= Solution
The annual, zonal-mean absorbed sunlight falls from the tropics towards the poles much more strongly than outgoing <thermal radiation>. Thus <Earth>'s top-of-atmosphere <radiation budget> has a tropical surplus and a polar deficit, with changes of sign around $35^\circ$ latitude in a basic sketch. The curves below are illustrative, rather than measured data; their area-weighted global means agree.
\Image[/past-exam-of-the-geophysical-fluid-dynamics-course-of-the-university-of-oxford/2025-radiation-balance.png]
{title=An illustrative latitude-dependent radiation budget and the required poleward energy transport}
In a statistically stationary <climate>, local <conservation of energy> requires the surplus to be exported. If $R(\phi)$ is the net downward radiative <heat flux> and $\mathcal H(\phi)$ the northward total <power> carried across the latitude circle, then
$$
\frac{d\mathcal H}{d\phi}=2\pi a^2R(\phi)\cos\phi,\qquad
\mathcal H(\phi)=2\pi a^2\int_{-\pi/2}^{\phi}R(\theta)\cos\theta\,d\theta.
$$
Here $a$ is <Earth>'s radius. The <terrestrial atmosphere> and <ocean circulation> carry this <energy> poleward by moving warm and cold fluid, including atmospheric transport of <latent heat>. Without this transport the assumed steady <radiation budget> could not persist: the tropics would warm and polar regions cool until the budget changed. The tropical atmospheric contribution includes the <Hadley cell>: rising warm air near the equator, poleward flow aloft, subtropical descent, and an equatorward lower return.
For a parcel moving without an axial <torque>, <conservation of angular momentum> applies in an inertial frame. The specific axial <angular momentum> at latitude $\phi$ is
$$
M=(\Omega a\cos\phi+u)a\cos\phi.
$$
“Initially at rest” means $u=0$ relative to the rotating surface at the equator, so $M=\Omega a^2$. Neglecting atmospheric height compared with $a$ gives
$$
\boxed{u(\phi)=\Omega a\left(\frac1{\cos\phi}-\cos\phi\right)
=\frac{\Omega a\sin^2\phi}{\cos\phi}.}
$$
With $\Omega=7.292\times10^{-5}\ \mathrm{s^{-1}}$ and $a=6.371\times10^6\ \mathrm m$, the eastward <velocity> at $30^\circ\mathrm N$ is
$$
\boxed{u(30^\circ)=134\ \mathrm{m\,s^{-1}}.}
$$
This is the ideal angular-momentum-conserving upper branch of the <Hadley cell>. Real <jet streams> exchange <angular momentum> through <eddies> and <friction>, so this estimate is not an observational claim about the actual wind speed.
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