Satellites and Binding Energy EX 11

It appears that the satellite is initially at rest and it is then launched with some speed while it is still very close to the surface of the Earth. Let's assume that, once it is launched, it is the gravitational interaction between the Earth and the satellite alone which plays the role in its motion. Therefore the energy will be conserved. We can determine the energy the satellite has from its orbit. The radius of the orbit is the sum of the radius of the Earth and the altitude of the satellite $r=6370km+50km=6420km=6.42\times 10^{6}m$. Note that in this case we used a more precise value of the radius of Earth because the altitude is so small. In this case, the energy in the orbit $E_{0}$ is

$E_{0}=-{\frac {1}{2}}{Gm_{E}m_{s} \over r}=-{\frac {1}{2}}{(6.67\times 10^{-11})(6\times 10^{24})(10) \over 6.42\times 10^{6}}=-3.12\times 10^{8}J$

The energy after the launch is the sum of the potential energy on the surface of the Earth:

$U_{s}=-{Gm_{E}m_{s} \over r}=-{(6.67\times 10^{-11})(6\times 10^{24})(10) \over 6.37\times 10^{6}}=-6.28\times 10^{8}J$

and its kinetic energy. Since the energy is conserved we can say that the energy in the orbit is equal to the energy after the launch and write $E_{0}=K+U_{s}$ or $-3.12\times 10^{8}J=K+(-6.28\times 10^{8}J)$. From this equation we obtain the kinetic energy $K=3.16\times 10^{8}J$. This yields the launching velocity of the satellite

$v=7.95\times 10^{3}m/s=7.95km/s$