acceleration due to gravity at earth surface is g .find effective value of acceleration due to gravity at a height of 32 km from sea level
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Answered by
15
Assume-
h= 32km = 32000m = 3.2 x10^4
R= 64km = 64000m = 6.4 x 10^4
g= 9.8m/s^2
We know that-
g'= g[R^2/(R+h)^2]
= 9.8(40.96 x. 10^8/92.16 x 10^8)
= 9.8 x 4.44
= 4.355m/s^2 ~ 4.4m/s^2
h= 32km = 32000m = 3.2 x10^4
R= 64km = 64000m = 6.4 x 10^4
g= 9.8m/s^2
We know that-
g'= g[R^2/(R+h)^2]
= 9.8(40.96 x. 10^8/92.16 x 10^8)
= 9.8 x 4.44
= 4.355m/s^2 ~ 4.4m/s^2
Answered by
13
The value of Acceleration due to gravity changes when we go to a higher altitude or go into a depth or go to a different latitude.
The change in acceleration due to gravity with altitude is given by,
g' = g ( R / R+ h)²
When h <<<< R,
Then g' = g ( 1 - h/R)
Radius of earth = 6400 km.
Since 32 <<< 6400,
g' = g ( 1 - h/R)
g' = g ( 1 - 32/6400)
g' = g ( 1 - 1/200)
g' = g ( 1 - 0.005)
g' = g ( 0.995)
g' = 0.995 g.
Therefore, Effective value of acceleration due to gravity at a height of 32 km from sea level is 0.995 g.
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