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Gravitational Satellite Angst

  1. Feb 13, 2008 #1
    A 220 kg satellite is in an approximatel circular orbit 640 km above the Earths surface.
    a) Speed of satellite and the period of its orbit?
    Vt^2 = GM/R
    = (6.67 x 10^-11)(6 x 10^24) / (6.37 x 10^6 + 640000) (That's radius of Earth plus height of satellite)
    = 7555.8 m/s
    T^2 = (4pi^2 / GM) x R^3
    = 4pi^2/ (6.67 x 10^-11)(6 x 10^24) x (7010000)^3 (again, radius of earth plus heigh of satellite)
    = 5829.3

    b) Total energy of satellite in its orbit?
    E = -1/2 (GMm/R)
    = -1/2 (6.67 x 10^-11)(6 x 10^24)(220) / 7010000)
    = 6279885877 J

    c) What is the angular momentum of the satellite about the centre of the Earth?
    L = Iw
    L = I (vt/R)
    = I (7555.8/7010000)


    I'm stuck now. Im pretty sure my first few answers are wrong anyway. I don't have much grasp of the concepts of this fancy gravitational stuff, I don't think. The question goes on....

    d) If the satellite loses 1.5 x 10^5 J per orbital revolution due to air resistance, determine the satellite's altitude and speed after its 1500th revolution.

    e) What is the angular momentum of the satellite about the centre of the Earth after the 1500th revolution? Has angular momentum about the centre of the Earth been conserved? If not, explain what has caused the change.

    Needless to say, I have no idea how to tackle these.
     
  2. jcsd
  3. Feb 13, 2008 #2

    mgb_phys

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    Science Advisor
    Homework Helper

    a, First start with Keplers law (T/2pi)^2 = r^3 / GM
    Well done for spotting that r is the height + radius of Earth!
    The speed is simple form the circumference of the orbit and period.

    b, Remember total energy is kinetic + potential.

    d, Use conservation of energy.
     
  4. Feb 13, 2008 #3
    A is ok. the second answer needs a unit.
    for B you need the kinetic energy as well, and the equation you use for the potential energy has E = 0 as R goes to infinity, E <0 everywhere else. you probably want to have E=0 at the surface of the earth.
    C angular momentum is m(v x R) (cross product of vectors) for circular motion it's simply
    mvr
     
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