Answer:
![1.69\cdot 10^(10)J](https://img.qammunity.org/2020/formulas/physics/high-school/h50i901f1rp12unu8729jzh3a6vzozvjas.png)
Step-by-step explanation:
The total energy of the satellite when it is still in orbit is given by the formula
![E=-G(mM)/(2r)](https://img.qammunity.org/2020/formulas/physics/high-school/27f2noqci8qgn9199psggweg7j2303f1gy.png)
where
G is the gravitational constant
m = 525 kg is the mass of the satellite
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:
![r=R+h=6370 km +575 km=6945 km=6.95\cdot 10^6 m](https://img.qammunity.org/2020/formulas/physics/high-school/d8utnm8t2gakf8hoqn7un9pxjf0hp04cva.png)
So the initial total energy is
![E_i=-(6.67\cdot 10^(-11))((525 kg)(5.98\cdot 10^(24) kg))/(2(6.95\cdot 10^6 m))=-1.51\cdot 10^(10)J](https://img.qammunity.org/2020/formulas/physics/high-school/bnh8iq2r5dzmmlhj0v3jnkgd7qchnumu7x.png)
When the satellite hits the ground, it is now on Earth's surface, so
![r=R=6370 km=6.37\cdot 10^6 m](https://img.qammunity.org/2020/formulas/physics/high-school/wt80lomkfgy0rdwgyuik12yk70vtlp0agf.png)
so its gravitational potential energy is
![U = -G(mM)/(r)=-(6.67\cdot 10^(-11))((525 kg)(5.98\cdot 10^(24)kg))/(6.37\cdot 10^6 m)=-3.29\cdot 10^(10) J](https://img.qammunity.org/2020/formulas/physics/high-school/o12ogbygpj4ilp315dq6nfg20pbb6qpewp.png)
And since it hits the ground with speed
![v=1.90 km/s = 1900 m/s](https://img.qammunity.org/2020/formulas/physics/high-school/h1l7hc5n9kska3hmupduochez4qvv7vrje.png)
it also has kinetic energy:
![K=(1)/(2)mv^2=(1)/(2)(525 kg)(1900 m/s)^2=9.48\cdot 10^8 J](https://img.qammunity.org/2020/formulas/physics/high-school/7ukbn8559diebfojekj0pa7tfrfkdu9ug5.png)
So the total energy when the satellite hits the ground is
![E_f = U+K=-3.29\cdot 10^(10)J+9.48\cdot 10^8 J=-3.20\cdot 10^(10) J](https://img.qammunity.org/2020/formulas/physics/high-school/wwk61va1q1pnxxgbuggqo91efgfw72fyd0.png)
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:
![\Delta E=E_i-E_f=-1.51\cdot 10^(10) J-(-3.20\cdot 10^(10) J)=1.69\cdot 10^(10)J](https://img.qammunity.org/2020/formulas/physics/high-school/coitkqblpi5otzajadw73nxry9gczo4z3a.png)