# Calculate analytically the time until two spheres meet due to gravitation [duplicate]

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.

• reformulate the problem in terms of a reduced mass and the difference between the two positions? Maybe that'll make things clearer
– Danu
Apr 21 '14 at 10:32
• I don't know what you mean analytically. Is it so hard to make a differential equation and solve it? Apr 21 '14 at 10:46
• 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. Apr 21 '14 at 11:11
• @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. Apr 21 '14 at 12:39

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 $$T=t(r_1+r_2).$$
• @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 $e=1$. Apr 23 '14 at 19:25