The Moon has a mean orbital speed of v = 1.03 kms-1, and a mean orbital distance of 384400 km. So the centripetal force is,
\({F}_{mo}\) = \(\frac{{v}^{2}}{{O}_{mo}}\) = 1.03^2/384400 = 0.00000275 kms-2
But the calculated \({F}_{mo}\) based on \({g}_{mo}\) = 0.00162 kms-2 is,
\({F}_{mo}\) = 7.34767309*10^22/(7.34767309*10^22+5.97219*10^24)*0.00162 =0.0000197 kms-2
which is too high, Moon's gravity should be adjusted to,
\({g}_{mo}\) = 0.00000275*(7.34767309*10^22+5.97219*10^24)/ (7.34767309*10^22)=0.000226 kms-2
At the apogee, orbital distance is 406740 km, based on the model developed,
\({v}_{mo1} + {v}_{mo2}\) = (406740*0.00000275)^(0.5) = 1.058 kms-1
And the perigee of 356410 km,
\({v}_{mo1} - {v}_{mo2}\) = (356410*0.00000275)^(0.5) = 0.990 kms-1
And so,
\({v}_{mo1}\) = (1.058+0.990 )/2 = 1.024 kms-1
Compare this with that derived from \(\triangle GPE\)
\({v}_{mo}\) = (3*0.000226*1738)^(0.5) = 1.085 kms-1
The Moon has only a small perpendicular initial velocity component of,
\({v}_{ip}\) = 1.085 - 1.024 = 0.061 kms-1
The other orthogonal component is
\({v}_{ir}\) = \({v}_{mo2}\) = 1.058 - 1.024 = 0.034 kms-1
Therefore the initial velocity of the Moon just before capture is,
\({v}_{i}\) = (0.061^2 + 0.034^2 )^(0.5) = 0.0698 kms-1
and the attack angle is tan-1 (-0.034/0.061) = -29.134 degrees to the perpendicular of the radial orbital line between earth and the Moon..
The Moon gravity has been adjusted downwards based on the centripetal force calculated from the measured data of mean speed and orbital distance. This has implication landing on the Moon. Gravity on the Moon it seems is very light. \({g}_{mo}\) = 0.000226 kms-2 .
The velocity of the Moon obtained from considering a decrease in \(GPE\), \(\triangle GPE\)
\(v\) = \(\sqrt{3{g}_{m}{r}_{m}}\) = (3*0.000226*1737)0.5 = 1.09 kms-2 which is consistent with the measured value of 1.03.
