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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.

2 Answers

5 votes
-- 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)
User MoeTi
by
8.4k points
5 votes

Answer:

1.4 x 10 7 m/h

Step-by-step explanation:

User StefanK
by
8.1k points