Question

Mountain pull. A large mountain can slightly affect the direction of “down” as determined by a plumb line. Assume that we can model a mountain as a sphere of radius R = 3.00 km and density (mass per unit volume) 2.6 × 103 kg/m3. Assume also that we hang a 0.500m plumb line at a distance of 3R from the sphere's center and such that the sphere pulls horizontally on the lower end. How far would the lower end move toward the sphere?

Answer 1

Assume

M be the mass of the large sphere,
m be the mass of the plumb bob,
G be the gravitational constant,
g be the acceleration due to gravity,
r be the density of the material of the large sphere,
T be the tension in the string,
a be the inclination to vertical of the plumb line,
s be the required distance
L be the length of the plumb line.

Then:
M = (4π/3) R^3 r
= (4π/3)(3 * 10^3)^3 (2.6 * 10^3)
=  2.941 * 10^14 Kg.

You can search for any mathematical expression, using functions such as: sin, cos, sqrt, etc. You can find a complete list of functions her

Resolving horizontally and vertically for the plumb line:
T cos(a) = mg ...(1)
T sin(a) = GMm / (3R)^2 ...(2)
s = L sin(a)
As a is very small, approximation is justified, and it is reasonable to replace sin(a) by tan(a):
s = L tan(a) ...(3)

Dividing (2) by (1):
tan(a) = GM / 9R^2 g

From (3):
s = LGM / 9R^2g
= (0.5)(6.67 * 10^(-11))(2.941 * 10^14) / 9(3 * 10^3)^2 (9.81)
= 9.6218* 10^(-6)m.