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

### Question Description

I’m studying for my Physics class and need an explanation.

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, Student has agreed that all tutoring, explanations, and answers provided by the tutor will be used to help in the learning process and in accordance with Studypool's honor code & terms of service. 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 UIUC Anonymous
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