Deriving the Rotation Period of a Planet from Density

[skullers_ab]

Homework Statement

It's a high school physics question that I remembered and need to understand for curiosity's sake. I'm stating the problem in my own words, so it is likely that I may be asking a wrong question. Here it is nonetheless:

Derive an expression that shows that the major variable affecting the rotation period of a planet is its density [assuming planet is a perfect sphere & mass is evenly distributed].

Homework Equations

Newton's law of gravitation: F = G*M*m/r^2
Centripetal force: F = m*v^2/r and v = 2*Pi*r/T
Density: rho = m/v

The Attempt at a Solution

Assuming the rotation is due to a centripetal force (which is due to the gravitational attraction of the surface mass and the rest of the planet's centre of mass) acting on the surface of the planet, I get this as my final expression for the period, T:

T = \sqrt{(3*Pi)/(G*rho)}
(I'm unable to get (3*Pi) and (G*rho) in brackets like this, for Latex)

The problem is that when I substitute the value of the density of Earth (5515 kg/m^3), I don't get anywhere near the actual 24 hours that I should.

 

[dulrich]

What number are you getting? I get about 1.4 days -- which seems pretty close to me based on the redumentary nature of the calculation...

BTW, the LaTeX is "\sqrt{\frac{3 \pi}{G \rho}}"

\sqrt{\frac{3 \pi}{G \rho}}

See https://www.physicsforums.com/showthread.php?t=386951

 

[D H]

Whoa! What centripetal force are you talking about here? (Hint: There is none.)

The Earth's rotation rate has the value that it does because the angular momentum of the Earth-Moon system is conserved.

 

[dulrich]

Is this a complete coincidence then? I wondered about that before I posted.

I think the assumption in the problem is basically modeling the Earth as an ideal gas which his held together by its own gravity. So gravity is providing the centripetal force that keeps the planet rotating in uniform circular motion.

The model is clearly ridiculous for the Earth (which is why I'm surprized the calculation is even close). But maybe it is useful in modeling the gas giants or the rotation of galaxies?

 

[skullers_ab]

Thank you for your responses folks.

@dulrich: The numerical value that I get for T is around 5062 seconds which are around 1.4 hours, not days. Please clarify if I've made an error in this. If not, then sadly it is quite off from 24 hours.
Also, thanks for the reference to LaTex usage, I'll keep that in mind.

@D H: Yes, I was doubtful about the same [the concept that centripetal force is involved]. I was starting to think the the planet's rotation was a combined effect of its angular momentum AND centripetal force [but now I guess, CF is totally out of the question]

I was reading more through Google, and read somewhere that based on the period of a satellite at the surface of a planet they can figure out the mean density of the planet. And then I came across this discussion:
http://www.spacekb.com/Uwe/Forum.aspx/astro/7649/Finding-minimum-rotation-period

From what I now understand, the formula I've stated is for calculating the minimum time period possible for a planet to rotate without breaking itself apart or the maximum velocity with which its surface can rotate. That would be the same as the velocity required to place a satellite in orbit near the surface of the Earth.

Hence, my understanding now is that if a satellite were to be placed in orbit at the surface, it would have a period of 1.4 hours [and a speed of 8km/s] for going around the earth. Am I correct in my inference?

Also, I managed to get my hands on the actual question which was in the paper. My mistake, I'd recalled the question incorrectly which led to my curiosity. Here it is:

The fastest possible rate of rotation of a planet is that for which the gravitational force on an object at the equator just provides the centripetal force needed for that rotation. The planet is assumed to be a sphere of uniform density. Show that the period is T = <br /> \sqrt{\frac{3 \pi}{G \rho}}<br />

 

[D H]

For one thing the coincidence isn't as strong as you think:

\sqrt{\frac{3\pi}{G\rho}} \approx 1.4\,\text{hours}

Good old google calculator: http://www.google.com/search?q=sqrt(3*pi/(G*earth+mass/(4/3*pi*(6378+km)^3)))

More importantly, a force (or rather, a torque) is not needed to keep the Earth rotating. The Earth has angular momentum due to its rotation. A torque would be needed to change the rotation rate.

 

[dulrich]

Yeah, my bad. Sorry about the confusion on the calculation.