How much below the surface of earth does g reduces by 36%of its value on surface of earth?
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We know that variation of g with depth ‘d’ is given as
g’ = g (1-(d/R)) (this is the first equation)
where g’ is the new acceleration due to gravity
d is the depth
and R is the radius
So if we say that new acceleration due to gravity is reduced to 36% that means
g’ = (36/100)g
Now substituting this value in the first equation we get
(36/100)g = g (1-(d/R))
so we have
36/100 = 1- (d/R)
simplifying we get………………….
d(depth) = R (1–0.36)
Putting the value of R and solving we get that the depth should be at
4096km
g’ = g (1-(d/R)) (this is the first equation)
where g’ is the new acceleration due to gravity
d is the depth
and R is the radius
So if we say that new acceleration due to gravity is reduced to 36% that means
g’ = (36/100)g
Now substituting this value in the first equation we get
(36/100)g = g (1-(d/R))
so we have
36/100 = 1- (d/R)
simplifying we get………………….
d(depth) = R (1–0.36)
Putting the value of R and solving we get that the depth should be at
4096km
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