Equation of the two body problem on a line: acceleration seems to be always negative

Question

This is almost not suitable for this forum but I just can't figure it out by myself:

In an exercise I have the following sentence:

Find the solution the two body problem restricted to a straight line:

$\ddot x(t)=-\frac{Gm}{x^2(t)},\ x(0)=r_0>0,\ \dot x(0)=v_0$

I don't have any problem with the question mathematically, but I was just wondering that from a physical point of view, the acceleration $\ddot x(t)$ that is written in a simplified form that comes from $\ddot x(t)=-\frac{Gmx(t)}{||x(t)||^3} = -\frac{Gmx(t)}{|x(t)|^{3}}=-\text{sign}(x)\frac{Gmx(t)}{x(t)^{3}}=-\text{sign}(x)\frac{Gm}{x(t)^{2}}$

Otherwise the acceleration is always negative, which is not how I imagine the world works.

Or am I wrong?

Answer 1

$$\ddot{x}=-\frac{Gm}{x^2}$$ Is the equation of the free-falling body, when it's dropped from $x(0)>0$. The acceleration is always negative because it's falling downwards, and it can't reach the $x<0$ half-plane. Of course, when $x(0)<0$, we would have $$\ddot{x}=\frac{Gm}{x^2}$$ And your derivation is correct from the vector form of Newton's gravitational law.

Answer 2

I think the answer to your question depends on your interpretation of the meaning of "$x$". You've apparently interpreted it as the Cartesian $x$-coordinate of one of the bodies relative to the centre of mass at the origin, in which case you're correct: the equation should be $\ \ddot x(t)= \frac{Gm|x(t)|}{x(t)^3}\ $ (or your equivalent expression).

However, if you take $\ x\ $ to be the distance of the body from the centre of mass (i.e. the polar $r$-coordinate) then the equation is correct as written. An indication that this might be what is intended is the use of the symbol "$r_0$" for the value of $\ x\ $ at $\ t=0\ $. A key point here is that $\ \ddot x(t)\ $ is not the acceleration as such (which is a vector quantity), but one of its components in a particular direction. In Cartesian coordinates, this direction remains fixed (viz. $\ \mathbf{i}\ $ for the $x$-component), whereas in polar coordinates, where $\ \ddot r\ $ is the component of the acceleration along the radius vector from the origin to the body, the direction, $\ \mathbf{u}_r=$$\mathbf{i}\cos\theta + \mathbf{j}\sin\theta\ $, varies.

In the case when the motion is in a straight line, what happens in polar coordinates when $\ r=0\ $ is that $\ \dot r\ $ is discontinuous and changes sign, being negative just before $\ r\ $ becomes $0$ and positive just after. The polar angle, $\ \theta\ $, is also discontinuous and either increases or decreases by $\ \pi\ $ radians. The acceleration $\ \ddot r\rightarrow-\infty\ $ as $\ r\rightarrow0\ $, but remains negative both before and after the instant when $\ r=0\ $.