Fastest Planet Rotation: Calculating Period using Volume & G

[john_simpson]

The fastest rate of rotation of a planet is that for which the gravitational force on matter at the equator just provides the centripetal force necessary for that matter to move with circular motion. Show that the period of rotation in this case is given by T= √3π/Gp where the planet is assumed to be a uniform sphere of density p (which has units of kgm^3(



Volume sphere = 4/3πr^3
Universal Gravitational Constant = 6.67 x10^-11



I truly have no idea where to start please help!

 

[Doc Al]

Imagine a small mass on the surface of the planet at the equator. What is its acceleration? What is the force on it? Apply Newton's 2nd law.

Hint: Review centripetal acceleration' and Newton's law of gravity.

 

[Moetasim]

Sure, let's break this down step by step. First, let's define some of the variables we have in this problem:

- T = period of rotation (in seconds)
- G = universal gravitational constant (in m^3/kg/s^2)
- p = density of the planet (in kg/m^3)
- r = radius of the planet (in meters)

The first step is to understand the concept of centripetal force. This is the force that keeps an object moving in a circular path. In this case, the gravitational force is acting as the centripetal force, pulling the matter at the equator towards the center of the planet.

We can use Newton's Second Law, F=ma, to relate the gravitational force and the centripetal force. The mass (m) of the matter at the equator is equal to the volume of that matter multiplied by its density (p). So we can write:

F = ma = pVrω^2

Where V is the volume of the matter, r is the radius of the planet, and ω is the angular velocity (in radians/second). Now, we can use the equation for gravitational force, F = GmM/r^2, where M is the mass of the entire planet. We can substitute this into our previous equation and solve for ω:

GmM/r^2 = pVrω^2

ω = √(GpV/M)

Now, we know that the period of rotation is equal to the time it takes for one full rotation, which is equal to 2π/ω. So we can write:

T = 2π/ω = 2π/√(GpV/M)

We also know that the volume of a sphere is equal to 4/3πr^3, so we can substitute this in for V:

T = 2π/√(Gp(4/3πr^3)/M)

Simplifying this further, we get:

T = √(3π/Gp)

And that's it! This is the equation for the period of rotation for the fastest rotating planet, assuming it is a uniform sphere of density p. I hope this helps!