= Solution
The diagram places the force centre on the circumference, not at the centre of the circle. Choose the polar axis through the circle's geometric centre. Its equation is $r=2a\cos\theta$, so $u=\sec\theta/(2a)$ and
$$
u''+u=\frac{\sec^3\theta}{a}=8a^2u^3.
$$
Substitute into <Binet's equation>:
$$
F(1/u)=-\frac{J^2u^2}{m}(u''+u)
=-\frac{8J^2a^2}{m}u^5.
$$
Consequently a <circular orbit through the force centre> is generated by
$$
\boxed{n=-5,\qquad C=\frac{8J^2a^2}{m},\qquad F(r)=-\frac{8J^2a^2}{mr^5}.}
$$
The result applies to the regular arc where $r>0$ and $J\ne0$. The force is singular at the point where the geometrical circle meets the origin, so the curve is not a regular periodic orbit continued through that point. Indeed, $U=-C/(4r^4)$ and this trajectory has zero total <energy>, with kinetic and negative potential terms diverging as the centre is approached.
\Image[/past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2019-central-force-orbits.png]
{title=Bounded radial motion for an attractive cubic force and the distinct offset-circle geometry produced by an inverse-fifth-power force}
Back to article page
