There are some apparently open questions in regard to this problem, and I'm not sure I understand exactly what you are doing above. So I will make a suggestion for a way to approach it. Fair warning - I am going to put my satellites into orbit with zero kinetic energy - in other words, I'm only going to oppose gravity and not set them orbiting around the planets. I choose to do this because your original answer suggested you are mostly interested in that part of the problem. Using the work-energy theorem and uniform circular motion, you could add the circular motion of the satellite.
Since we can create stable orbits "anywhere" above a planet, let's assume a constant orbital distance relative to the radius of the body. If the body has a radius of $R$, let's assume each satellite orbits at a radius of $r_s=\gamma R$, where $\gamma$ is some constant. The ISS orbits at like 300 km, so $\gamma \sim (6300+300)/6300 \sim 1.05$ in that case.
Now the work required to launch a satellite into that orbit comes from the force of thrust, which is needed to oppose the force of gravity:
$$F_t(p_i)=\frac{GM(p_i)m}{r^2}$$
Here I am writing "$p_i$" as "planet $i$", so $M(p_i)$ is the mass of planet $i$, I hope that is not too confusing. $m$ is the mass of our satellite and $G$ is the gravitational constant. The work required is the integral of the force from the radius of the planet to the radius of the orbit:
$$W(p_i)=\int_{R(p_i)}^{r_s(p_i)} F_t(p_i) dr=GM(p_i)m\int_{R(p_i)}^{\gamma R(p_i)}\frac{dr}{r^2}$$
Notice that this illustrates one of the problems with your solution - since the force depends on the radius, you cannot simply write $\Delta W=F\Delta x$, you have to take the integral to get the right answer.
I hope this gives you enough to get started - if you still have problems I will add more details. The answer will depend on the value of $\gamma$ that you choose. This was not the only possible way to parametrize the problem. I think another reasonable way would have been to shoot for a constant orbital velocity - but you might run into some problems with flying inside a planet if you choose poorly!
Additionally, don't be worried about large numbers - the energy of a baseball being thrown by a professional pitcher is like 100 J, so launching a spaceship should easily be more than a million times that, right? Putting a baseball ($m$=150 g) into orbit (as above) takes around $10^8$ J. Expect big numbers!
Multiply the equation by $dx/dt$ and integrate once to get
$$
\frac{1}{2}\Bigl(\frac{dx}{dt}\Bigr)^2=\frac{1}{x}+\frac{v_0^2}{2}-\frac{1}{x_0}.
$$
Then
$$
\frac{dx}{dt}=\pm\sqrt{\frac{2}{x}+v_0^2-\frac{2}{x_0}}\,,
$$
a first order equation in separated variables whose solution is
$$
\int_{x_0}^x\frac{dz}{\sqrt{\frac{2}{z}+v_0^2-\frac{2}{x_0}}}=\pm t.
$$
The integral can be computed explicitly, although I do not think you can find a closed form for $x$ in terms of $t$.
Best Answer
Yes you're modeling it wrong. Gravity is an attractive force, so if the mass is at r = 0, and r is nonnegative, then the gravity force should point towards zero, i.e. negative.
$$ - \frac{GMm}{r^2} = m\ddot r$$
(This assumes $m\ll M$ so we don't need to consider reduced mass. See the other answers for the more general case.)
But now Alpha even refuses to solve it. That's fine, because Mathematica can't solve many things. The usual approach is to notice that
$$ \frac{d^2r}{dt^2} = \frac{dv}{dt} = \frac{dv}{dr}\frac{dr}{dt} = v \frac{dv}{dr} $$
to convert that 2nd-order ODE to a 1st-order ODE:
$$ -GM \frac{dr}{r^2} = v \,dv $$
This gives the solution v(r).
$$ \frac{GM}r + \text{constant} = \frac{v^2}2 \qquad (*) $$
But recall $v = dr/dt$. So we have another 1st-order ODE to solve, which gives r(t). The solution should look like that “hell of answer” because of the square root.
Note: If you rearrange the terms you should see that Equation (*) is just conservation of energy.