physics help please.?

Question

Suppose a rotating spherical body such as a planet has a radius r and a uniform density p, and the time required for one rotations is T. At the surface of the planet, the apparent acceleration of a falling object is reduced by the acceleration of the ground out from under it. Derive an equation for the apparent acceleration of gravity,g, at the equator in terms of r,p,T, and G.

Applying the equation above, what fraction is the apparent weight reduced at the equator compared to the poles, due to the Earth's rotation?

From the equation above, derive an equation giving the value of T for which the apparent acceleration of gravity becomes zero, such as objects can spontaneously drift off the surface of the planet. Show that T only depends on p, and not on r.

Answer 1

The gravitational attraction is given by Fg = GMm/r^2, where M = mass of the planet = 4/3 * π * r^3 * p, so

Fg = G*(4/3)*π*r^3*p*m/r^2 = G*(4/3)*π*r*p*m The gravitational acceleration of the object ag is Fg/m =

ag = G*(4/3)*π*r*p

The acceleration due to rotation ar = w^2*r, where w = angular velocity = 2*π/T, so

ar = 4*π^2*r/T^2

The net apparent acceleration is then ag - ar =

g = G*(4/3)*π*r*p - 4*π^2*r/T^2

g = 4*π*r*[G*p/3 - π/T^2]

when apparent acceration g = 0

G*p/3 = π/T^2

T = √[3*π / G*p]