Acceleration due to gravity at earth's surface
is 'g' m/s2. Find the effective value of acceleration
due to gravity at a height of 32 km from sea
level : (R = 6400 Km)
(1) 0.5 g m/s2
(2) 0.99 g m/s?
(3) 1.01 g m/s2 (4) 0.90 g m/s?
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Answer:
(2) 0.99m/s²
Explanation:
we know,
effective acceleration due to gravity at a height h from earth's surface is equal to
g'= g× R²/(R+h)²
R=6400 , h=32
using binomial theory , we get
g'=g [ 1- 2h/R]
》g' = g [1- 2×32/6400]
》g'=g [1-1/100]
》g'=g [1- 0.1]
》g'=g×0.99
》g'= 0.99
answer is 0.99m/s²
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