I have been trying to determine, given the position of a point mass an initial distance $x_0$ from the surface of a spherically symmetric body with mass $M$ and radius $R$, the position of the point mass after a time $t$ has passed. Using Newton's law of gravitation (since the body is spherically symmetric we may treat it as a point mass) and the principle of conservation of energy I have set up the following differential equation: $$-\frac{GM}{R+x_0}+\frac{v_0^2}{2}=-\frac{GM}{R+x(t)}+\frac{\dot x^2(t)}{2},x(0)=x_0$$ Assuming that I set this up correctly (it's been a bit since I studied any mechanics), I now need to solve this differential equation.
Attempt: We can rearrange this equation a bit: $$GM\left(\frac{1}{R+x}-\frac{1}{R+x_0}\right)+\frac{v_0^2}{2}=\frac{\dot x^2}{2}$$ $$\frac{GM(x_0-x)}{(R+x)(R+x_0)}+\frac{v_0^2}{2}=\frac{\dot x^2}{2}$$ $$\frac{2GM(x_0-x)+v_0^2(R+x)(R+x_0)}{2(R+x)(R+x_0)}=\frac{\dot x^2}{2}$$ $$\dot x=\pm\sqrt\frac{2GM(x_0-x)+v_0^2(R+x)(R+x_0)}{(R+x)(R+x_0)}$$ $$\dot x\sqrt\frac{(R+x)(R+x_0)}{2GM(x_0-x)+v_0^2(R+x)(R+x_0)}=\pm1$$ Now this is in a position to be integrated; the integration is messy but it does yield a general solution. However, trying to find a particular solution poses problems. Mathematica will not return a solution with the given initial condition. Is there a method of finding a particular solution to this differential equation (or an approximation)?
The solution to the problem is not a familiar function of $t$. Here, I present a series representing the starting motion; it probably does not converge when $t$ is big enough to crash into $R+x=0$.
You can make the problem (assuming $v_0 = 0$) a bit more tractable by the following scaling transformations:
Work with $u = MG(R+x)$ adn $\tau = MGt$. This rescaling makes the equation into $$\frac{d^2 u}{d\tau^2} = - \frac1{u^2}$$ with $u'(0) = 0$ and $u(0) = MG(R+x_0) \equiv u_0$.
Then transform again to a time variable $s = \frac{\tau}{u_0^{3/2}$.
When you do this, and expand as a series by writing $$ u(s) = u_0 \sum a_n s_n $$ you find that $a_n = 0$ for odd $n$ (this is not surprising, because if you throw the object upward, the motion is symmetric about the time it comes to a stop). Of course, $a_0 = 1$. For even terms, $$ a_{2n} = - \frac{D_n}{(2n)!} $$ where $D_n$ refers to the sequence presented in OEIS as sequence A120419, where the sequence with positive instead of negative coefficients is related to the soluction of an integral equation.