A 450 kg satellite is in a circular orbit at an altitude of 525 km above the Earth's surface. Because of air friction, the satellite eventually falls to the Earth's surface, where it hits the ground with a speed of 2.00 km/s. How much energy was transformed to internal energy by means of friction?
For "orbit", centripetal acceleration = gravitational acceleration, or
v^2/r = GM/r^2
v^2 = GM/r
and KE = 1/2*m*v^2
= GMm/2R
and GPE = -GMm/R
so Total Energy = -GMm/2r
so initially total energy
= -6.67*10^-11 * 450*5.98*10^24 / 2*(450+6400)*10^3
TE = -1.310*10^10 J
leter GPE will be
= 6.67*10^-11 * 450*5.98*10^24 / (6400)*10^3
= 2.804*10^10
KE = 1/2*m*v^2 = 1/2 * 450*(2000)^2 = 9*10^8 J
so by conservation of energy
Initail energy = Final energy
-1.310*10^10 = -2.804*10^10 + 9*10^8 + lost energy
lost energy = 1.403*10^10 J
Me = 5.98*10^24 kg
Re = 6370 km = 6.37*10^6 m
h = 525 km = 0.525*10^6 m
intial potential energy U1 = -G*Me*m/(Re+h)
intial kinetic energy k1 = 0.5*m*V^2
= 0.5*m*(G*Me/(Re+h))
initial total mechanical energy, TE1 = U1 +
K1
= -0.5*G*Me*m/(Re+h)
=
-0.5*6.67*10^-11*5.98*10^24*450/((6.37+0.525)*10^6)
= -1.3*10^10 J
fial potential energy U2 = -G*Me*m/Re
= -6.67*10^-11*5.98*10^24*450/6.37*10^6
= -2.818*10^10 J
fial kinetic energy k1 = 0.5*m*V^2
= 0.5*450*(2000)^2
= 0.09*10^10 J
final total mechanical energy, TE2 = U2 + K2
= -2.728*10^10 J
workdone by friction = TE2 - TE1
= -2.728*10^10 - (-1.3*10^10)
= -1.428*10^10 J
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