## Shape of a planet spinning very quikly

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?

 PhysOrg.com physics news on PhysOrg.com >> Promising doped zirconia>> New X-ray method shows how frog embryos could help thwart disease>> Bringing life into focus
 Recognitions: Gold Member Science Advisor Staff Emeritus 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".
 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.

## Shape of a planet spinning very quikly

 Quote by TheBigK1d I haven't taken calculus yet and don't really know how to calculate what its shape might be.
You have to add the potential of gravity and centrifugal force. The shape of the planet is a surface of constant combined potential.

Recognitions:
Homework Help
 Quote by 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.
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 ge(1 + α sin2λ + β sin4λ), where ge 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.

 Quote by 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?
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.

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

 Tags centripetal force, gravity, planet