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+### Statement
+
+$2.7.39.$ [Insert the problem statement]
+
+### Solution
+
+### Statement
+
+$2.7.39^*.$
+a. It is known that the tidal deformation of the Earth itself and oceanic tides slow down the Earth's rotation. Explain how the torque required for this is generated.
+b. The solar atmospheric tide on Earth reaches its maximum two hours before the Sun passes through the zenith. Does this tide contribute to or hinder the slowing down of the Earth's daily rotation?
+
+### Solution
+
+<b>Part a)</b>
+
+The Earth rotates faster than the Moon orbits it (one rotation per $\sim 24$ hours for Earth, versus one orbit per $\sim 27.3$ days for the Moon). Due to the viscosity of the oceans and friction within the Earth's crust, the tidal bulges do not perfectly align with the Moon; instead, they are dragged forward in the direction of the Earth's rotation. As a result, the Moon's gravitational pull acts asymmetrically on these bulges, creating a retarding torque.
+
+The Moon exerts a stronger gravitational force on the closer tidal bulge (which is shifted forward) than on the farther one. This creates a net torque directed against the Earth's rotation. Consequently, the Earth's rotation slows down, and the Moon's orbit expands (due to the conservation of angular momentum).
+
+<b>Part b)</b>
+
+If the solar atmospheric tide peaks before the Sun reaches the zenith, it means the tidal bulge in the atmosphere is shifted forward in the direction of the Earth's rotation. The Sun exerts a stronger gravitational pull on this leading part of the tide, creating a torque that accelerates the Earth's rotation (since the tide is effectively being "pulled" forward).
+
+This is the opposite of the lunar tidal braking. Thus, this atmospheric tide hinders the slowing down of the Earth's daily rotation (though only slightly, as the solar tide is significantly weaker than the lunar one).
+
+#### Answer
+
+[Insert a concise answer or boxed result]