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We know that the value of 'g' varies with altitude. As we go up from earth surface or go down into earth's interior, the value decreases. To find the weight of the body at a height of 3R

1. you can find the effective g at height 3R and multiply it with mass. Or
2. we know that the gravitational force is inversely proportional to the distance between two objects. Since other parameters except the distance is constant, just find the ratio between the forces which will be equal to the ratio of square of the distances.

initial distance = R
final distance = R + 3R = 4R
initial force = 144 N
final force = F

 \frac{F}{144} = ( \frac{R}{4R} )^{2} = \frac{1}{16}\\ \\ \Rightarrow F=144 \times \frac{1}{16} \\ \\ \Rightarrow F= \boxed{9N}
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At any height mass is always constant
w = mg = 144N -----------------(1)
let g represent acceleration due to earth at the surface of earth and g' is at a height  of 3R.
at the surface of earth
g = GM/R² ----------------(2)         where G is constant, M is mass of earth, R is the Radius of earth.
at a height of h = 3R
g' = GM/(R+h)² = GM/(R+3R)²
g' = GM/16R² -----------------------------(3)
equation (2) divided by equation (3)
g/g' = 16
g'/g = 1/16
let the weight at a height of 3R is w'
w' = mg'  ----------------(4)
equation (4) divided by (1)
w'/w = g'/g
w' = 144×1/16
w' = 9N
hence at a height of 3R weight is 9N.


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