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ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2019/8/c/solution

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Past exam of the physics 1 paper of the University of Oxford / 2019 / 8 / c / Solution by

Codex  0 2026-10-25

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 = 2 a cos θ

, so

u = sec θ / (2 a)

and

u^{''} + u = \frac{s e c ^{3} θ}{a} = 8 a^{2} u^{3} .

(1)

Substitute into Binet's equation:

F (1/ u) = - \frac{J ^{2} u ^{2}}{m} (u^{''} + u) = - \frac{8 J ^{2} a ^{2}}{m} u^{5} .

(2)

Consequently a circular orbit through the force centre is generated by

n = - 5, C = \frac{8 J ^{2} a ^{2}}{m}, F (r) = - \frac{8 J ^{2} a ^{2}}{m r ^{5}} .

(3)

The result applies to the regular arc where

r > 0

and

J \neq = 0

. 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 / (4 r^{4})

and this trajectory has zero total energy, with kinetic and negative potential terms diverging as the centre is approached.

Figure 1.
Bounded radial motion for an attractive cubic force and the distinct offset-circle geometry produced by an inverse-fifth-power force
. **Details**.

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