<span title="Gravitation!!!!!!!!!!!!!!!!!!!!!!!!!!!!?">Gravitation!!!!!!!!!!!!!!...</span>

Question

The fastest possible rate of rotation of a planet is that for which the gravitational force on material at the equator barely provides the centripetal force needed for the rotation. Show then that the corresponding shortest period of rotation is given by

T =square root of ( 3π/ Gρ)

where ρ is the density of the planet, assumed homogeneous.

Answer 1

there are a few steps to this...

consider the forces on an object at the equator...there is the normal force acting out, the grav force acting toward the center, and these two forces sum to equal the centripetal force

in terms of Newton&#039;s second law, these are written:

N-F(grav)=-mv^2/R

or N=F(grav)-mv^2/R

when F(grav)=mv^2/R the normal force is zero, and the object no longer has contact with the Earth, so the condition we need to find is when F(grav)=mv^2/R

remember that F(grav)=GmM/R^2

equating, we get:

GmM/R^2 = mv^2/R

which leads to

v^2=GM/R   (1)

now, v is the linear speed of the object, and is equal to the distance traveled divided by time

the time to make one rotation is P (the period) and the distance is just the circumference of the Earth, or 2 pi R, so we have

v=2 pi R/P

or v^2=4 pi^2 R^2/P^2

using this expression for v^2 in equation (1) above:

4 pi^2 R^2/P^2 = GM/R

which yields

P^2=4 pi^2 R^3/(GM) (2)

now, if the Earth is a sphere, the mass of a sphere is

M= density x volume = rho x 4/3 pi R^3

substitute this expression for M into (2) above and you will get your answer of P=sqrt[3 pi/G rho]