Olbers’ paradox argues that if the universe is infinite, eternal and static then the night sky should be completely covered by stars. Suppose the universe is spherical, eternal and static. The stars are all spherical with the same radius and light does not reflect back from the edge of the universe. Also suppose the very many stars are randomly distributed; light scatters from the surface of a star but then travels in straight rays. Furthermore, assume stars take up a fraction of 1 in 1029 of the volume of the universe and, from the center of the universe looking out, stars appear to cover one part in a trillion of the sky. Can you estimate the radius of the universe in units of the radius of a star? Can you estimate how many stars there are? If the radius of the stars was 10 times greater but took up the same total volume, what fraction of the sky would be covered in stars?
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The Shape of Melting Ice
In this puzzle a cube of ice completely melts in exactly one hour. Throughout the melting the temperature inside the entire ice cube remains uniform, just a tiny amount below the freezing point. The external environment maintains a temperature just above the freezing point. The heat transfer rate is uniform across the surface of the cube. Under the same conditions how long would the same amount of ice shaped like a regular tetrahedron take to melt? What about shaping it like a torus with a major radius twice its minor radius? Is there any specific shape that would take the longest time to melt? Is there any specific shape that would take the shortest time to melt?
Observe that for a given volume V the instantaneous melting is proportional to the surface area. For a cube the edge length b = V(1/3) and therefore the surface area is S = 6 V(2/3). A regular tetrahedron with edge length l, has surface area S = (l2) √3 and volume V = (l3) / (6√2). So, for a tetrahedron S = 6 (3(1/6)) V(2/3), or 3(1/6) = 1.20094… as much surface area as a cube with the same volume. Both the cube and the tetrahedron will shrink but maintain the same shape as they melt. So the tetrahedron melts 3(1/6) = 1.20094… times as fast for each progressively smaller volume V. Therefore from the initial volume to volume 0, the tetrahedron takes 3(-1/6) = 0.832683… hours or 49 minutes and 58 seconds to melt.
A torus has S = 4 π2 r R and V = 2 π2 r2 R, where R is the major radius and r is the minor radius. However, as the torus melts uniformly, R remains the same, but r shrinks and the torus shrinks to a thin ring and then disappears. The minor radius r decreases at a uniform rate per time, since S = dV/dr and dV/dt being proportional to S together imply that dr/dt is constant. In fact r will decrease at one half the rate per time as the edge b of the melting ice cube decreased. This is so because for the torus dV/dt = S dr/dt and for the cube dV/dt = S d(b/2)/dt, and in both cases dV/dt is proportional to S. If initially R = 2r, then V = 4 π2 r3 = b3 and therefore r = (4 π2) (-1/3) b = (0.293684…) b. So, the torus takes only 2 (0.293684…) = 0.587368… hours or 35 minutes and 15 seconds to melt.
The shape with the least area per volume will take the longest time to melt. This is a sphere where S = 4 π r2 and V = (4/3) π r3. For the sphere S = (6(2/3)) (π(1/3)) V(2/3) and therefore it melts (π/6)1/3) = 0.805996 times as fast as the cube. So the sphere takes (π/6)(-1/3) = 1.2407… hours or 74 minutes and 27 seconds to melt.
For arbitrary shapes the surface area for a given volume can be made arbitrarily great. For example, a box with a square base with edge l and height l/k has V = (l3)/k and S = 2 l2 +4 (l2)/k. So, in this case S = (2 k(2/3) + 4 k(-1/3) )V(2/3). Since S > k(2/3) V(2/3), S can be made as large as desired by increasing k to make k(2/3) as large as desired. So, the box can be made as broad and shallow as needed to melt as quickly as desired.
AR Puzzle Editor Jon Evans is president of Convergent Actuarial Services, Inc. in Delray Beach, Florida.
