To account for orbital eccentricity, it is proposed that Earth speed is the superposition of 2 velocity components
\({V}_{e1} - {V}_{e2}\) = (\({F}_{e}\)*perihelion)^(0.5) = (0.000005886*147098290)^(0.5) = 29.42 kms-1
where the velocities subtracted and
\({V}_{e1} + {V}_{e2}\) = (\({F}_{e}\)*aphelion)^(0.5) = (0.000005886*152098232)^(0.5) = 29.92 kms-1
where the velocities added. And so,
\({V}_{e1}\) = (29.42+29.92)/2 = 29.67 kms-1 and
\({V}_{e2}\) = 29.92 - 29.67 = 0.25 kms-1
The smaller component is rotating with same period as the first larger component and the two sum in a vector manner. Imagine the smaller vector rotating at the tip of the larger vector, as the latter sweeps through in circular motion.
An exaggerated plot is shown below.
The following equations were plotted:
((149.560)^2-x^2)^(0.5)
-((149.560)^2-x^2)^(0.5)
30*cos((pi/2)*x/149.56)+((149.560)^2-x^2)^(0.5)
30*cos((pi/2)*x/149.56)-((149.560)^2-x^2)^(0.5)
The first larger velocity is in circular motion as a result of a drop in GPE and under the constrain of Orbs not increasing without energy input. This larger component is of two parts, one due to \(\triangle GPE\) and the other, a parallel velocity component of the initial velocity \({V}_{i}\) when Earth was captured by the Sun.
\({V}_{e1}\) = \(\sqrt{2*\triangle GPE} + {V}_{ip} \) = 13.69 + \({V}_{ip}\) = 29.67 kms-1
\({V}_{ip}\) = 29.67 - 13.69 = 15.98 kms-1
The second smaller velocity is the perpendicular component of \({V}_{i}\).
\({V}_{ir}\) = 0.25 kms-1
\({V}_{i} =\sqrt{{V}^{2}_{ip} + {V}^{2}_{ir}}\) = ((0.25)^2 + (15.98)^2)^(0.5) = 15.98 kms-1
Earth was captured at the part of the orbit where there was no deviation from the circular path, in the direction of the final orbital path. At that point the velocity component pointing towards the Sun was 0.25 kms-1. This conclusion comes from examining the situation after the capture.
Earth was capture by the Sun with an initial velocity of 15.98 kms-1. At an attack angle of tan-1\((\frac{{V}_{ir}}{{V}_{ip}}\)) = tan-1(-0.25/15.98) = -0.896 degrees to perpendicular of the orbital line joining Earth and the Sun.
After Earth's capture, both \({V}_{e1}\) and \({V}_{e2}\) rotate with the same period, as they are both in the same gravitational field, under the same set of conditions. The larger rotating vector is simulated by a circle and the smaller periodic vector simulated by the cosine function.
I love it when I make things simpler...
