A satellite of mass 205 kg is launched from a site on Earth's equator into an orbit at 200 km above the surface of Earth.

Question

(a) Assuming a circular orbit, what is the orbital period of this satellite?
s
(b) What is the satellite's speed in its orbit?
m/s
(c) What is the minimum energy necessary to place the satellite in orbit, assuming no air friction?
J

Answer 1

200*10^3 = .2*10^6 m high
R earth = 6.38*10^6
so R = 6.58 * 10^6 meters
G = 6.67384 × 10-11 m3 kg-1 s-2
earth mass = Me = 5.97*10^24 kg

m v^2/R = G Me m /R^2

v^2 = G Me/R
= 6.67 * 5.97 /6.58 * 10^(-11+24-6)
= 6.05 * 10^7 = 60.5 * 10^6
so
v = 7.78*10^3 = 7,780 m/s (PART B)

C = 2 pi R = 2 pi *6.58*10^6
= 41.3 * 10^6 meters circumference
so T = 41.3*10^6/7.78*10^3
= 5.31 * 10^3 seconds (PART A) = 1.47 hours

Work done to increase r from Re to R + Ke

potential energy = U call it 0 at infinity
U = -G Me m/r
U at R = -G Me m/R
U at Re = -G Me m /Re
U at R - U at Re = G Me m /Re-G Me m/R
= G Me m(R-Re)/(R Re)
then Ke = (1/2)m v^2
= (1/2) m G Me/R
sum of U + Ke =
G Me m(R-Re)/(R Re) + (1/2) G Me m Re/(R Re)
= (G Me m )(R-Re+.5Re)/(R Re)
= (G Me m)(R-.5Re)/(R Re)
since R and Re are about the same really
= about (1/2)G Me m /R