acceleration due to gravity on surface of earth is given as
[tex]g = \frac{GM}{R^2}[/tex]
now at the surface of earth we know its value will be g = 9.8 m/s^2
it is given that we need to find the weight at height three earth radii above the surface
so we can say
[tex]g' = \frac{GM}{(R + 3R)^2}[/tex]
[tex]g' = \frac{GM}{16R^2}[/tex]
now from above equation we can say that here the acceleration due to gravity is 1/16 times of the gravity at surface
So as we know that the weight is measured as
[tex]W = mg[/tex]
if gravity is decreased by 1/16 times then weight will also decrease by the same factor
So weight will be measured as
[tex]W' = \frac{1}{16}* 90 = 5.625 lb[/tex]
So its weight above the surface will 5.625 lb at that height
