Question

The depth d at which the value of acceleration due to gravity becomes 1/n times the value at the surface is ?

Answer 1

Let us consider that acceleration due to gravity at earth's surface is $g_0$

then, acceleration due to gravity in depth d from earth's surface is given by, $g=g_0\left(1-\frac{d}{R}\right)$

where , R is radius of the earth .

a/c to question,

The depth d at which the value of acceleration due to gravity becomes 1/n times the value at the surface.

e.g., $g=\frac{1}{n}g_0$

so, $\frac{1}{n}g_0=g_0\left(1-\frac{d}{R}\right)$

or, $\frac{1}{n}=1-\frac{d}{R}$

or, $d=R\left(1-\frac{1}{n}\right)$

hence, depth from the earth's surface is $d=R\left(1-\frac{1}{n}\right)$

Answer 2

Answer:

Let us consider that acceleration due to gravity at earth's surface is  

then, acceleration due to gravity in depth d from earth's surface is given by,  

where , R is radius of the earth .

a/c to question,

The depth d at which the value of acceleration due to gravity becomes 1/n times the value at the surface.

e.g.,  

so,  

or,  

or,  

hence, depth from the earth's surface is

Explanation: