Let me begin by sincerely thanking all of you for your comments and direction. It is proving most valuable in raising my level of competence as regards the physics and math involved.
Steamking: Thanks...I now surmise (I believe correctly) that force is measured in Newtons, whereas energy is measured in Joules. I got Joules because I didn't square the meter unit in the denominators, because I thought that you can't do such things as compute or exponentially increase or diminish the value of units, whereas this can be done with scalars, but I must have been wrong about this. I reason that energy is force applied to a given distance, or is this work? What is the arcane distinction then between energy and work? Is work the kinetic species of energy, whereas energy measured in Joules could have many different varieties? Finally: if I do arrive at some number of Newtons, how do I convert this to Joules?
haruspex: You give sound advice regarding the utility of remaining symbolic initially. So much so that I realize I need to drop the lunar radius component latterly subtracted in parantheses from the ##r_{\text{moon}}## term. I also realized that the calculation should involve simply a distance between the centers of mass already mentioned. So to answer your question in symbolic terms, the relationship of ##r_{\text {Earth}}## to ##r_{\text {Moon}}## is:
##r_{\text {Earth}}~ =## (distance in meters from Earth COM to Moon COM) - ##r_{\text {Moon}}##.
gneill: the whole system is not meant to be treated very realistically, but rather is grossly simplified such that the moon and Earth are not rotating, are stationary in respect to each other, and that there are no other systems astrophysically. Also, the path of travel is in a straight line. Perhaps I neglected to mention these simplifying assumptions previously. You bring to my attention an interesting detail though. Would however the force necessary in my very simplified system be less than the necessary force of a real space vehicle in a real system, or the same (when all outside gravitational forces are disregarded)?
Now I come to resume my struggle to solve this problem...
At any given point along the path, the total gravitational force exerted upon this object will be considered as:$$F_{\text {Moon}} - F_{\text {Earth}} = F_{\text{gravity}}$$
...using the lunar center of mass as the frame of reference. Given the much greater mass of the Earth compared to the moon, over this path there will be a far greater net force exerted by the Earth. And so, if I was to simply integrate across the entire distance, I would end up with a negative number, indicating that no force would be necessary to fall to the Earth from the Moon's surface, which is obviously wrong. I only have to make the object overcome the gravitational pull of the Moon, and then can fall to the Earth after reaching a point of equilibrium betwixt these differing gravitational forces. To calculate the force necessary to overcome the lunar gravity and reach this point of equilibrium, I must first compute where this point is on the flightpath. To do so, I set the gravitational force equal to 0 in the former equation:
$$F_{\text {Moon}} - F_{\text {Earth}} = 0$$
And then add ##F_{\text {Earth}}## to both sides, thus generating:
$$F_{\text {Moon}} = F_{\text {Earth}}$$
Next I restate this equation by putting in all the variables of these force functions:
$$G \left(\frac {(mass_{\text{moon}})(mass_{\text{object}})}{r_{\text{moon}}^2} \right)= G \left(\frac {(mass_{\text{earth}})(mass_{\text{object}})}{r_{\text{earth}}^2} \right)$$
This equation may however be algebraically simplified by eliminating the numbers that occur on both sides of the equation to render:
$$\frac {mass_{\text{moon}}}{r_{\text{moon}}^2} = \frac {mass_{\text{earth}}}{r_{\text{earth}}^2}$$
restating ##r_{\text{earth}}## in terms of ##r_{\text{moon}}##yields (with the phrase: "COMCOM" indicating the distance between the centers of mass for the Earth and Moon):
$$\frac {mass_{\text{moon}}}{r_{\text{moon}}^2} = \frac {mass_{\text{earth}}}{((\text{COMCOM}) - r_{\text{moon}})^2}$$
A little algebra and simplification is applied to isolate ##r_{\text{moon}}## thusly:
$$\frac {\text{COMCOM}}{\sqrt{\frac{\text{mass}_{\text{Earth}}}{\text{mass}_{\text{Moon}}}} +1} = r_{\text{balance point}}$$
Next, known and aforementioned values are plugged into the formula to thus give us (these values were listed above in the initial entry of this thread):
$$\frac {\text{392,580,569_m}}{\sqrt{\frac{5.972 * 10^{24}_{}\text{kg}}{7.34767309 × 10^{22}_{}\text{kg}}} +1} = r_{\text{balance point}}$$
Which computes to: ##39,197,693.5716_m## away from the Moon's COM. This will form the upper boundary of my integration, with the lower end being the starting point at the lunar surface or ##1,737,287_m##.
Taking my previously established formula of gravitational force at any point ##F_{\text {Moon}} - F_{\text {Earth}} = F_{\text{gravity}}##, I incorporate all the relevant variables to render it into a complete form viz:
$$\left(G \left(\frac {(mass_{\text{moon}})(mass_{\text{object}})}{r_{\text{moon}}^2} \right)\right) - \left( G \left(\frac {(mass_{\text{earth}})(mass_{\text{object}})}{(\text{COMCOM} - r_{\text{moon}})^2} \right)\right) = F_{\text{gravity}}$$
Now to integrate...
