Inverse Square Attraction

Imagine two point-like particles experiencing mutual attraction.

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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How do we solve this differential equation? If you’re not interested in the details, click here to skip to the result.

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}}

cataclysm

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.

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