If a satellite is in a sufficiently low orbit, it will encounter air drag from t

Question

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 r 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) A satellite with mass 2540kg is initially in a circular orbit a distance 290km above the earth's surface. Due to air drag, the satellite's altitude decreases to 220km . Calculate the initial orbital speed.

v= 7730 m/s

(c) Calculate the initial mechanical energy. E= ?7.59*1010 J

(g) Calculate the work done by the force of air drag. W=?

Explanation / Answer

a) vi = sqrt(GM/(R+h)

vi = sqrt(6.67e-11*5.97e24)/(6380000+290000)

vi = 7730 m/s

b)

vf = sqrt(GM/(R+h)

vf = sqrt(6.67e-11*5.97e24)/(6380000+220000)

vf = 7770 m/s

dv = Vf - vi = 7770-7730 = 40 m/s

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c) Ei = 0.5*m*vi^2 - GMm/r1 = 0.5*GMm/r1-GMm/r1 = -GMm/2r1 = -(6.67e-11*5.97e24*2540)/(6380000+290000) = -1.52*10^11 J

d) dK = KEf - KEi = 0.5*m*vf^2-0.5*m*Vi^2

dK = 0.5*m*(vf^2-vi^2)

dK = 0.5*2540*((7770*7770)-(7730*7730)) = 7874*10^5 J

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e) dU = -GMm*((1/r2) - (1/r1))

dU = -(6.67e-11*5.97e24*2540)*((1/(6380000+220000))-(1/(6380000+290000)))

dU = 1.608*10^9

f) dE = dU + dk = 7.8756*10^8 J

g) W = -dE = -7.8756*10^8 J

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