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

 Physics Tutor: None Selected Time limit: 1 Day

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

...
Jul 21st, 2014
...
Jul 21st, 2014
Dec 9th, 2016
check_circle
Mark as Final Answer
check_circle
Unmark as Final Answer
check_circle
Final Answer

Secure Information

Content will be erased after question is completed.

check_circle
Final Answer