If we combine Newton's second law of motion i.e. $F=m\ddot{x}$ and Newton's law of gravity i.e., $$ F=G\frac{mM}{x^2}, $$ where $x$ is distance, we obtain the following equation:
$$\ddot{x}=MG\frac{1}{x^2}$$ My question is how do we solve this differential equation for $x$ ? I see no ordinary way of doing it. I have tried using Laplace transformation but apparently there exists no transformation for $\frac{1}{x^2}$ so I'm left with nothing to use.
On
First, there should be a minus in the front of your RHS. But, moving on ...
When you have an acceleration dependent only on position, a typical technique is to transform the acceleration term like so:
$\displaystyle \frac{d}{dt}(\frac{dx}{dt})=\frac{dx}{dt}\frac{d}{dx}(\frac{dx}{dt})=\frac{1}{2}\frac{d}{dx}(\frac{dx}{dt})^2$ So, dropping messy constants etc.
$\displaystyle \frac{1}{2}\frac{d}{dx}(\frac{dx}{dt})^2 = -\frac{1}{x^2}$
or
$\displaystyle \frac{1}{2}(\frac{dx}{dt})^2 = \frac{1}{x} + E_0$
where $E_0$ is the initial energy of the system $v(0)^2/2-1/x(0)$. This should be more tractable.
On
Hint: Set $p(x)=x'(t)$ so you'll have $x''(t)=p(t)p'(t)$ so $$x''=\frac{MC}{x^2}\to pp'=\frac{MC}{x^2}$$ and so $$\int pdp=\int \frac{MC}{x^2}dx$$
On
You can trudge through the general solution (stipulating that $x=x(t)$) to
$$ x^{''} = f(x) $$
as
$$ \int [c_{1} +2 \int f(x) dx]^{-1/2} dx = c_{2} \pm t $$
In your case (with $f(x)=MG/x^{2}$) the outer integral gets a bit messy, but still doable. Also, $c_1$ and $c_2$ are the integration constants you'll need to determine with your boundary conditions.
Hope this helps some.
Cheers,
Paul Safier
$$ \ddot x=\frac{GM}{x^2}\quad\Longrightarrow\quad \ddot x\dot x=\frac{GM\dot x}{x^2}\quad\Longrightarrow\quad \frac{1}{2}(\dot x^2)^{\bf\dot{}} =-\left(\frac{GM}{x}\right)^{\!\bf\dot{}}, $$ which implies that $$ \frac{1}{2}\dot x^2 =-\frac{GM}{x}+c, $$ for some positive constant $c$. Hence $$ \dot x=\pm\sqrt{2c-\frac{2GM}{x}}. $$