Two Particles Falling Together Under Gravity

Two point masses \(m_1\) and \(m_2\) start at rest a distance \(R\) apart in empty space, with no forces other than their mutual gravity. How long do they take to collide?

Reducing to one equation

Put both masses on the \(x\)-axis, with \(m_1\) on the left and \(m_2\) on the right. Newton’s second law for each body gives

\[\begin{aligned} m_1 \ddot{x}_1 &= G \frac{m_1 m_2}{\left(x_2-x_1\right)^2}, \\ m_2 \ddot{x}_2 &= -G \frac{m_1 m_2}{\left(x_2-x_1\right)^2}, \end{aligned}\]

so the accelerations are

\[\ddot{x}_1 = G \frac{m_2}{\left(x_2-x_1\right)^2}, \qquad \ddot{x}_2 = -G \frac{m_1}{\left(x_2-x_1\right)^2}.\]

Subtracting, and writing \(M=m_1+m_2\) and \(r=x_2-x_1\) for the separation, we get the standard result

\[\ddot{r}=-\frac{GM}{r^2}.\]

The two bodies meet at their centre of mass, which stays fixed.

Energy integral

Multiplying by \(\dot r\) and integrating in time from the start, when \(r = R\) and \(\dot r = 0\),

\[\frac{d}{dt}\left(\frac{1}{2}\dot{r}^2\right)=-\frac{GM}{r^2}\, \dot{r} \quad\Longrightarrow\quad \frac{1}{2}\dot{r}^2 = \int_{R}^{r} -\frac{GM}{r'^2} \, dr' = GM\left(\frac{1}{r}-\frac{1}{R}\right).\]

The separation decreases, so we take the negative root:

\[\frac{dr}{dt} = -\sqrt{2GM\left(\frac{1}{r}-\frac{1}{R}\right)}.\]

Time to collision

Separating variables,

\[\begin{aligned} \sqrt{\frac{Rr}{R-r}}\, dr &=-\sqrt{2GM}\, dt, \\ \int_{R}^{0}\sqrt{\frac{r}{R-r}}\, dr &= -\sqrt{\frac{2GM}{R}}\; T . \end{aligned}\]

To evaluate the integral, substitute \(r = R\sin^2\phi\), so that \(dr = 2R\sin\phi\cos\phi\, d\phi\) and \(\sqrt{r/(R-r)} = \tan\phi\):

\[\int_{0}^{R}\sqrt{\frac{r}{R-r}}\, dr = 2R\int_0^{\pi/2} \sin^2\phi \, d\phi = \frac{\pi R}{2}.\]

Hence \(\frac{\pi R}{2} = \sqrt{\frac{2GM}{R}}\, T\), and the time to collision is

\[T= \frac{\pi}{2} \sqrt{\frac{R^3}{2GM}} .\]

A check with Kepler’s third law

A head-on fall is the limit of a Kepler ellipse with eccentricity \(e \to 1\), a degenerate ellipse of semi-major axis \(a = R/2\). The fall takes half of an orbital period:

\[\frac{1}{2} \cdot 2\pi\sqrt{\frac{a^3}{GM}} = \pi\sqrt{\frac{R^3}{8GM}} = \frac{\pi}{2}\sqrt{\frac{R^3}{2GM}},\]

in agreement with the direct calculation.

Credits

This calculation follows discussions on Physics Forums and Physics Stack Exchange.

Changes on 25 October 2026

What was wrong. Nothing in the result. One block of equations was garbled when it was typeset.

What changed. The set-up is explained more carefully, and the answer is checked against Kepler’s third law for a radial orbit. Previously titled “Two Particles Collision”.