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:

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

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

$E_i=-\frac{1}{2}\frac{G{M}_{E}m}{r}=-\frac{1}{2}\frac{(6.67\times 10^{-11})(6\times 10^{24})(10^4)}{3.94\times 10^7}=-5.1\times 10^{10}J$

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

$E_f=-\frac{1}{2}\frac{G{M}_{E}m}{r}=-\frac{1}{2}\frac{(6.67\times 10^{-11})(6\times 10^{24})(10^4)}{4.16\times 10^7}=-4.8\times 10^{10}J$

The boosters must provide the difference between these two energies.

$\mathrm{\Delta }E=E_f-E_i=-4.8\times 10^{10}-(-5.1\times 10^{10})=3\times 10^{9}J$