Gravitation Physics Help, Best answer to first right!?

Question

Identical masses m hang from strings on a balance at the surface of Earth. The strings have negligible mass and differ in length by h. Assume that Earth is spherical, with density of 5.5 g/cm^3.

a) Show that the difference ∆W in the weights, due to one mass being closer to Earth than the other, is 8πGpmh/3

b) Find the difference in length that will give a ratio ∆W/W = 1 x 10^-6, where W is either weight

Answer 1

the force of the earth on one is

GmM/r^2

the force on the other is GmM/(r+h)^2

taking the difference

W2-W1 is equivalent to finding the derivative

dW/dr=(GmM/r^2)

or dW=-2GmM/r^3  dr (1)

recognizing that the density of a sphere is

rho=M/V = M/(4/3 pi r^3)

we can rewrite M/r^3 in (1) as

M/r^3=4/3 pi rho

so that (1) becomes:

dW=-2Gm(4/3 pi rho) dr

since dr=h, we get finally, the difference in weights, dW

dW=8pi Gm rho h/3

to find dW/W, divide the expression just above by W=GmM/r^2 to get:

dW/W = 8 pi rho h/3/[M/r^2]

use a consistent set of units, find the mass and radius of the Earth, set dW/W=10^-6 and solve for h