Question

At what depth from the earth's surface the value of acceleration due to gravity reduces by 1%

Answer 1

In terms of depth d from the surface of the earth , this formula becomes,

g(d)=[g(Re)/Re]][Re-d]=g(Re)[1-(d/Re)]…………..(1)

Here, Re is radius of the earth=6400 km.

g(Re) is value of g on the surface of the earth=9.8m/s^2

Depth for g(d)=g(Re)/2:

Using equation (1),

g(Re)/2=g(Re)[1-(d/Re). Therefore,

(1/2)=1-(d/Re)…………..(2). Then,

d/Re=1/2 or d=Re/2

Thus,

d=(4000)/2=2000km………….(3)

Depth for g(d)=g(Re)/4:

For this, replace 1/2 in equation (2) by 1/4.

Therefore,

1/4=1-(d/Re) or

d/Re=3/4. So,

d=(3/4)Re=(3/4)(4000)=3000 km.

Answer is 3000km