x^2 + 0.012 K/(239-x)^2 where K is a positive constant and x is the distance of the object from the center of the earth, measured in thousands of miles. How far from the center earth is the "dead spot" where no net gravitational force acts of the upon the object? (Express your answer to the nearest thousand he distance to the moon is approximately 240,000 miles.)Explanation / Answer
F=(-K/x2)+(0.012K/(239-x)2)
Net gravitational force is 0
0=(-K/x2)+(0.012/(239-x)2)
0= (-(239-x)2 + .012x2)
.012x2= (239-x)2
x=215miles , 268 miles