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Homework Help: Check my work on Geosynchronous orbits?

  1. May 19, 2009 #1
    1. The problem statement, all variables and given/known data
    NASA has a mission to put a satellite into an equatorial geosynchronous orbit, so that its orbital motion will keep it above longitude 150 as the Earth rotates beneath it. This is done in three steps:
    1) The satellite is placed into a circular equatorial orbit 220 miles above the surface

    2) The satellite's propulsion system is used to increase its velocity in order to place it into an elliptical transfer orbit with perigee 220mi and apogee at the distance of the circular geosynchronous orbit.

    3) The velocity of the satellite at apogee is changed in order to insert it into the circular geosynchronous orbit

    Assuming that the exhaust velocity of the satellite rocket is 2500m/sec and the final mass in geosynch. orbit is 676kg, calculate the mass of fuel expended by the satellite's rocket in order to raise it from its 220 mile circular orbit

    2. Relevant equations
    V= sqrt GMEarth/r
    K= 1/2 MSatellite V2
    U= -GMEMSatellite
    K + U = E
    solving for v:
    V=sqrt GMEarth( 1/a + 2/r)


    3. The attempt at a solution
    220mi*1.6km/mi= 354 km
    Vlow orbit= 7907 m/s

    Radius of geosynchronous orbit is 4.23*107m
    Appogee=Radius of geosync
    Perigee=Radius of Earth+354000 = 6.73*106m
    2a=Apogee + Perigee = 2.45*107m

    Vp= sqrt GME ( 1/a + 2/r ) = sqrt GME (1/2.45*107 + 2/6.73*106 ) = 11626 m/s

    Va= sqrt GME ( 1/2.45*107 + 2/4.23*107 )
    = 5936 m/s

    VGeosync = sqrt GME/r = 3079 m/s

    I think up to that point is correct. I found some help online about the next part and I get a seemingly close answer. Can someone explain how/why this works, or a method that does work if this one is incorrect?

    [tex]\Delta[/tex]V1 = Vp - Vlow orbit = 3719 m/s
    [tex]\Delta[/tex]V2 = Vgeo sync - Va = -2857
    [tex]\Delta[/tex]Vtotal = [tex]\Delta[/tex]V1 + [tex]\Delta[/tex]V2 = 862 m/s

    862 = 2500*ln(M0/676)
    .3488 = ln(M0) - ln(676)
    M0 = 954.3 kg

    so 954.3 - 676 = 278.3kg fuel burned
    Last edited: May 19, 2009
  2. jcsd
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