Given a cylindrically shaped space vehicle (D = Im, height is 0.7 m, average den

Question

Given a cylindrically shaped space vehicle (D = Im, height is 0.7 m, average density is I.I g/cm3) with a flat solar cell panel on an arm (mass of 32 kg, effective moment arm is 1.5 m, effective average area facing normally toward sun is 0.6 m2) in a set of essentially frictionless bearings and in a low orbit at 160 km altitude with sunlight being received, on the average, about 60% of the period:
(a) Compute the maximum solar pressure-caused torque and the angular displacement this would cause during I day if not corrected.
(b) Using the data from the atmospheric table in Appendix 2 and an arbitrary average drag coefficient of 1.0 for both the body and the flat plate, compute the drag force and torque.

Explanation / Answer

Accosding to the given probelm,

a)p = 4 . 5 × 10 6 cos [ ( 1 k s ) cos + 0 . 67 k d ]

p = 4.5*10^-6 * cos(0)*((1-0.25)*cos(0)+0.67(0.01))

p = 4.5*10^-6*1*(0.7567)

p = 3.405*10^-6 N/m^2

T = p*A*l T = 3.405*10^-6 * 0.6 * 1.5

T = 3*10^-6 N

Density = Mass*Volume

Volume oF a Cylinder = pi*h*r^2 V = 3.14*0.7*0.5^2

V = 0.55 m^3

P = m/v 1.1g/cm3 = 550000m M = 605000 g or 605 kg Ma = ½ * m*r^2 Ma = ½ * 605 * 0.5^2 Ma = 75.6 kg/m ² = Ma*a 3*10^-6 = 75.6a

Angle = 3.97*10^-8 ²heta = ½ *a*t^2

Angular speed = a*t W = 3.97*10^-8 * 3600

Angular speed = 1.43*10^-4 degree/s = 2.5*10^-5 rad/s = 9*10^-3 rad/hr

b)Drag = Cd * ((p*V^2)/2)*A

A = 0.6 + (0.7*1)

A = 1.3

Radius of Earth = 6.37*10^6 m

Mass of Earth = 5.98*10^24 kg

R = Rearth+Height = 6.53*10^6

V = U = Square Root (G*M/r)

V = Square Root ((6.67*10^-11*5.98*10^24)/6.37*10^6)

V = Square Root (626163.3)

V = 791.3 m/s

D = 1 * ((3*10^-9*791.3^2)/2)*1.3

D = 9.4*10^-4 x 1.3

D = 0.00122

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