The Moon and Earth system is not subjected to any forces, conservation of energy still holds. And since GPE=0, we still have,
\({v}^{2}_{t}+{v}^{2}_{s}={c}^{2}\)
But from previously, we have seen that gravity is acting outwards on the Moon side and orbital radius, Orbs does not increase because of energy needed. So, if Orbs is to remain constant and \({v}_{s}\) is not zero then there must be a force acting as the centripetal force that keeps the Moon in circular motion. Where's the force that's with you?
Look at
\(\cfrac { dGPE(Orbs) }{ dOrbs } =-{ g }_{ m }e^{ -\cfrac { Orbs }{ { r }_{ m } } }+{ g }_{ m }e^{ (\cfrac { x-Orbs }{ { r }_{ m } } ) }\)
Consider this,
\(d GPE(Orbs)= {F}_{ob} d(Orbs) \)where
\({ F }_{ ob }={ g }_{ m }(e^{ (\cfrac { x-Orbs }{ { r }_{ m } } ) }-e^{ -\cfrac { Orbs }{ { r }_{ m } } })\)
Integrating both sides,
\(\int d GPE(Orbs) =GPE(Orbs)=\int{F}_{ob} d(Orbs) \)
We get an energy term on the L.H.S, and an integration along distance on the R.H.S. This suggests that Fob is a force acting along Orbs. From
\(\cfrac{d GPE(Orbs)}{d Orbs} > 0\)
incremental work done along Orbs is positive, and since the initial energy expression GPE, is the integration of gravity (made negative, a reasoned but arbitrary negative sign) to give a positive expression along x, this suggests that Fob is also in the direction of gravity. Fob is pointing towards earth. Fob is the centripetal force.
From \(\cfrac{{v}^{2}}{{O}_{m}}\)= force needed per unit mass to keep an object in circular motion at radius Om and speed v, we have
\(v=\sqrt { {O}_{m}{ g }_{ m }(e^{ (\cfrac { x-Orbs }{ { r }_{ m } } ) }-e^{ -\cfrac { Orbs }{ { r }_{ m } } })}\)
Theoretically, the body is parked at the lowest GPE point, where GPE=0. It is through this point that the Fob acts. This figure is also the orbit of the Moon measured from the surface of Earth. But Om is measured from the center of the earth, that is Om = x + re.
