# Inverse Square Attraction

Imagine two point-like particles experiencing mutual attraction.

The masses of the two particles are $m_1$ and $m_2$, and their respective positions are $\bold{r}_1$ and $\bold{r}_2$.

Suppose the force on particle 1 due to particle 2 is

$\displaystyle \bold{F}_{12}=-\frac{\alpha(\bold{r}_1-\bold{r}_2)}{|\bold{r}_1-\bold{r}_2|^3}$

and, from Newton’s third law, the force on particle 2 due to particle 1 is

$\displaystyle \bold{F}_{21}=-\bold{F}_{12}=\frac{\alpha(\bold{r}_1-\bold{r}_2)}{|\bold{r}_1-\bold{r}_2|^3}$

where $\alpha$ is a positive constant. A force of this form is said to obey an inverse square law, since

$\displaystyle |\bold{F}_{ij}|\propto\frac{1}{|\bold{r}_i-\bold{r}_j|^2}$

Two familiar examples of forces that take this form are gravitation between masses and the electrostatic force between opposite charges.

First we’ll answer the question: what is the equation of motion governing the particles’ separation?

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From Newton’s second law,

$\bold{F}_{12}=m_1\ddot{\bold{r}}_1$

$\bold{F}_{21}=m_2\ddot{\bold{r}}_2$

where $\ddot{x}$ denotes the second derivative of $x$ with respect to time. By equating these expressions for the forces with those above, we can say that

$\displaystyle \ddot{\bold{r}}_1=-\frac{\alpha}{m_1}\frac{\bold{r}_1-\bold{r}_2}{|\bold{r}_1-\bold{r}_2|^3}$

$\displaystyle \ddot{\bold{r}}_2=\frac{\alpha}{m_2}\frac{\bold{r}_1-\bold{r}_2}{|\bold{r}_1-\bold{r}_2|^3}$

hence

$\displaystyle \ddot{\bold{r}}_1-\ddot{\bold{r}}_2=-\alpha\Bigg(\frac{1}{m_1}+\frac{1}{m_2}\Bigg)\frac{\bold{r}_1-\bold{r}_2}{|\bold{r}_1-\bold{r}_2|^3}$

This equation can be tidied up by defining the separation vector $\bold{r}$ as

$\bold{r}=\bold{r}_1-\bold{r}_2$

and the system’s reduced mass $\mu$ through

$\displaystyle \frac{1}{\mu}=\frac{1}{m_1}+\frac{1}{m_2}$

So

$\displaystyle \ddot{\bold{r}}=-\frac{\alpha}{\mu}\frac{\bold{r}}{|\bold{r}|^3}$

Force convenience we define

$\bold{r}=r\hat{\bold{r}}$

where $r=|\bold{r}|$ is the physical distance between the two masses and $\hat{\bold{r}}$ is a vector of unit magnitude which is parallel to $\bold{r}$. This simplifies the equation of motion to

$\displaystyle \ddot{r}\hat{\bold{r}}=-\frac{\alpha}{\mu}\frac{\hat{\bold{r}}}{r^2}$

assuming the vector $\hat{\bold{r}}$ does not vary with time (the two particles lie on a common line for all time). So, dispensing with the vectorial aspect,

$\displaystyle \ddot{r}=-\frac{\alpha}{\mu}\frac{1}{r^2}$

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We’re first going to use a very useful trick using the chain rule:

$\displaystyle \ddot{r}=\frac{d\dot{r}}{dt}$

$\displaystyle \ddot{r}=\frac{d\dot{r}}{dr}\frac{dr}{dt}$

$\displaystyle \ddot{r}=\frac{d\dot{r}}{dr}\dot{r}$

This changes the independent variable from $t$ to $r$. Substituting this expression into the equation of motion makes it directly integrable:

$\displaystyle \frac{d\dot{r}}{dr}\dot{r}=-\frac{\alpha}{\mu}\frac{1}{r^2}$

$\displaystyle \mu\dot{r}\,d\dot{r}=-\frac{\alpha}{r^2}\,dr$

$\displaystyle \frac{1}{2}\mu\dot{r}^2=\frac{\alpha}{r}+c$

where $c$ is a constant of integration to be determined from initial conditions. If we stipulate that the two particles are initially at rest and separated by a distance $r_0$, then

$\displaystyle c=-\frac{\alpha}{r_0}$

$\displaystyle \frac{1}{2}\mu\dot{r}^2=\alpha\Bigg(\frac{1}{r}-\frac{1}{r_0}\Bigg)$

We rearrange for the derivative of $r$, getting ready to integrate again:

$\displaystyle \frac{dr}{dt}=-\sqrt{\frac{2\alpha}{\mu r_0}\Big(\frac{r_0}{r}-1\Big)}$

It’s important that we take the negative root; if the force is attractive, the distance between the two particles must be decreasing.

