The astronaut orbiting the Earth in Figure P4.32 is preparing to dock with a Westar VI satellite. The satellite is in a circular orbit 470. km above the Earth's surface, where the free-fall acceleration (centripetal acceleration) is 8.19 m/s2. Take the radius of the Earth as 6370 km. Determine the speed of the satellite in km/h. THis is a circular motion problem where a = v^2/r can be used right? 8.19 m/s^2 converted to km/h^2 = 1061.42 km/h^2 sqrt(1061.424*(470+6370) = 2694.46 my answer was 2694.46; 2690 but the system is telling me that I'm off by a multiple of ten. What am I doing wrong?