5. Many people assume air resistance acting on a moving object will always make

Question

5. Many people assume air resistance acting on a moving object will always make the object slow down. It can, however, actually be responsible for making the object speed up. Consider a 100-kg Earth satellite in a circular orbit at an altitude of 200 km. A small force of air resistance makes the satellite drop into a circular orbit with an altitude of 100 km (a) Calculate its initial speed. (b) Calculate its final speed in this process. (c) Calculate the initial energy of the satelliteEarth system. (d) Calculate the final energy of the system. (e) Show that the system has lost energy and find the amount of the loss due to friction. (f) What force makes the satellites speed increase? You will find a free-body diagram to be useful in explainin userul in explaining your answer.

Explanation / Answer

(A) for satellite moving in circular path:

Gravitational force = centripetal force

G M m / r^2 = m v^2 / r

v = sqrt[ G M / r ]

v = sqrt[ (6.67 x 10^-11 x 5.972 x 10^24 ) / (6371000 + 200000)]

v = 7785.87 m/s

(B) when r = 6371000 + 100000 m

v = sqrt[ G M / r ]

v = sqrt[ (6.67 x 10^-11 x 5.972 x 10^24 ) / (6371000 + 100000)]

v = 7845.8 m/s

(C) total energy = KE + PE

= m v^2/2 - G M m/r

and GM/r = v^2

TE = m v^2 /2 - m v^2 = - m v^2 / 2
initial energy = - 100 x 7785.87^2 / 2 = - 3.031 x 10^9 J

(D)final TE = - 100 x 7845.8^2 / 2 = - 3.078 x 10^9 J

(E) final energy is less then initial energy.

energy lost = TEi - TEf = - 3.031 x 10^9 - (- 3.078 x 10^9 )

= 4.68 x 10^7 J

(f) as distance decreases, gravitational force increases.

so centripetal force increases( by factor of square)

and centrietal force = m v^2 / r

hence v have to increase.

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