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bonanova
Daedalian Member

 Posted: Mon Aug 06, 2012 8:02 am    Post subject: 1 Imagine a large circular and planar skating pond say 100m in diameter. The ice is a frozen inviscid fluid [frictionless to ice skates]. A perpendicular can be dropped from the center of the rink to the center of the earth. Thus all objects on the rink are pulled by gravity toward its center. At what initial circumferential speed would a skater continue in circular orbit of the pond? What would be his orbital period? Neglect irritating details like air friction and Earth's inhomogeneity. Assume if you like gravity's acceleration to be 9.8 m/s 2 and the diameter of the earth to be 40,000km. Assume anything else that's reasonable and at least mildly relevant._________________ Vidi, vici, veni.
Elethiomel
Daedalian Member

 Posted: Mon Aug 06, 2012 8:49 am    Post subject: 2 How did an inviscid fluid freeze into a planar skating pond? Would it not follow the curvature of the Earth? (Edit: Does this qualify as an irritating detail?)
bonanova
Daedalian Member

Posted: Mon Aug 06, 2012 9:24 am    Post subject: 3

 Elethiomel wrote: How did an inviscid fluid freeze into a planar skating pond? Would it not follow the curvature of the Earth? (Edit: Does this qualify as an irritating detail?)

Yeah, that's irritating. Assume a gigantic milling machine ... etc.
_________________
Vidi, vici, veni.
novice
No harm. Pun intended!

 Posted: Mon Aug 06, 2012 10:06 am    Post subject: 4 My answer: The skater's centripetal acceleration (towards the center of the rink) is determined by the radial component of the force of gravity acting on the skater. This component is determined by the angle between the perpendicular to the ice and the line from the skater to the center of the earth. Call this angle alpha, we then have tan(alpha) = 0.1km / 40000km, so alpha = atan(1/400000). The radial component of gravity's acceleration is then g * sin(alpha) = g * tan(alpha) for very small alpha = g * tan(atan(1/400000)) = g/400000. The orbital velocity is determined from centripetal acceleration by the formula a = v*v/r, so v = sqrt(a * r) = sqrt(g * r / 400000) = sqrt(g / 8000). That means an initial circumferential speed of 35 mm per second would let the skater enter a circular orbit. One orbit would have a length of PI*d = v * t, so the orbital period t = PI*d/v = PI*100/0.035 = 8971 seconds = 2.49 hours.
bonanova
Daedalian Member

 Posted: Mon Aug 06, 2012 8:55 pm    Post subject: 5 If we replace the person on skates, which might supply their own lateral force, by a smooth rock; would that change the answer? What if we painted the name Vostok I on it.?_________________ Vidi, vici, veni.
Zag
Tired of his old title

 Posted: Mon Aug 06, 2012 10:21 pm    Post subject: 6 I feel pretty sure that Coriolis force is going to be more significant than gravity in this case. Consider how huge the circle has to be so that the around the edges the ice has a noticeable slope to them to someone standing on it. BTW, I had thought originally that this was going to lead to a discussion of Lagrangian points, which you should look up if you don't know what they are.
bonanova
Daedalian Member

 Posted: Thu Aug 09, 2012 4:10 pm    Post subject: 7 What might catch one's interest is that the correct answer here also obtains for low earth satellite orbits and round trips on a gravity train._________________ Vidi, vici, veni.
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