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

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

Answer 2

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.