To determine the period of Mars' rotation, we start by understanding the concept of synchronous orbits. In a synchronous orbit, the orbital period of a satellite matches the rotational period of the planet it orbits. This means that the period of Mars, denoted as \( T_{\text{Mars}} \), is equal to the orbital period of the satellite.

We can derive the relationship using the formula for synchronous orbits, which is given by:

\[T_{\text{sat}}^2 = \frac{4\pi^2 r_{\text{sync}}^3}{G M}\]

Here, \( T_{\text{sat}} \) is the orbital period of the satellite, \( r_{\text{sync}} \) is the radius of the synchronous orbit, \( G \) is the gravitational constant (\( 6.67 \times 10^{-11} \, \text{m}^3/\text{kg} \cdot \text{s}^2 \)), and \( M \) is the mass of Mars (\( 6.42 \times 10^{23} \, \text{kg} \)). To isolate \( T_{\text{Mars}} \), we rearrange the equation:

\[T_{\text{Mars}} = \sqrt{\frac{4\pi^2 r_{\text{sync}}^3}{G M}}\]

Next, we need to find the radius of the synchronous orbit, \( r_{\text{sync}} \). We can use the satellite's velocity, which is given as \( v_{\text{sat}} = 1450 \, \text{m/s} \). The relationship between the velocity of a satellite and the radius of its orbit is expressed as:

\[v_{\text{sat}} = \sqrt{\frac{G M}{r}}\]

Squaring both sides gives us:

\[v_{\text{sat}}^2 = \frac{G M}{r_{\text{sync}}}\]

Rearranging this equation allows us to solve for \( r_{\text{sync}} \):

\[r_{\text{sync}} = \frac{G M}{v_{\text{sat}}^2}\]

Substituting the known values into this equation, we find:

\[r_{\text{sync}} = \frac{(6.67 \times 10^{-11}) (6.42 \times 10^{23})}{(1450)^2} \approx 2.04 \times 10^7 \, \text{m}\]

Now that we have \( r_{\text{sync}} \), we can substitute it back into the equation for \( T_{\text{Mars}} \):

\[T_{\text{Mars}} = \sqrt{\frac{4\pi^2 (2.04 \times 10^7)^3}{(6.67 \times 10^{-11}) (6.42 \times 10^{23})}}\]

Calculating this gives us:

\[T_{\text{Mars}}^2 \approx 7.83 \times 10^9\]

Taking the square root results in:

\[T_{\text{Mars}} \approx 88470 \, \text{seconds} \approx 24.6 \, \text{hours}\]

This calculation reveals that the rotation period of Mars is approximately 24.6 hours, demonstrating how the properties of synchronous orbits and satellite velocity can be utilized to derive significant planetary characteristics.