physics

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.
(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

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asked by Dallas
  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

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    posted by Damon

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