# An impractical but difficult problem I stumbled upon

1. Sep 13, 2011

### bigmoneyhat

I'm only a high school physics/calculus student, so bear with me.

The problem is as such: There are two planetary masses, with no pre-existing velocity or acceleration that exist in a vacuum with no other forces being acted on them other than their gravity. What I want to find is how long it would take for these two masses to collide.

I've tried working on this problem myself for a while. So any help would be appreciated.

2. Sep 13, 2011

### rcgldr

3. Sep 14, 2011

### rcgldr

Call the spheres A and B, and let the origin of the coordinate system lie at the midpoint of the line segment defined by their centres.
Calling the sphere centres' positions as functions of time

$$x_{A} (t) , x_{B} (t) , x_{A} (0) = -5 , x_{B} (0) = 5$$

respectively, we define the distance function between them as:

$$D(t)=x_{B} (t) - x_{A} (t) , D(0) = D_{0} = 10$$

Setting up Newton's 2nd law for both, we get, with unit masses:

$$\frac{d^{2}x_{A}}{dt^{2}} = \frac{G}{D^{2}}$$

$$\frac{d^{2}x_{B}}{dt^{2}} = -\frac{G}{D^{2}}$$

whereby the equation for d(t) is readily derived:

$$\frac{d^{2} D}{dt^{2}} = -\frac{2G}{D^{2}} (*)$$

We assume that the initial velocities are 0, i.e

$$\frac{dD}{dt}\mid_{t=0}=0$$

Let us multiply (*) with dD/dt:

$$\frac{d^{2}D}{dt^{2}}\frac{dD}{dt}=-\frac{2G}{D^{2}}\frac{dD}{dt}$$

Integrating both sides from t=0 to some arbitrary t-value, taking due notice of the initial conditions, yields:

$$\frac{1}{2}(\frac{dD}{dt})^{2}=\frac{2G}{D}-\frac{2G}{D_{0}}$$

multiplying with two, taking the square root and remembering that D(t) will be decreasing, we get the diff. eq:

$$\frac{dD}{dt} = -\sqrt{ \frac{4G}{D_{0}}} \sqrt{ \frac{D_{0} - D} {D}}$$

This is a separable diff.eq; we write:

$$\sqrt{ \frac{D}{D_{0} - D}} dD = -\sqrt{ \frac{4G}{D_{0}}} dt$$

We now remember that when they spheres make contact, D(T)=2, where T is the time we're looking for! Thus, we get the equation for T, integrating both sides:

$$\int_{10}^{2} \sqrt{ \frac{D}{D_{0}-D}} dD = -\sqrt{ \frac{4G}{D_{0}}} T$$

or equivalently:

$$T = \sqrt{ \frac{D_{0}}{4G}} \int_{2}^{10} \sqrt{ \frac{D}{D_{0}-D}}dD$$

In order to crack that integral, let us set:

$$u=\sqrt{\frac{D}{D_{0}-D}}\to{D}=D_{0}-\frac{D_{0}}{1+u^{2}}$$

$$dD=\frac{D_{0}2u}{(1+u^{2})^{2}}du$$

The limits are

$$D=10\to{u}=\infty,D=2\to{u}=\frac{1}{2}$$

We thereby get the expression for T in u:

$$T = D_{0} \sqrt{ \frac{D_{0}}{G}} \int_{ \frac{1}{2}}^{ \infty} \frac{u^{2}du} {(1+u^{2})^{2}}$$

doing integration by parts.

$$\int \frac{1}{1+u^{2}}du = \frac{u}{1+u^{2}} + \int \frac{2u^{2}}{(1+u^{2})^{2}}du$$

$$\int \frac{u^{2}}{(1+u^{2})^{2}}du = \frac{1}{2}( \arctan(u) - \frac{u}{1+u^{2}})+C$$

and

$$T = \frac{10^{ \frac{3}{2}}} {2 \sqrt{G}} ( \frac{ \pi}{2} - \arctan( \frac{1}{2}) + \frac{2}{5})$$

Last edited: Sep 14, 2011