How much energy was transformed into internal energy by means of air friction?

Physics
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A 600 kg satellite is in a circular orbit at an altitude of 450 km above the Earth's surface. Because of air friction, the satellite eventually falls to the Earth's surface, 

Jul 21st, 2014

For orbit

Centripetal force = Gravitational acceleration

Fc = Fg                     

mv^2/r = GmM/r^2

v^2/r = GM/r^2     or

v^2 = GM/r

Now

K.E = 1/2mv^2

K.E = GmM/2r

G.P.E = -GmM/r

So Total Energy Will be

T.E = K.E + G.P.E

       = GmM/2r - GmM/r

       = -GmM/2r

Where     G = 6.674e-11 Nm^2/kg^2         m = 600 kg        M = 5.98e24 kg          

r = r + R  where           r = 450e3 m     and          R = 6.371e3 m

Now Putting the values, we get

T.E = -GmM/2r

       = -(6.674e-11)(600)(5.98e24) / (2(450 + 6.371)e6m )

T.E = -23946.312 / 912.742

T.E = -0.00262e11 J

And G.P.E = 375.8e11 J 

        K.E = 2.66e11 J

So T.E = K.E + G.P.E + Lost Energy

-0.00262e11 = 2.66e11 - 375.8e11 + Lost Energy

-0.00262e11 = -373.14e11 + Lost Energy

Hence 

           Energy Transformed or Lost Energy = 373.13738e11 J

                                or                                  = 37.3e12 J

                               or                                   = 37.3e3 GJ

Jul 21st, 2014

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Jul 21st, 2014
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Jul 21st, 2014
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