A satellite of mass 210 kg is launched from a site on Earth's equator into an orbit at 210 km above the surface of Earth.

Question

(a) Assuming a circular orbit, what is the orbital period of this satellite?

(b) What is the satellite's speed in its orbit?

(c) What is the minimum energy necessary to place the satellite in orbit, assuming no air friction?
J

Answer 1

For (a) and (b), the satellite mass will not matter. The orbit distance from the center of the earth is R = 6378 + 210 = 6588 km. Speed is
V = 2 pi R/P
where P is the period.
Get the period from

G M/R^2 = V^2/R

G M = V^2*R = 4 pi^2 R^3/P^2

Solve for P

M is the earth's mass

Energy required = (1/2) m V^2 + GMm[1/Re - (1/(Re + 211,000m)]

m is the satellite mass

There is a small additional term (+ or - that takes into account the rotational speed of the earth at the launch point. That gives the rocket launcher a bit of a boost, especially at the equator.

Answer 2

I need a better explanation than that, and an answer to show you are right. thanks