$\displaystyle -\sqrt{\frac{\mu r_0}{2\alpha}}\frac{dr}{\sqrt{\frac{r_0}{r}-1}}=dt$

$\displaystyle -\sqrt{\frac{\mu r_0}{2\alpha}}\int_r^{r_0}\frac{dr}{\sqrt{\frac{r_0}{r}-1}}=t$

This integral looks scary, but can be solved by the substitution

$r=r_0\cos^2\theta$

The differential is

$dr=-2r_0\sin\theta\cos\theta\,d\theta$

Substituting this in, we find

$\displaystyle \sqrt{\frac{\mu r_0}{2\alpha}}\int_{\theta(r_0)}^{\theta(r)}\frac{2r_0\sin\theta\cos\theta}{\sqrt{\sec^2\theta-1}}d\theta=t$

After tidying up all the trigonometric terms, we’re left with

$\displaystyle 2r_0\sqrt{\frac{\mu r_0}{2\alpha}}\int_{\theta(r_0)}^{\theta(r)}\cos^2\theta\,d\theta=t$

This is a comparatively non-terrifying integral. We use the half-angle formula to get rid of the cosine squared:

$\displaystyle r_0\sqrt{\frac{\mu r_0}{2\alpha}}\int_{\theta(r_0)}^{\theta(r)}(1+\cos 2\theta)\,d\theta=t$

We integrate:

$\displaystyle r_0\sqrt{\frac{\mu r_0}{2\alpha}}\Big[\theta+\frac{1}{2}\sin 2\theta\Big]_{\theta(r_0)}^{\theta(r)}=t$

We use the double angle formula to make re-substitution easier:

$\displaystyle r_0\sqrt{\frac{\mu r_0}{2\alpha}}[\theta+\sin\theta\cos\theta]_{\theta(r_0)}^{\theta(r)}=t$

In terms of the variable $r$ this is

$\displaystyle r_0\sqrt{\frac{\mu r_0}{2\alpha}}\Bigg[\arccos\Bigg(\sqrt{\frac{r}{r_0}}\Bigg)+\sqrt{\frac{r}{r_0}}\sqrt{1-\frac{r}{r_0}}\Bigg]_{r_0}^r=t$

$\displaystyle r_0\sqrt{\frac{\mu r_0}{2\alpha}}\Bigg(\arccos\Bigg(\sqrt\frac{r}{r_0}\Bigg)+\sqrt{\frac{r}{r_0}\Big(1-\frac{r}{r_0}\Big)}\Bigg)=t$

This is the time taken for the two particles initially separated by distance $r_0$ to reach a separation $r$.

When do the particles collide? This occurs when the particles’  separation $r$ is equal to 0:

$\displaystyle t(0)=t_0=r_0\sqrt{\frac{\mu r_0}{2\alpha}}\arccos 0$

$\displaystyle t_0=r_0\sqrt{\frac{\mu r_0}{2\alpha}}\frac{\pi}{2}$

$\displaystyle t_0=\sqrt{\frac{\mu \pi^2 r_0^3}{8\alpha}}$

Using the definition of the reduced mass $\mu$, this is

$\displaystyle t_0=\sqrt{\frac{\pi^2 r_0^3}{8\alpha}\frac{m_1 m_2}{m_1+m_2}}$

This equation is fun to play with.

For example, let’s say the two objects are attracted via the gravitational force, in which case the constant of proportionality $\alpha$ is given by

$\alpha=Gm_1 m_2$

where $G$ is the gravitational constant. Then

$\displaystyle t_0=\sqrt{\frac{\pi^2 r_0^3}{8G(m_1+m_2)}}$

Here’s an example: if suddenly stopped in its orbit, how long would it take the Earth to fall into the Sun?

The relevant numbers are

$m_1=6\times 10^{24}\,\text{kg}\quad (\text{mass of Earth})$

$m_2=2\times 10^{30}\,\text{kg} \quad(\text{mass of Sun})$

$r_0=1.5\times 10^{11}\,\text{m}\quad(\text{radius of Earth's orbit})$

$G=6.7\times 10^{-11}\,\text{m}^3\,\text{kg}^{-1}\,\text{s}^{-2}$

and they yield

$t_0=5.6\times 10^6 \,\text{seconds}$

$t_0\approx 2.2 \,\text{months}$

This result does not take into account the finite radius of the Earth and Sun, which are both negligible compared to one astronomical unit. I suppose the better question would be: what would you do with these two and a bit months?

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Here’s another example. Consider two neutrons separated by one metre. How long would they take to collide under gravitation? The answer is

$t_0=75\,\text{million millennia}$

This timescale exceeds the estimated age of the universe.

Now instead consider a proton and an antiproton initially separated by a metre. These particles have almost exactly the same mass as the neutrons, but they carry equal and opposite charges, both equal in magnitude to that of an electron. The coefficient $\alpha$ for the Coulomb force is

$\displaystyle \alpha=-\frac{q_1 q_2}{4\pi\epsilon_0}$

where $q_1$ and $q_2$ are the electric charges of the two particles and $\epsilon_0$ is the permittivity of free space. If $q_1$ and $q_2$ have opposite signs, $\alpha$ is positive (the force is attractive) and so we can use the equation for $t_0$. If we plug in the appropriate values, the time taken for the proton and antiproton to collide is

$t_0=2.1\,\text{seconds}$

This demonstrates the staggering difference in strength between the gravitational and electrostatic forces. Bigger is electricity!

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Similar methods can be used to solve a related problem, that of motion under an inverse-square repulsive force. Interacting charges of the same sign experience this kind of force. Also, a body in space experiences this kind of force due to radiation pressure from nearby stars. The relevant integral in this case is

$\displaystyle \sqrt{\frac{\mu r_0}{2\alpha}}\int_r^{r_0}\frac{dr}{\sqrt{1-\frac{r_0}{r}}}=t$

which is almost identical to the one we saw before, save a few sign changes. You can use exactly the substitution, but instead of integrating cosine squared you’ll end up with

$\displaystyle t\propto\int \sec^3\theta\,d\theta$.

which is a bit meaner.