Satellites and Binding Energy EX 13

The shuttle is in an orbit of radius $r$ which 5% smaller than rg. Therefore the radius of this orbit $r=0.95r_{g}$.

First we have to determine the radius in the geosynchroneous orbit using Kepler's Law:

$r=\left({GM_{E}T^{2} \over 4\pi ^{2}}\right)^{\frac {1}{3}}=\left({(6.67\times 10^{-11})(6\times 10^{24})(8.64\times 10^{4})^{2} \over 4\pi ^{2}}\right)^{\frac {1}{3}}=4.16\times 10^{7}m$

Then $r=0.95r_{g}=0.95\times 4.16\times 10^{7}=3.94\times 10^{7}m$ and the energy of the shuttle in this orbit is

$E_{i}=-{\frac {1}{2}}{GM_{E}m \over r}=-{\frac {1}{2}}{(6.67\times 10^{-11})(6\times 10^{24})(10^{4}) \over 3.94\times 10^{7}}=-5.1\times 10^{10}J$

To get into the geosynchroneous orbit the shuttle needs to have the energy

$E_{f}=-{\frac {1}{2}}{GM_{E}m \over r}=-{\frac {1}{2}}{(6.67\times 10^{-11})(6\times 10^{24})(10^{4}) \over 4.16\times 10^{7}}=-4.8\times 10^{10}J$

The boosters must provide the difference between these two energies.

$\Delta E=E_{f}-E_{i}=-4.8\times 10^{10}-(-5.1\times 10^{10})=3\times 10^{9}J$