Solving the gravitation problem

by John Church

I posed a gravitation problem in the May Sidereal Times. The problem was to find the time that it would take for two nonmagnetic cubes, 4 cm on an edge, with masses of 1 kilogram each, originally placed with their centers 1 meter apart on a perfectly frictionless horizontal surface, to touch under the sole effect of their mutual gravitation. The right-hand cube is fixed in place and the left-hand cube is free to move towards its neighbor. Assume that there are no nearby masses that could perturb the experiment. The Earth’s vertically-downward gravitational pull is assumed not to have any effect on purely horizontal and frictionless motion.

Here’s my solution. Newton’s Law of Universal Gravitation says that the force attracting these two cubes to each other will be directly proportional to the product of their masses (i.e. 1 kg x 1 kg) and inversely proportional to the square of their separation, which starts off at 1 meter and then begins to decrease. The constantly changing acceleration of the left-hand cube is then specified by Newton’s Second Law of Motion: acceleration equals force divided by mass. Putting all this together, the result is a differential equation (DE). The gravitational constant G is taken as 6.67 x 10^-11 in the SI system of units (meters, kilograms, and seconds).

Now comes calculus. We need to solve this DE to find the separation as a direct function of time. In this case, we need to specify two initial conditions. The first condition is that the initial separation is 1 meter and the second is that the initial velocity of the movable cube is zero.

I have an application called Mathematica that is very good at solving DE’s, but it couldn’t solve this one in terms of known math functions because the equation is nonlinear. Nonlinear in mathematics means that the dependent variable (the separation in this case) is raised to a power other than 1. The separation is squared due to the basic nature of gravitation, i.e. the inverse square law.

Nonlinear DE’s are notoriously hard to solve except in special cases. This reflects the real world, in which most things are nonlinear. Results are usually not directly proportional to causes. This is exemplified in the old saying, “For want of a nail the shoe was lost, for want of a shoe the horse was lost,” etc.

Back to gravitation. Mathematica was able to solve this DE numerically in terms of what’s called an Interpolating Function. Mathematica also made a graph showing the position of the left-hand cube as a function of time, starting from t = 0 when the left-hand cube is released. As expected, the movable cube hardly moves at all for a very long period of time, but it gradually speeds up and the effect feeds back on itself as the separation decreases.

Using the Interpolating Function, one can find the moment when the two 4-cm cubes will just touch by finding the center-to-center distance at various times in seconds and zeroing in by successive approximations. The cubes will touch when this distance is 4 cm, i.e. half the diameter of each cube.

The answer that I found was 162,737 seconds. How does this relate to my anagram “To find us a pretty egg in Ohio, eh?” Anagram solution: This many seconds is about “One point eight eight four days.” It would take nearly two full days for the movable cube to finally touch the fixed one after its short journey. This would be a very boring experiment to watch, even if it could actually be set up and carried out.

I also found how fast the movable cube was moving just before it touched the stationary one. This was about 1 millimeter per minute, which would be barely perceptible with a strong magnifying glass. There wouldn’t be much of a bang at the end. The Big Bang would be a lot more interesting. Talk about nonlinearity !

This entry was posted in September 2019, Sidereal Times and tagged . Bookmark the permalink.

1 Response to Solving the gravitation problem

  1. John Church says:

    Woops, I made a computational mistake. The correct time for the moving cube to touch the stationary one should be 1.569 days, not 1.884 days. I am currently using the math for some other interesting gravitational examples. More about this later. — John C.

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