 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-2/8b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

r

is measured in the rotating frame, its inertial velocity is

\dot{r} + ω \times r

. The masses producing the gravitational field are stationary in this frame, so their potential energy is a time-independent function

V (r)

. Thus kinetic minus potential energy gives the stated Lagrangian

L = \frac{m}{2} (\dot{r} + ω \times r)^{2} - V (r) .

Put

v = \dot{r} + ω \times r

. Then

\frac{\partial L}{\partial r ˙} = m v, \frac{\partial L}{\partial r} = - m ω \times v - \nabla V .

Substitution in the Euler-Lagrange equation yields

m (\ddot{r} + 2 ω \times \dot{r} + \dot{ω} \times r + ω \times (ω \times r)) = - \nabla V .

p = m (\dot{r} + ω \times r),

H (r, p, t) = \frac{p ^{2}}{2 m} - ω (t) \cdot (r \times p) + V (r) .

\dot{r} = \frac{p}{m} - ω \times r, \dot{p} = - ω \times p - \nabla V .

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