I'll integrate first for the moon's gravity, and then the Earth's. Represented symbolically, it appears thusly:
$$\left(\int_{\text{start}}^{\text{balance point}}\!G \left(\frac {(mass_{\text{moon}})(mass_{\text{object}})}{r_{\text{moon}}^2} \right)dr\right) - \left(\int_{\text{start}}^{\text{balance point}}\! G \left(\frac {(mass_{\text{earth}})(mass_{\text{object}})}{(\text{COMCOM} - r_{\text{moon}})^2} \right)dr\right) = F_{\text{to reach balance point}}$$
The constants in the numerators and G can all be multiplied together and taken outside the integral to be multiplied by the result later:
$$\left((G)(mass_{\text{moon}})(mass_{\text{object}})\int_{\text{start}}^{\text{balance point}}\! \left(\frac {1}{r_{\text{moon}}^2} \right)dr\right) - \left((G)(mass_{\text{earth}})(mass_{\text{object}})\int_{\text{start}}^{\text{balance point}}\! \left(\frac {1}{(\text{COMCOM} - r_{\text{moon}})^2} \right)dr\right) = F_{\text{to reach balance point}}$$
The integrals themselves are fairly simple - being quickly looked up in a table to be;
$$(G)(mass_{\text{moon}})(mass_{\text{object}})\int_{\text{start}}^{\text{balance point}}\! \left(\frac {1}{r_{\text{moon}}^2} \right)dr = -\frac{(G)(mass_{\text{moon}})(mass_{\text{object}})}{r_{\text{moon}}}_{\text{balance point}} +\frac{(G)(mass_{\text{moon}})(mass_{\text{object}})}{r_{\text{moon}}}_{\text{start}}$$
for the first, and guessing and checking to generate:
$$(G)(mass_{\text{earth}})(mass_{\text{object}})\int_{\text{start}}^{\text{balance point}}\! \left(\frac{1}{(\text{COMCOM} - r_{\text{moon}})^2} \right)dr = \frac{(G)(mass_{\text{earth}})(mass_{\text{object}})}{-r_{\text{moon}} + COMCOM}_{\text{balance point}} - \frac{(G)(mass_{\text{earth}})(mass_{\text{object}})}{-r_{\text{moon}} + COMCOM}_{\text{start}}$$
for the second.
Now that I've determined what these formulas of integration symbolically are, I plug in the values for ##F_{\text{moon}}## and compute:
$$(G)(mass_{\text{moon}})(mass_{\text{object}}) = (6.67408 *10^{-11}_{~~~~~~~m^3 s^{-2}\text{kg}^{-1}})(7.34767309 * 10^{22}_{~~~\text{kg}})(61.14_{\text{kg}}) = 2.9982419 * 10^{14}_{~~~m^3 s^{-2}\text{kg}}\Rightarrow$$
$$ -\frac{2.9982419 * 10^{14}_{~~~m^3 s^{-2}\text{kg}}}{39,197,693.5716_m}_{\text{balance point}} +\frac{2.9982419 * 10^{14}_{~~~m^3 s^{-2}\text{kg}}}{1,737,287_m}_{\text{start}} = -7,649,026.30726_{\text{Joules}} + 172,581,841.688_{\text{Joules}} = 164,932,815.38_{\text{Joules}} = F_{\text{moon}}$$
Next I similarly plug in and compute values for ##F_{\text{earth}}##:
$$(G)(mass_{\text{earth}})(mass_{\text{object}}) = (6.67408 *10^{-11}_{~~~~~~~m^3 s^{-2}\text{kg}^{-1}})( 5.972 * 10^{24}_{~~~\text{kg}})(61.14_{\text{kg}}) = 2.436894 * 10^{16}_{~~~m^3 s^{-2}\text{kg}}\Rightarrow$$
$$ \frac{2.436894 * 10^{16}_{~~~m^3 s^{-2}\text{kg}}}{-39,197,693.5716_m + 392, 580, 569_m}_{\text{balance point}} -\frac{2.436894 * 10^{16}_{~~~m^3 s^{-2}\text{kg}}}{-1,737,287_m +392, 580, 569_m}_{\text{start}} = 68,959,029.5855_{\text{Joules}} - 62,349,645.2985_{\text{Joules}} = 6,609,384.287_{\text{Joules}} = F_{\text{earth}}$$
Now that I have integrated both functions of ##F_{\text{moon}}## and ##F_{\text{earth}}## as well as computed their values over the range of the limited integral, I simply need to subtract one value from another according to the aforementioned equation of ##F_{\text {Moon}} - F_{\text {Earth}} = F_{\text{gravity}}##...
$$164,932,815.38_{\text{Joules}} - 6,609,384.287_{\text{Joules}} = 158,323,431.093_{\text{Joules}}= F_{\text{gravity}}$$
...Which is the necessary energy to reach this point of equilibrium.
Ladies and gentlemen: I'm going to pause here for now, due to the following reasons; I wish to give you all an opportunity to scrutinize my work thus far and determine if there are any necessary corrections and thus forestall compounding errors and/or building upon faulty intellectual/mathematical foundations as it were, but also because I'm tired.
So here is my attempt at a solution for now. Stay tuned for the next episode...
