Question 949291
The circumference of the orbit is {{{(pi)diameter}}} so
{{{Circumference=(pi)(8.46*10^7) m}}}
This is the distance traveled in one period (23.93 hrs)
rate=distance/time={{{(pi)(8.46*10^7)m/23.93hrs}}}={{{(1.111*10^7) m/hr}}}
{{{(1.111*10^7) m/hr}}}*{{{(1 hr/3600 sec)}}}=3085.14 m/sec ANSWER The orbital speed of the satellite is 3085.14 m/sec.