\({F}_{ob}=\cfrac{d{GPE}_{m}}{d Orbs}=-{ g }_{ m }e^{ -\cfrac { Orbs }{ { r }_{ m } } }+{ g }_{ m }e^{ (\cfrac { x-Orbs }{ { r }_{ m } } ) }= {g}_{m}\)
A force develops that opposes an increase in Orbs. It acts on the system and is shared by both the Moon and Earth. It is proposed that it is divided among the two bodies in proportion to their masses.
\({F}_{e} + {F}_{m} = {F}_{ob}\)
\({F}_{m}=\cfrac{{M}_{m}}{{M}_{e}+{M}_{m}}{F}_{ob}\), and
\({F}_{e}=\cfrac{{M}_{e}}{{M}_{e}+{M}_{m}}{F}_{ob}\)
and acts as a centripetal force, keeping Orbs constant.
\({F}_{m}\) = 7.34767309*10^22/(7.34767309*10^22+5.97219*10^24)*0.00162
\({F}_{m}\) = 0.0000197 kms-2
This force acts as centripetal force, therefore
\(\cfrac{{v}^{2}_{m}}{{o}_{m}} = {F}_{m}\)
\({v}_{m}\) = ((384400+6371)* 0.0000197)^(0.5) = 2.77 kms-1
This answer is consistent with the value obtained from considering a decrease in \(GPE\), \(\triangle GPE\)
where v=2.90 kms-1.
Both theoretical answer are consistent.
Both theoretical answer are consistent.
