Calculate the velocity of a 1650kg satellite that is in a circular orbit of 4.2 x 10^6m above the surface of a planet which has a radius 3.95 x 10^5 m and the period of the orbit is 2.0 hours.

-- We're going to be talking about the satellite's speed. 
"Velocity" would include its direction at any instant, and
in a circular orbit, that's constantly changing.

-- The mass of the satellite makes no difference.

Since the planet's radius is  3.95 x 10⁵m  and the satellite is
orbiting  4.2 x 10⁶m  above the surface, the radius of the
orbital path itself is

                               (3.95 x 10⁵m) + (4.2 x 10⁶m)

                     =        (3.95 x 10⁵m) + (42 x 10⁵m)

                     =           45.95 x 10⁵ m

The circumference of the orbit is  (2 π R) =  91.9 π x 10⁵ m.

The bird completes a revolution every 2.0 hours,
so its speed in orbit is

                                     (91.9 π x 10⁵ m) / 2 hr

                        =        45.95 π x 10⁵  m/hr  x  (1 hr / 3,600 sec)

                        =           0.04 x 10⁵      m/sec

                        =              4 x 10³      m/sec  

                                     (4 kilometers per second)

Bsame force but in the same direction