## A gravitational challenge

by John Church

The AAAP has members who enjoy science and math challenges. I invite all who are interested to try their hands at this one.

Consider two identical, homogeneous, nonmagnetic cubes, each having a mass of 1 kg, measuring 0.04 meters (4 cm) on an edge. One of these cubes is securely fixed in place on a horizontal, inert, and perfectly frictionless surface. The other cube, initially at rest and 1 meter to the left of the fixed cube (center to center), is free to slide towards the fixed cube under the sole influence of their mutual gravitation. Such cubes would have a density of 15.6 g/cm3 or about that of a 70-30 alloy of gold and silver. Assume that there are no significant masses nearby to perturb the experiment. Assume also that any possible movement of the fixed cube is negligible.

(Note, in memory of Henry Cavendish and his pioneering work on gravitation, you could also use far less costly lead cubes of different masses and sizes, but cost is no object here.)

The cubes would touch when their center-to-center distance had been reduced to 0.04 m. How long would it take for this to happen? How fast would the movable cube be going just before contact? Take the gravitational constant (G) as 6.67 x 10-11 m3 kg-1 sec-2 . Present a graph to show the position of the movable cube as a function of time.

Here is a diagram of the setup. My word processor does not have a square symbol to represent a cube, so a capital “O” will have to do:

Movable cube

O—————————————————–OO Fixed cube

Center is initially 1 meter from Cube centers are 0.04 meter from

center of fixed cube. each other when they touch.

Time = 0 Speed = 0 Time = ? Speed = ?

A Canadian friend of mine (“eh?”) came up with an anagram of my solution for the length of time that this would take. Ignore the comma and the question mark.

To find us a pretty egg in Ohio, eh?

In the spirit of Galileo, I can show priority by giving the solution to the anagram. We are not in Ohio or Canada of course, let alone Italy, but this particular egg (a lost Easter egg?) is perhaps not unattractive. Finders keepers!

Possibly one or more of you can find time and speed solutions early enough to make the June Sidereal Times. (If not then, perhaps for the Midsummer edition.) However, as they say on final exams, you must show your work, including the graph. If you can demonstrate that your solutions are more egg-xact than mine, then congratulations: you win and I get egg on my face, so to speak.

You get extra credit if you also decipher the anagram successfully. Fair warning – there are many possible solutions and you should do the math first. One entry per challenger please. There are anagram solvers on the Internet, but I don’t think they will help you very much; I tried them already.

If nobody has found solutions by the time of our September meeting, I can give a 10-minute talk with my answers. Good luck!

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