Question

An object weighs 100 newtons on earth's surface when it is moved to a point one earth radius above earth's surface it will weigh

Answer 1

25 newtons.  
Since the force of gravity varies inversely with the square of the distance. At distance R, the force is 100 Newtons. So we now have a distance 2R. So  
(1) 100 = M/R^2
 (2) X = M/(2R)^2 
 Solve equation (1) for M
 100 = M/R^2
 100R^2 = M 
 Substitute the value for M into equation (2) and solve
 X = M/(2R)^2
 X = (100R^2)/(2R)^2
 X = (100R^2)/(4R^2)
 X = (100)/(4)
 X = 25 
 So the object at the higher altitude will weight 25 newtons.

Answer 2

Let
 M=mass of earth
 Object of mass m is at a distance r from center of earth, where r is greater than radius R of earth. 
 G=Universal gravitational constant
 F=Gravitational force on the object 
 F = GMm/r^2 
 But object's acceleration is 'g*'
 then 
 F=mg*
 mg* = GMm/r^2
 The acceleration due to gravity = g* = GM/r^2 
 At surface of earth = g =GM/R^2 
 g*/g =(R/r)^2 
 Distance from center of earth of a point one Earth radius above Earth's surface=r =2R 
 r = 2R 
 g*/g =(R/2R)^2 
 g*/g =(1/2)^2 
 g*/g =1/4 
 Weight at a point one Earth radius above Earth's 
surface=mg*.....(a) 
 Weight at Earth's 
 surface=mg=100 N....(b) 
 Divide equation (a) by (b)
 Weight at a point one Earth radius above Earth's 
surface/weight on surface = mg*/mg =g*/g=1/4
 Weight at a point one Earth radius above Earth's 
surface=(1/4)*100=25 N 
 Weight at a point one Earth radius above Earth's 
surface=25 N