Question

show that weight of all bodies is zero at earth

Answer 1

weight(W) of a body depends on two things. one is "mass"(m), and the other is "acceleration due to gravity"(g).

the relation between them is given by  

W=m*g

since we cant do much with the constant mass of the body, the only way to find out about the weight at the center of the earth would be to find out about the "g" at the center of the earth.

The gravity at a radius r depends only on the mass inside the sphere of radius r; all the contributions from outside cancel out. This is a consequence of the inverse-square law of gravitation. Another consequence is that the gravity is the same as if all the mass were concentrated at the center of the Earth. Thus, the gravitational acceleration at this radius is

where G is the gravitational constant and M(r) is the total mass enclosed within radius r. If the Earth had a constant density ρ, the mass would be M(r) = (4/3)πρr^3  and the dependence of gravity on depth would be

g at depth d is given by g'=g(1-d/R) where g is acceleration due to gravity on surface of the earth, d is depth and R is radius of Earth. If the density decreased linearly with increasing radius from a density ρ0 at the centre to ρ1 at the surface, then ρ(r) = ρ0 − (ρ0 − ρ1) r / re, and the dependence would be

now since ur depth is "0", when you substitute r=0 in the above equation, you will see that g=0 at center of the earth.

hence the weight of a body too will become 0.

Answer 2

wrong the weight of object is different