Here summarize some orbits of motion that I encountered in life experience.

Orbit of double encircling motion

On Feb 28th, I met a problem in a chat group that why any point on a ball encircling a larger ball with half radius has an orbit of a straight line. After some calculation, I find this kind of complex motion can be explained with a super position of two circling motions as the above this shows:

As this figure shows, a small circle of radius r is circling around the larger circle with radius $R = 2r$ on the outer side. However, strange thing happens as the point A on the small circle moving in a straight line along AA’, when the smaller circle is encircling the larger circle.

$r = R/2$

A simpler relation can be draw from this motion that the length of curve $l_{AB} = R\theta = R\Omega t = r\phi = r\omega t$, thus we have the relation between two angular velocity as: $\frac{\omega}{\Omega} = \frac{R}{r} = 2$. However, if we look from the coordinate $x'y'$, we can find that the angular velocity of the small circle is only half that of the original value as: $\omega' = \frac{1}{2}\omega = \Omega$. And we know that only two parallel coordinates could be add up directly. Thus we can split the motion of point A on the small circle in two parts.

The first is the motion of the center of the small circle $P(X,Y)$, taking the center of the larger circle O as the start point. The orbit equation is:

$\begin{array}{l} X = r cos(\Omega t)\\ Y = r sin(\Omega t) \end{array}$

Second is the motion of the point $A(x_A, y_A)$ around the small circle heat $P(X, Y)$. Suppose the relative position of point A in the $x'y'$ coordinates is $A(x', y')$, then we have:

$\begin{array}{l} x' = r cos(-\omega' t)\\ y' = r sin(-\omega' t) \end{array}$

Notice that the two circular motions are moving in the opposite direction, thus the two their angular velocity is different in sign (one positive, one negative). Where $\omega' = \frac{1}{2}\omega = \Omega$, then the real motion of point A(x, y) in the ‘xy’ coordinates is the super position of the two motions as:

$\begin{array}{l} x = X + x' = r cos(\Omega t) + r cos(-\omega' t) = R cos(\Omega t)\\ y = Y + y' = r sin(\Omega t) + r sin(-\omega' t) = 0 \end{array}$

From this relation we can clearly see that the point A(x, y) is moving on the x axis in a straight line.

If the radius of the small ball is not half of the radius of the larger ball. The trajectory of a point on the small ball will not be a straight line. There are two cases. However the angular velocity in the $x'y'$ will be the same:

$r < R/2$

In this case, the motion equation becomes:

$\begin{array}{l} x = X + x' = (R/2 + r)cos(\Omega t)\\ y = Y + y' = (R/2 - r)sin(\Omega t) \end{array}$

This the parameter equation of the ellipse $\frac{x^2}{(R/2 + r)^2 } + \frac{y^2}{(R/2 - r)^2} = 1$

$r > R/2$

In this case, $R/2 - r < 0$. However, it can be easily seen that the orbit equation in the case has the same expression as the case for $r < R/2$. Which is: $\frac{x^2}{(r + R/2)^2 } + \frac{y^2}{(r - R/2)^2} = 1$.