Imagine you have two homogeneous spheres with the same diameter of $d=0.1 m$. They have the same mass $m = 1 kg$. The distance between the centers of mass is $r= 1 m$. Their electrical charge shall be disregarded. At $t=0$ the spheres do not have any relative motion to each other. Due to gravitation they will accelerate and start moving towards each other. After some time they will touch each other.

How to calculate analytically the time it takes the two spheres to meet each other. I'm not interested in a numerical solution.

I have already tried several ways but I don't get to a solution.

Imagine that the 2 spheres have different masses and diameters. $m_{1}=2 kg$, $m_{2}=5 kg$, $d_{1}=0.03 m$, and $d_{2}=0.3m$. How to calculate analytically when and where the 2 spheres are going to meet?

How do you calculate the second problem taking the theory of relativity into account? I know that it will not change the result that much but I am interested in the mathematical solution.

  • $\begingroup$ reformulate the problem in terms of a reduced mass and the difference between the two positions? Maybe that'll make things clearer $\endgroup$
    – Danu
    Commented Apr 21, 2014 at 10:32
  • $\begingroup$ I don't know what you mean analytically. Is it so hard to make a differential equation and solve it? $\endgroup$
    – evil999man
    Commented Apr 21, 2014 at 10:46
  • $\begingroup$ See e.g. Wikipedia. Possible duplicates: physics.stackexchange.com/q/3534/2451 , physics.stackexchange.com/q/14700/2451 , physics.stackexchange.com/q/19813/2451 , and links therein. $\endgroup$
    – Qmechanic
    Commented Apr 21, 2014 at 11:11
  • $\begingroup$ @Awesome: By analytically, I mean to solve it with an equation or something like that instead of solving it numerically with a computer. How to solve this differential equation? Unfortunately I have no idea how to solve $d^2 s / d^2 t = G m / r^2$. Concerning my second problem I don't even know how to formulate the differential equation since the masses aren't equal. $\endgroup$
    – user50224
    Commented Apr 21, 2014 at 12:39

1 Answer 1


Qmechanic basically answered it in a comment, but to reiterate, the Wikipedia article on free-fall details this exact scenario: the time as a function of separation becomes $$t(y)=\sqrt{\frac{y_0^3}{2G(m_1+m_2)}}\left(\sqrt{\frac{y}{y_0}\left(1-\frac{y}{y_0}\right)}+\cos^{-1}\left(\frac{y}{y_0}\right)\right)$$ where $y_0$ is the initial separation. Letting the radii be $r_1,r_2$, this yields a collision time $$T=t(r_1+r_2).$$

  • $\begingroup$ That looks good. Where are they going to collide? $\endgroup$
    – user50224
    Commented Apr 23, 2014 at 6:36
  • 1
    $\begingroup$ @user50224: In the first case, 108367 second, in the second case, 58107 second. $\endgroup$ Commented Apr 23, 2014 at 11:49
  • $\begingroup$ Thanks for your answer,but I didn't mean when the are going to collide but where. Do you know a formula for that as well? $\endgroup$
    – user50224
    Commented Apr 23, 2014 at 19:08
  • 1
    $\begingroup$ @user50224: Sorry, I misread your original comment. The total linear momentum is zero at all times. Therefore, the collision configuration will have the same center of mass as the initial configuration. So that's where it'll be. If you're interested in where the surfaces collide, you can get that as well. $\endgroup$ Commented Apr 23, 2014 at 19:11
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    $\begingroup$ @user50224: I plugged it into Mathematica, as << PhysicalConstants` y0 = 1 Meter; m1 = 1 Kilo Gram; m2 = 1 Kilo Gram; y = 0.1 Meter; G = GravitationalConstant; Convert[Sqrt[y0^3/( 2 G (m1 + m2))] (Sqrt[y/y0 (1 - y/y0)] + ArcCos[y/y0]), Second]. I gave a link to the derivation in my answer already; the derivation follows from it being a special case of a Keplerian orbit with eccentricity $e=1$. $\endgroup$ Commented Apr 23, 2014 at 19:25

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