 # How much energy was transformed into internal energy by means of air friction? Anonymous
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### Question Description

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,

Bilal_Mursaleen
School: UIUC   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

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