G'day Art,
neufer wrote:
Which (although wrong for so many reasons) is pretty close to my prior calculation of 3h 50m:
I added italics for the words which triggered my special attention. I'm always eager to learn from my mistakes and errors. Would you be so kind to point me what is wrong and for which reasons? Is it the assumptions? Is it the numbers, which i did not add up correctly?
Next you focus the attention on the Kuiperbelt object named Santa, since it is an example of a fast rotating object.
The wiki reports:
M = 4.2E21 kg
r1 = 1.96E6 m
r2 = 1.518E6 m
r3 = 0.996E6 m
g = 0.44 m/s²
T = 14075 s
Unfortunately the wiki does not report around which axis Santa is rotating.
First calculate the accelleration of gravity at the surface of Santa, when it is
not rotating. This is application of Newtons law of attraction between two masses.
a1 = GM/(r1)^2 = 6.67E-11 x 4.2E21/(1.96E6)^2 = 0.0729 m/s²
a2 = GM/(r2)^2 = 6.67E-11 x 4.2E21/(1.518E6)^2 = 0.1215 m/s²
a3 = GM/(r3)^2 = 6.67E-11 x 4.2E21/(0.996E6)^2 = 0.2823 m/s²
Neither of these values are equal to 0.44 m/s² as mentioned in the wiki.
Since r3 is about half the size of r1, the accelleration at r3 is four times as large as at r1. The numbers proove it, so not likely an numerical error. Since it does not reproduce the value of 0.44 m/s² in the wiki, there must be an error in my formula or in the data.
Assuming Santa is an ellipsoid with three different axes, its volume is
V = 4 π x r1 x r2 x r3 / 3.
Filling in the numbers: V = 1.241E19 m³ Suppos Sante is made of solid ice, than its mass would be 850 times its volume: 1E22 kg. Nevertheless it is assumed to have a soli d rock core, so it must be havier. With the wiki mass, its density is 340 kg/m³ and that is a lot lighter than water. Other references mention that Santas mass is 32% of Pluto, which produce a value in accordance with the value mentioned in the wiki, e.g.
http://www.gps.caltech.edu/~mbrown/2003EL61/
Now calculate the centrifugal accelleration for the three radii, when Santa is
rotating:
c1 = ω²r1 = 4 π² x 1.96E6 / 14075² = 0.39 m/s²
c2 = ω²r2 = 4 π² x 1.518E6 / 14075² = 0.30 m/s²
c3 = ω²r3 = 4 π² x 0.996E6 / 14075² = 0.20 m/s²
In this case c3 is half the value of c1, since ci is proportional to the radius. The centrifugal accellerations are considerable compared to the accelleration of gravity at the surface of Santa.
The application of the formula for the surface acceleration of gravity produces 9.8 m/s² for the earth, which is quite natural. So i trust this formula. It is also reported in
http://en.wikipedia.org/wiki/Surface_gravity.
In the formula for the calculation of the centrifugal force isn't an error either. It is a textbook result, see e.g.
http://en.wikipedia.org/wiki/Centrifugal_force
Combination of ai and ci lead to the following result:
- If Santa is rotating around r3, c1 > a1 and c2 > a2. An object placed at the equator of Santa would launch itself. It should be anchored to stay on the surface.
If Santa is rotating around r2, c1 > a1 and c3 < a3. At the equator at the most elongated point an object will be launched, at the least elongated point it is stable.
If Santa is rotating around r1, c2 > a2 and c3 < a3. At the equator at the most elongated point (1518 km) an object will be launched, at the equator at 996 km it is stable.
Using the value of the wiki (0.44m/s²) an object will remain on the equatorial surface.
So Art, please be so kind and point me where i made an error in these calculations
By the way, it is not my intention to launch a theory of any kind, i just like do some simple numerical excercises to get a grip on the numbers involved.