How Strong is Gravity?

If $x$ is the distance from the centre of each sphere to their centre of mass (so the two centres are $2x$ apart), then $m\ddot{x} \; = \; -\frac{Gm^2}{4x^2}$ where $m = \tfrac43\pi R^3\rho$ is the mass of each sphere. Thus $\tfrac12v^2 \; = \; \frac{Gm}{4x} - \frac{Gm}{2D} \; = \; \frac{Gm(D-2x)}{4Dx}$ where $v$ is the speed of each sphere, so that $\frac{dx}{dt} \; = \; v \; = \; -\sqrt{\frac{Gm(D-2x)}{2Dx}}$ Thus the time until the spheres collide is $\begin{aligned} T & = \; \int_R^{\frac12D}\sqrt{\frac{2Dx}{Gm(D-2x)}}\,dx \; = \; \sqrt{\frac{D^3}{Gm}}\int_{\sin^{-1}\sqrt{\frac{2R}{D}}}^{\frac12\pi} \sin^2\theta\,d\theta \\ & = \; \tfrac12\sqrt{\frac{D^3}{Gm}}\left[\tfrac12\pi - \sin^{-1}\sqrt{\frac{2R}{D}} + \frac{\sqrt{2R(D-2R)}}{D} \right] \end{aligned}$ With the given values, $T = 16061.7$ seconds, or $\boxed{4.46158}$ hours.

Two-body problems are most easily studied with the concept of 'reduced mass', which considerably simplifies the calculations.

In this model we could just say

• there is attraction by a single fixed field due to the reduced mass $μ=\frac{m_1m_2}{m_1+m_2}$
• only one body is moving, under the influence of this  gravitation field
• the vector difference of position just corresponds to the position of the moving body relative to the fixed mass μ
• kinetic energy of the system corresponds to kinetic energy of just the moving body, as if it had the reduced mass

First calculate the mass, $m=(4/3)πR^3ρ = 32982 kg$ and $μG=\frac{1}{2}mG=1.10062×10^-6 m^3s^{-2}$ .

Finding the time for point masses, until distance is 0 If we had a point mass, we could say that its is in an an extremely elliptical orbit (ε=1).

Orbital time only depends on the length of the (semi)major axis, so can be found via a circular orbit: A full circular period takes $T=2πr/v$ , and we can find v from setting the gravitational acceleration $μG/r^2$ equal to the centripetal acceleration $v^2/r$ so that $v=\sqrt{μG/r}$ . Hence $T=2π \sqrt{r^3/μGr}$ .

This same period would apply to an extremely elliptical orbit (ε=1), where one of the foci is at the starting point, while the other is at the end point. Time to impact $t$ then corresponds to only a quarter period, $t=π/2 \sqrt{r^3/μG}$ . So the major axis of our 'ellipse' is 10m and for r we need to use the semimajor axis (r=5) to find $t=π/2 \sqrt{r^3/μG}=16740.0 \text{ seconds}$ But this would be the time if they were point masses, balls with radius 1 will meet before that, namely when the separation of their centers equals 2 m.

Estimating the overestimation (the time difference if they approach only to distance 2) Next consider the energy of the whole system. Since potential energy is less naturally modeled using the reduced mass, I switch back to the model with two moving masses. The gravitational potential energy of the masses together is $-m^2G/r$ , and the kinetic energy of the two masses together is $mv^2$ . So $v^2-mG/r$ is a constant, which is $-mG/10$ , given the initial condition. This means that at impact, when the centers have separation of 2 meters, $v^2-mG/2=mG/10$ . So at impact, each of the masses is moving at $v=\sqrt{0.4 mG}=9.38 ×10^{-4}m/s$ . If their centers continued moving at this speed, they would meet in 1066 s, but they would accelerate further. So 1066 is an upper limit to our overestimation and hence the actual time to impact is somewhere between 15674 and 16740 seconds.

Alas, this is not accurate enough for three-digit precision, so I resorted to a numerical approach, which gave $16062 s$ , which is $\boxed{4.462 h}$ . The impact speed calculated above and the total time to reach D=0 were both duplicated with this numerical approximation (the latter at 16741), which gives confidence that the accuracy is within a second.

But I really should have done the integral, like Mark Hennings did. Nice job Mark!