= Solution
Assume an inviscid homogeneous ocean, no horizontal <pressure gradient>, and no <wind stress>. Horizontal uniformity then removes advective acceleration, and the parcel equations on a <beta plane> are
$$
\dot u-f(y)v=0,
\qquad
\dot v+f(y)u=0,
\qquad
f(y)=f_0+\beta y,
$$
with $u=\dot x$ and $v=\dot y$.
Since $\dot u=f(y)\dot y$, integration from the stated initial data gives
$$
\boxed{
\dot x=u=f_0y+\frac12\beta y^2}.
$$
This is conservation of the parcel's zonal canonical momentum: its zonal speed records the meridionally accumulated <Coriolis acceleration>. Multiplying the two momentum equations by $u$ and $v$ and adding gives
$$
\frac d{dt}\frac{u^2+v^2}{2}=0.
$$
Thus <kinetic energy> and speed are constant:
$$
\boxed{\dot x^2+\dot y^2=V^2}.
$$
Eliminating $\dot x$ gives the required <ordinary differential equation>
$$
\boxed{
\dot y^2=V^2-
\left(f_0y+\frac12\beta y^2\right)^2}.
$$
For $f_0,\beta>0$, the equator is at $y_e=-f_0/\beta$. There
$$
f_0y_e+\frac12\beta y_e^2
=-\frac{f_0^2}{2\beta}.
$$
The parcel must encounter a meridional turning point before reaching the equator if it is to remain strictly in the Northern Hemisphere. The energy relation therefore requires
$$
\boxed{
V<\frac{f_0^2}{2\beta}},
$$
or $\beta V/f_0^2<1/2$. Equality is the limiting trajectory that reaches the equator with zero meridional speed.
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