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Satellite in low-eart orbit

  1. Apr 2, 2008 #1
    [SOLVED] satellite in low-eart orbit

    1. The problem statement, all variables and given/known data

    If a satellite is in a sufficiently low orbit, it will encounter air drag from the earth's atmosphere. Since air drag does negative work (the force of air drag is directed opposite the motion), the mechanical energy will decrease. If E decreases (becomes more negative), the radius of the orbit will decrease. If air drag is relatively small, the satellite can be considered to be in a circular orbit of continually decreasing radius.

    A satellite with mass 2760 is initially in a circular orbit a distance 330 above the earth's surface. Due to air drag, the satellite's altitude decreases to 225 . Calculate the initial orbital speed.

    a) find the initial orbital velocity - got this to be 7710m/s
    b) find the change in velocity - got the 2nd velcoity to be 7765m/s so common sense says deltav = v2-v1 = 7765-7710 = 55m/s but it's wrong and I have no idea why, I'll show you what I did
    d) calculute the change in gravitational potential energy - again how I don't see how I'm getting this wrong


    2. Relevant equations

    U = -GMm/r

    K = 1/2mv^2

    E = K+U

    Fg = GMm/r^2

    v = (GM/r)^1/2

    M = 5.97x10^24kg - mass of earth

    R = 6.38x10^6 - radius of earth

    r = (height above planet + R)

    m = 2760kg - mass of satellite

    3. The attempt at a solution

    r1 = 330km(1000m/km) = 3.3x10^5m+R = 6.71x10^6

    v1 = (GM/6.71x10^6)^1/2 = 7710 - correct

    r2 = 225km(1000m/km) = 2.25x10^5+R = 6.6x10^6

    v2 = (GM/6.6x10^6)^1/2 = 7765

    change in v = v2-v1 = 7765-7710 = 55m/s - wrong (don't know how)

    U1 = -GMm/r1 = -1.64x10^11

    U2 = -GMm/r2 = -1.67x10^11

    change in U = U2-U1 = -1.67x10^11+1.64x10^11 = -3x10^9J - wrong, I have no idea what's wrong, I know I'm using the right formulas, if anybody can point me in the right direction I'd really appreciate it.
     
    Last edited: Apr 2, 2008
  2. jcsd
  3. Apr 2, 2008 #2

    D H

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    Staff Emeritus
    Science Advisor

    You did fine. The result at which you arrive is just a bit contrary to common sense. The result might make a bit more sense if you look at the total specific energy (specific means "divided by mass"): [itex]1/2v^2-GM/r[/itex]. Since [itex]v^2=GM/r[/itex] for a circular orbit, the total specific energy is just [itex]-1/2\,GM/r[/itex] for a circular orbit. Thus the energy becomes more negative as the radius decreases. In other words, energy is lost as the orbital altitude decreases.
     
  4. Apr 2, 2008 #3
    thanks, what you wrote will work for total mechanical energy (which is a part of the question, I know I can get that right if I can just do these parts I'm having trouble with).

    if my 2nd velocity and potentional energies are correct, why do my delta's turn out wrong?
     
  5. Apr 2, 2008 #4
    tried

    GM/2(-1/r2+1/r1) for the total mech energy, didn't get it.
     
  6. Apr 2, 2008 #5
    edit: part b) is actually supposed to be the increase in velocity, but I don't see how that would change it much, my delta kinetic energy entered was correct so I'm sure I have the correct 2nd velocity.
     
  7. Apr 2, 2008 #6

    D H

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    Staff Emeritus
    Science Advisor

    Since you say your results are "wrong", what are the "right" answers? One source of error is that you were a bit too loose with your rounding. In some places you only expressed your results to two decimal places. You inherently lose accuracy when you subtract two numbers that are quite close to one another.
     
  8. Apr 2, 2008 #7
    I'm not sure what the right answers are, I'm using masteringphysics and they're probably just being really precise as you say, they only give the correct answer when all the tries are done.

    I don't see how 55 could be wrong for the velocities though.

    edit: yeah I think you're right on the rounding I tried a ballpark 60 on the deltav and got it, I'll see if I retry my calculations again with more sig figs.
     
    Last edited: Apr 2, 2008
  9. Apr 2, 2008 #8
    okay, so far so good, just got my change in potential energy and my change in total mechanical energy correct, now the only thing left is work done by air drag, edit: which by conservation of energy turned out to be the same as mechanical energy:

    -1.3x10^9J, thanks for the help DH!
     
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