An object is taken to a height above the surface of earth at a distance 45R from the centre of the earth. Where radius of earth, R=6400km. The percentage decrease in the weight of the object will be
(A)36%✓ Correct
(B)50%
(C)64%
(D)25%
Answer: (A)
Solution
At distance r=45R from Earth's center, the height above the surface is h=45R−R=4R.
The effective gravity at this height:
geff=(1+Rh)2g=(1+41)2g=(5/4)2g=2516g
Percentage change in weight:
ggeff−g×100=g2516g−g×100=25g−9g×100=−36%
The weight decreases by 36%.
JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Single correct option