The masses of the two particles are and , and their respective positions are and .
Suppose the force on particle 1 due to particle 2 is
and, from Newton’s third law, the force on particle 2 due to particle 1 is
where is a positive constant. A force of this form is said to obey an inverse square law, since
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?
From Newton’s second law,
where denotes the second derivative of with respect to time. By equating these expressions for the forces with those above, we can say that
This equation can be tidied up by defining the separation vector as
and the system’s reduced mass through
Force convenience we define
where is the physical distance between the two masses and is a vector of unit magnitude which is parallel to . This simplifies the equation of motion to
assuming the vector does not vary with time (the two particles lie on a common line for all time). So, dispensing with the vectorial aspect,
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:
This changes the independent variable from to . Substituting this expression into the equation of motion makes it directly integrable:
where 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 , then
We rearrange for the derivative of , getting ready to integrate again:
It’s important that we take the negative root; if the force is attractive, the distance between the two particles must be decreasing.
This integral looks scary, but can be solved by the substitution
The differential is
Substituting this in, we find
After tidying up all the trigonometric terms, we’re left with
This is a comparatively non-terrifying integral. We use the half-angle formula to get rid of the cosine squared:
We use the double angle formula to make re-substitution easier:
In terms of the variable this is
This is the time taken for the two particles initially separated by distance to reach a separation .
When do the particles collide? This occurs when the particles’ separation is equal to 0:
Using the definition of the reduced mass , this is
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 is given by
where is the gravitational constant. Then
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
and they yield
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?
Here’s another example. Consider two neutrons separated by one metre. How long would they take to collide under gravitation? The answer is
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 for the Coulomb force is
where and are the electric charges of the two particles and is the permittivity of free space. If and have opposite signs, is positive (the force is attractive) and so we can use the equation for . If we plug in the appropriate values, the time taken for the proton and antiproton to collide is
This demonstrates the staggering difference in strength between the gravitational and electrostatic forces. Bigger is electricity!
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
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
which is a bit meaner.