(a) What is the speed of a 231 kg satellite in an approximately circular orbit 6

Question

(a) What is the speed of a 231 kg satellite in an approximately circular orbit 646 km above the surface of Earth?

b) What is the period? min Suppose the satellite loses mechanical energy at the average rate of 2.0 105 J per orbital revolution. Adopt the reasonable approximation that the trajectory is a "circle of slowly diminishing radius". c) Determine the satellite's altitude at the end of its 1700th revolution. m

d) Determine its speed at this time. m/s

(e) Determine its period at this time. min

(f) What is the magnitude of the average retarding force on the satellite? N

Explanation / Answer

mass of satellite m = 231 kg height h = 646 km Radious of earth R = 6400 km

g = 9.8 m/s/s

gravitational acceleration at a height h gh = g/(1+h/R)^2 = 8.085 m/s/s

for the satellite to be in circular motion the gravitaionls force is equal to centrifugal force

mv2/(R+h) = mgh ----------------(1)

v2 = 8.085*(6400+646)*10^3 = 56966910

v= 7547.64 m/s = 7.6 km/s

anglar speed w = v/r

period T = 2pi/w = 2pir/v = 2*3.14*7046/7.6 = 5.83x10^3 s

Energy lost per revolution = 2x105 = 210 J

energy lost in 1700 revolutions = 357000J

it is equal to the change in KE

if v1 and v2 are the inital and final KE

0.5*(v1^2-v2^2)*m =E

v2^2 = v1^2-2*E/m = (7.6x10^3)^2 -2*357000/231

V2 = 7599.8 m/s

The change in velocity is negligible and the period remains almost same

h is change in height then

mgh = 357000J

h = 357000/(231*8.085) = 191 m

altitude = 646.0 - 0.0191 = 645.98 km

average retarding force F = 357000/191 = 1869 N

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