Question

The radius of the earth is about 6370Km. An object of mass 30Kg is taken to a height of 230Km above the surface of earth. (a) What is the mass of the body (b) What is the acceleration to gravity at this height (c) What is the weight of the body at this height.

Answer 1

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%