Question

Calculate the percentage decrease in the weight of a body when it is taken 32 km below the surface of the Earth. [Given:R e]​

Answer 1

Answer:

Acceleration due to gravity at a depth is given by

g'=g(1-d/R)

g'= g(1-32/6400)

g'=g(1-1/200)

g'-g=(1/200)g

(g'-g)/g =1/200

If m is the mass of the body,then mg and mg' will be the weight of the body on the surface of the earth and at a depth of 32km below the surface of the earth,then,

% decrease in weight ={(mg'-mg)/mg}x 100

={(g'-g)/g}x100

=(1/200) x 100=0.5%

or

Here d= 32 km R= 6400 km

weight of body at depth d is mg`

=mg(1-d/R)%

decrease in weight =( mg- mg`)/mg × 100= 32 / 6400×100 = 0.5%

you can do any one of there because both the sum are same