Imagine a straight tunnel through the centre of a uniform Mars (g = 3.7 m/s² at the surface, R =…

Step 1. The force. At distance $r$ from the centre: $F = -\dfrac{mg}{R}r$, in proportion to the displacement and toward the centre: SHM.

Step 2. Period. $\omega = \sqrt{\dfrac{3.7\ \text{m/s}^2}{3.39 \times 10^{6}\ \text{m}}} = 1.045 \times 10^{-3}\ \text{rad/s}$, so $T = 6014\ \text{s} = 100.2\ \text{min}$; one way is half of that, $50.1\ \text{min}$.

Step 3. (b) Largest speed $= \omega R = \sqrt{gR} = 3.54\ \text{km/s}$.