Let Harry’s position be , and that of the pursuing bludger be . The bludger’s velocity is at all times parallel to the vector joining the bludger to Harry, and we will assume for simplicity that it flies at constant speed . Its instantaneous velocity may then be expressed as
where represents the magnitude of vector .
Suppose that, to evade the bludger, Harry simply flies in a circle of radius at constant speed in the plane. His position may then be described by
where we have taken without loss of generality.
The bludger’s equation of motion is practically insoluble for arbitrary . That being said, we can make a pretty good guess as to what the bludger’s steady-state motion will be, if such a state exists; the bludger will also move in a circle, trailing Harry by some fixed distance.
Let’s use the ansatz
The phase is included to account for the fact that the bludger’s motion will lag Harry’s slightly.
To solve for and , proceed as follows:
1) Write out the components of the bludger’s equation of motion, and take their ratio to obtain an expression for .
2) Given that the bludger moves in a circle, find an expression for in terms of and .
3) Equate your two expressions, and use our ansätze for and to generate an equation containing only the variable .
4) Rearrange your equation to find an expression for the angle in terms of the radii and , and hence in terms of and . Write down an expression for in terms of the same variables.
5) If , by what distance are Harry and the rogue bludger separated? What happens if ?
For part (2), differentiate the Cartesian equation of a circle; doing so implicitly will give the desired result fastest.
For part (4), you may (or may not) want to make use of the compound angle formulae and . If we have made a reasonable guess, will drop out of the equation entirely.