Answer:
[tex]1.69\cdot 10^{10}J[/tex]
Explanation:
The total energy of the satellite when it is still in orbit is given by the formula
[tex]E=-G\frac{mM}{2r}[/tex]
where
G is the gravitational constant
m = 525 kg is the mass of the satellite
[tex]M=5.98\cdot 10^{24}kg[/tex] is the Earth's mass
r is the distance of the satellite from the Earth's center, so it is the sum of the Earth's radius and the altitude of the satellite:
[tex]r=R+h=6370 km +575 km=6945 km=6.95\cdot 10^6 m[/tex]
So the initial total energy is
[tex]E_i=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24} kg)}{2(6.95\cdot 10^6 m)}=-1.51\cdot 10^{10}J[/tex]
When the satellite hits the ground, it is now on Earth's surface, so
[tex]r=R=6370 km=6.37\cdot 10^6 m[/tex]
so its gravitational potential energy is
[tex]U = -G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24}kg)}{6.37\cdot 10^6 m}=-3.29\cdot 10^{10} J[/tex]
And since it hits the ground with speed
[tex]v=1.90 km/s = 1900 m/s[/tex]
it also has kinetic energy:
[tex]K=\frac{1}{2}mv^2=\frac{1}{2}(525 kg)(1900 m/s)^2=9.48\cdot 10^8 J[/tex]
So the total energy when the satellite hits the ground is
[tex]E_f = U+K=-3.29\cdot 10^{10}J+9.48\cdot 10^8 J=-3.20\cdot 10^{10} J[/tex]
So the energy transformed into internal energy due to air friction is the difference between the total initial energy and the total final energy of the satellite:
[tex]\Delta E=E_i-E_f=-1.51\cdot 10^{10} J-(-3.20\cdot 10^{10} J)=1.69\cdot 10^{10}J[/tex]
