Asatellite orbiting the moon very near the surface has a period of 110 min. use this information, together with the radius of the moon r=1.74×106m, to calculate the free-fall acceleration on the moon's surface.

1.58 m/s^2 For this problem, going to assume that the satellite is VERY close to the surface of a rather unusually smooth moon. In fact, I'm going to assume the satellite is almost skimming the surface, mere millimeters away from it. The relationship between acceleration and velocity for a body in orbit is a = v^2/r Now we've been given r, and we still need v. Since velocity is defined as distance over time, we need to calculate the distance which will be 2 pi r. So 2 * 3.14159 * 1.74x10^6 m = 10932742 m Therefore v will be 10932742 / (110 * 60) = 10932742/ 6600 = 1656 m/s Substituting into the equation for a, gives us a = (1656 m/s)^2/1.74x10^6 m a = 2742336 m^2/s^2 / 1.74x10^6 m a = 1.576055 m/s^2 Rounding to 3 significant figures gives 1.58 m/s^2 Note: If you wish a fun way of learning about orbital mechanics, you could do far worse than playing Kerbal Space Program. As the creator of XKCD, Randall Munroe, says. https://xkcd.com/1356/

26.4 is about the right time

Yes