Thursday, December 29, 2022

Planetary Motion Demonstrated With Simple Spin Model

The construction and use of a simple centripetal force model turned out to be one of the most successful elements of the CXC astronomy course taught in the Caribbean. Students were to use a simple straw or glass tubing and insert a string through it with a mass  m (e.g. rubber ball or other) at one end and one or more weights at the other.  This is shown below:

The CXC astronomy Centripetal Force Apparatus

 In the experiment, the students were to carry out several trials, altering the weights each time and the length of the free string undergoing revolution. They were to note these quantities (the weights, defined in newtons (N), were designated the tension T) and also time the complete rotations made in 20 cycles – thereby obtaining the time for one. The view from above with the tension T labeled (which the students were also required to do) is shown here:

                                      The Tension force in the string and Fc.

We expected the students to be able to somehow infer that the force providing the centripetal force was the tension T in the string (created by the suspended weights) in the same way the centripetal force associated with a given planet was provided by its own centripetal force arising from the gravitational force of attraction of the Sun. Most students never made the connection, though it was often cited!

Realizing the content would be difficult, Janice and I wrote a special ‘Discovering the Stars’ column (in The Barbados Advocate) timed to coincide with the teaching of the section. It included the diagram below:

                                 Diagram used to describe centripetal force in more detail.

Of course, even this was overly simplified since to grasp what’s really happening one needed a more detailed force diagram quantitatively showing the v-vectors at the tangents and their changes:


From this diagram, the key aspect students needed to grasp is that if the motion is uniform then the only way that the centripetal acceleration (ac = v2/ r) arises is via the change in direction of the velocity vector, v. Thus, the acceleration is: D v/r   or:  (vv)/ r, but the magnitude of each vector is: |v| = rq/ t = r w. By similar triangles one would obtain:

D v/v  = s/ r  and   D v = v(s/r)  but s = (rq)/ t

So: D v = v(q/ t) = vw

And since: w = v/r then:

a c = D v/ r = vw/ r = = v2/ r

And: GMm/ r 2   =    m v2/ r

The gravitational force of attraction is equal to the centripetal force. Students were at least partially able to apply this to the case of the planets. In other words, they used their models to reason qualitatively that the Sun provided what the weights did in their model: the force that attracts the planets with an acceleration toward the center of their orbits.  

See Also:

Selected Questions - Answers From All Experts Astronomy Forum (Abolish Kepler's 2nd Law?)


Selected Questions -Answers From All Experts Astronomy Forum (Model Planet And Moon Orbits)

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