Question

how much below the surface of the earth does the acceleration due to gravity become 70% of its value at the surface of earth ?

Answer 1

g'=g (1-d/R)
g/g'= 70/100
7/10= 1-d/R
d/ r = 1-7/10
d/R= 3/10
d=3R/10
d=3×6400/10 (R=rad of earth=6400km)
d=1920km
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Answer 2

Let the depth at which acceleration due to gravity ( g) becomes 70% or 7g/10 = d.

We know that
$g_{d} = g(1 - \frac{d}{r} )$

Given
$g_{d} = \frac{7g}{10}$

Also, R = 6400 km [ Radius of earth ]

7g/10 = g [ 1 - d / 6400]

7/10= 1-d/6400

d/6400 = 1 - 7/10

d/6400 = 3/10

d =$3 \times 6400 \div 10$

d= 1920 km.

Therefore,
$1920 \: \: \: km$ below the surface of the earth does the acceleration due to gravity become 70% of its value at the surface of earth