[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.