Shape of a planet spinning very quikly

TheBigK1d

I had an idea of a planet that was spinning fast enough that at its equator, the outward force from the centripetal force would almost equate the planet's own gravity. However, this would change as you got closer and closer to the poles. I was just wondering what this planet might look like. Unfortunately, I haven't taken calculus yet and don't really know how to calculate what its shape might be.

Any help?

Janus

It would be a very oblate spheroid. Hal Clement created something like this for his 1954 novel "Mission of Gravity". He even wrote an essay entitled "Whirligig World" in which he describes how he worked out the particulars of his planet "Mesklin".

DaveC426913

Indeed, Saturn exhibits this effect most of all our planets.

It is the least dense of all the planets - less dense than water - and its day is less than 11 hours.

Its equatorial circumference is about 10% larger than its polar circumference.

A.T.

You have to add the potential of gravity and centrifugal force. The shape of the planet is a surface of constant combined potential.

haruspex

Sure, but that's a hard problem. The gravitational potential depends, in turn, on the shape.
http://seismo.berkeley.edu/~rallen/e...ctures/L16.pdf gives the strength of gravity at latitude λ as g_e(1 + α sin²λ + β sin⁴λ), where g_e is the value at the equator, but does not indicate how α and β depend on rate of spin/oblateness.
http://en.wikipedia.org/wiki/Equator...cal_expression provides an expression as a function of spin, but says the formula is only valid for small deviations from the spherical.

ImaLooser

The polar axis must be at least 56% of the equatorial diameter. After that the planet would be nonaxisymmetric. I don't know what would happen then, but it wouldn't be good.

TheBigK1d

Thanks for the help - I'll check out the book you mentioned