Gravitational Effect of a Fast Rotating Gas Giant

[bigbywolfe]

I started playing around with the Star Hero star system creation relying mostly on random rolls. I ended up with one of my planets being a Gas Giant (mass 85) circling a M0V (hot Red Dwarf) and it has the Fast Rotation Anomaly and only has an 8 hour day. There is a note in the Fast Rotation description about how very large planets with fast spin may have perceptibly weaker gravity at the equator due to the centrifugal effect and there is an example using Earth. The example problem's solution is a decimal (.0009 meters per second squared, or .009%). For my gas giant the sum was 1.04. Am I correct in reading that the planet would be 100% reduction and thus a full 1G lighter at the equator? That seems extreme.

 

EDIT: Or is it 100% of the planet's actual gravity (which in this case in only 1.5)? That seems even more extreme. I'm starting to get the feeling I am not understanding something in this Rotation Anomalies section...

[Cancer]

That's not an extreme example; Saturn has a rotation period of 10.5 hours. You might try working through an example case of Saturn (the mass, radius, rotation speed, etc. are in Wikipedia) and see if you can get the numbers for a "known case". Minor deviations aren't important, just see of you're in the right order of magnitude.

[shuddemell]

You know, I couldn't get his numbers to work out correctly, so I must be missing something... From Star Hero the formula would be [(Mass^.333*500)/(3600*Radius)]^2.... Which according to my calculations works out like this: If Earth's Mass is 1, and the period of rotation is 24 hours, then this is [(1^.333*500)/(3600*24)]^2 = (.005787)^2= .000033, not the .0009 as listed.... what am I missing?

 

BTW, I cross posted this for Mr. Long to answer as well.

[Tasha]

Now I am no physisist, but wouldn't centrifigal force have no real effect on Gravity? Gravity is driven by mass alone. From what I can see a fast rotating gas giant would still have crushing gravity, but have most of it's mass concentrated around it's equator (Basically flattening out the planet, or squishing it making it shorter pole to pole vs it's width around the equator.

[Xavier Onassiss]

Now I am no physisist' date=' but wouldn't centrifigal force have no real effect on Gravity? Gravity is driven by mass alone. From what I can see a fast rotating gas giant would still have crushing gravity, but have most of it's mass concentrated around it's equator (Basically flattening out the planet, or squishing it making it shorter pole to pole vs it's width around the equator.[/quote']

 

Consider the extreme example: what if a planet rotated so fast its equator moved at orbital velocity. You'd experience micro-gravity standing on the ground at the equator. (And the planet probably wouldn't last very long!) Most fast rotating planets are just less extreme examples of this.

[shuddemell]

Well, remember gravity is driven by both masses involved, so if you have a counteracting force, in this case the centrifugal force, then APPARENT gravity would be lessened. Actual gravity is still the same (that is on earth 9.8 m/s^2). What you are really determining is the FORCE felt from gravity, which of course, isn't gravity itself. It might be useful to remember that gravity is a type of acceleration, and since in spinning objects the direction of the velocity is constantly changing, then that is a type of acceleration as well, and they are in opposite directions, so they are essentially counteracting each other.

[Cancer]

I'm looking for a good discussion of rotational flattening but not finding one here at home ... looks like it'll take a trip to the office and library.