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? (Take $R=6400km$)

Answer 1

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, $R=6400km$ (Radius of earth)

$\frac{7g}{10}=g(1-\frac{d}{6400})$

$\frac{7}{10}=1-\frac{d}{6400}$

$\frac{d}{6400}=1-\frac{7}{10}$

$\frac{d}{6400}=\frac{3}{10}$

$d=\frac{3}{10}\times 6400$

$d=1920km$

Therefore,

1920km below the surface of earth does the acceleration due to gravity become 70% of its value at the surface of earth.