1. **State the problem:** We need to find the linear velocity $\upsilon$ at the surface of Mercury, given its radius $r = 2.44 \times 10^6$ m and angular velocity $\omega$ in rad/s. 2. **Formula used:** The linear velocity $\upsilon$ at the surface of a rotating sphere is related to its angular velocity $\omega$ and radius $r$ by the formula: $$\upsilon = r \times \omega$$ 3. **Explanation:** This formula comes from the relationship between angular and linear motion. The linear velocity is the distance traveled per unit time along the circular path at the surface, which equals the radius times the angular velocity. 4. **Intermediate work:** Substitute the given radius into the formula: $$\upsilon = (2.44 \times 10^6) \times \omega$$ Since $\omega$ is given in rad/s, multiply directly. 5. **Final answer:** The linear velocity at Mercury's surface is $$\boxed{\upsilon = 2.44 \times 10^6 \times \omega \text{ m/s}}$$ This means once you know $\omega$, multiply it by $2.44 \times 10^6$ to get the linear velocity in meters per second.