Value of g on the surface of earth is 9.8 m//s^(2). Find its (a) at height h = R from the surface , (b) at depth d = (R)/(2) from the surface . (R = radius of earth)
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Answer:
a) At a height h=R
g'=g(1-2h/R)
=g(1-2R/R)
=g(1-2)
g'=-g
b)At a depth d=R/2
g'=g(1-d/R)
=g(1-R/2/R)
=g(1-1/2)
=g(1/2)
g'=g/2
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The value of g at h = R is g' = g / 4 and At d = R / 2 is g" = g / 4 = 9.8 / 4 m/s^2
Explanation:
Given data:
g = 9.8 m/s^2
(a) at height h = R from the surface
At h = R
g' = g / (1 + h / R)^2
g' = g / 4
(b) at depth d = (R)/(2) from the surface
At d = R / 2
g" = g ( 1 - d / R)
g' = g / (1 + R/R')
g' = g / 4 = 9.8 / 4 m/s^2
The value of g at h = R is g' = g / 4 and At d = R / 2 is g" = g / 4 = 9.8 / 4 m/s^2
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