Let’s push Mercury nearby Sun…
if Sun was without Spin then Mercury, subject only to an Aether flowing radially towards the Star, would describe a trajectory on a well-defined plane, trajectory that, dependently on its initial momentum, could possibly have a sad ending…
Sun has instead its own rotation…
the orbit of Mercury is then conditioned by a Spin force that pushes the Planet tangentially, giving it a rotational movement around the Star.
Let’s suppose that the Mercury’s first orbit is on a plane passing through the Sun’s axis of rotation… Mercury, moving closer to the equatorial plane of Sun, will be subject to an increasing Spin force that will bend its orbital plane, i.e. the Planet will uniform its movement of revolution to the rotation of the Star.
Then the orbital “plane” of Mercury, similarly to a coin that, in search of its balance, rolls oblique on a plane, progressively reduces its inclination with respect to the equatorial plane of Sun, until to coincide with this.
Result: Mercury revolves on the equatorial plane of Sun, on a circular orbit.
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page extracted from the eBook Theory of Spherical Vortices
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