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)