# Exact distance travelled by an object due to gravity only

I am aware of the fact that for two point masses in space, the time that it will take for them to collide is, T=$$\pi \sqrt{\frac{r_i^3}{8GM}}$$, where M is the sum of the 2 bodies' masses, $$r_i$$ is the distance between them and I'm assuming that the 2 objects were initially at rest and nothing else is affecting the system. However I would like to know the distance travelled by the objects. I'm assuming a 1D line where the first object, $$m_1$$ is on the origin initially and the second object, $$m_2$$ is $$r_i$$ units to the left of the first object (initially).

If I denote $$x_1$$ as the distance travelled by the first object with reference to the origin, and $$x_2$$ as the distance travelled by the second object to the left (so $$x_2$$ is negative), then I get these equations:

$$x_1'' (t) = G\frac{m_2}{(r_i-x_1+x_2)^2}$$ $$x_2'' (t) = -G\frac{m_1}{(r_i-x_1+x_2)^2}$$ $$x_1 (0) = 0$$ $$x_2 (0) = 0$$ $$x_1 (T) = r_i + x_2 (T)$$ Note that the distance between the 2 objects, $$r=r_i-x_1+x_2$$

I don't know how to solve those by themselves but of course I can get to the point: \begin{align*} r''(t) = - \frac{GM}{r^2} \end{align*} Defining v (t)=r' (t): \begin{align*} \frac{dv}{dt} & = - \frac{GM}{r^2} \\ \frac{dv}{dr} \frac{dr}{dt} & = - \frac{GM}{r^2} \\ v \frac{dv}{dr} & = - \frac{GM}{r^2} \\ \int_{0}^{v}v dv & = -GM \int_{r_i}^{r} \frac{1}{r^2} dr \\ \frac{v^2}{2} & = GM ( \frac{r_i-r}{rr_i}) \\ v & = \pm \sqrt{2GM}\sqrt{ \frac{r_i-r}{rr_i}} \\ \sqrt{ \frac{rr_i}{r_i-r}} \frac{dr}{dt} & = \pm \sqrt{2GM} \end{align*}

I don't know what to do from here. I also thought of the law of the conservation of energy so I came up with this: \begin{align*} x_1\sum F_1 & = x_2 \sum F_2 \end{align*}

Essentially I just want to know what $$x_1 (T)$$ is.

Since they are initially at rest and since there are no external forces we know that the center of mass does not change. When they collide then they must be located at the center of mass. Therefore, the distance they travel is simply the distance to their center of mass:

$$r_1(T)=r_2(T)= \frac{m_1 r_1(0) + m_2 r_2(0)}{m_1+m_2}$$

so the distance is $$x_1 = r_1(T)-r_1(0)$$

Where $$r_1(t)$$ and $$r_2(t)$$ are the object's positions at time $$t$$ (it is not necessary for $$r_1(0)=0$$).

• Ok just to clarify, in your answer $r_2(t)$ is the second object's position relative to the origin, while in my post $x_2(t)$ is the second object's position relative to it's initial position, right? Also if I wanted to generalize this to spherical objects with radii R, then the final position (of the centre of mass of the object) would be $r_1(T)-r_1(0)-R$, right? Commented Apr 5, 2021 at 14:59
• Yes $r_2(t)$ is the second object's position relative to the origin. For non-point objects they will collide at some $t<T$. The center of mass will still be given by the expression above, but they will collide before reaching it.
– Dale
Commented Apr 5, 2021 at 15:46
• So would this picture describe the collision or would it be different? imgur.com/a/iO7RNx5 Commented Apr 6, 2021 at 10:34
• The center of mass will probably wind up inside the larger object. It is possible but very unlikely for it to end up at the surface.
– Dale
Commented Apr 6, 2021 at 11:36
• May I ask why that is the case? Commented Apr 6, 2021 at 12:43