What horizontal speed must be given to a body at the…
What horizontal speed must be given to a body at the equator, in the direction of the Earth's rotation, so that it moves parallel to the Earth's surface (i.e. keeps its height)? At the equator $g = 9.78\ \mathrm{m/s^2}$ and the Earth's radius is $R = 6378.4\ \mathrm{km}$.
Explanation of Method
Speed needed to circle the Earth at its surface.
Step 1: Condition
To stay at the same height, the body must move on a circle of radius $R$ around the Earth's centre. Gravity must supply exactly the centripetal acceleration: $\frac{v^2}{R} = g \;\Longrightarrow\; v = \sqrt{gR}$.
Step 2: Numbers
\[ v = \sqrt{9.78 \cdot 6.3784 \cdot 10^6}\ \mathrm{m/s} = \sqrt{6.238 \cdot 10^7}\ \mathrm{m/s}. \]
\[v \approx 7.90\ \mathrm{km/s} \approx 28\,400\ \mathrm{km/h}\]
This is the first cosmic speed (orbital speed at the surface, air resistance neglected).
Result
\[\boxed{\,v \approx 7.90\ \mathrm{km/s} \approx 28\,400\ \mathrm{km/h}\,}\]
Remark
Notes. The original value 7904 m/s comes from a small multiplication slip (the product is $6.238 \cdot 10^7$, not $6.241 \cdot 10^7$). Why "in the direction of the Earth's rotation"? The ground at the equator already moves east at about 465 m/s, so launching eastward needs less speed relative to the ground: a more careful calculation that separates gravity from the rotation effect contained in $g = 9.78$ gives about 7.45 km/s relative to the ground. The simple answer $\sqrt{gR} \approx 7.9\ \mathrm{km/s}$ is what the problem expects.