Question

4)How much below the surface of the earth the acceleration due to gravity
i)reduce to 36 %
ii)reduce by 36 % of its value on the surface of the earth
Radius of earth=6400km

Answer 1

Explanation:

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} )g

d

=g(1−

r

d

)

Given

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

d

=

10

7g

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 103×6400÷10

d= 1920 km.

Therefore,

1920 \: \: \: km1920km below the surface of the earth does the acceleration due to gravity become 70% of its value at the surface of earth