Satellites and Binding Energy EX 4

(a) The initial energy is

$E={\frac {1}{2}}U=-{\frac {1}{2}}{Gm_{E}m_{s} \over r}=-{\frac {1}{2}}{(6.67\times 10^{-11})(6\times 10^{24})(50) \over 2\times 6.4\times 10^{6}}=-7.8\times 10^{8}J$

(b) First we have to determine the energy of the satellite in the new orbit. It is 20% less than its initial energy. Here we have to think carefully because the energy is negative. What does it mean to loose 20%? Imagine that you owe $100 and you lose some additional money. Your debt will increase! If you lose 20% of your money you may say that your debt increases by 20% to $120 or by a factor of 1.2. Similarly, the energy in the new orbit will be $E=1.2\times (-7.8\times 10^{8})=-9.4\times 10^{8}J$. We will use the equation for the energy and isolate $r$ from this equation:

$E=-{\frac {1}{2}}{Gm_{E}m_{s} \over r}$

$r=-{\frac {1}{2}}{Gm_{E}m_{s} \over E}$

$r=-{\frac {1}{2}}{(6.67\times 10^{-11})(6\times 10^{24})(50) \over -9.4\times 10^{8}}=1.1\times 10^{7}m$

Note that the satellite descended from the initial distance of $r=2R_{E}=1.28\times 10^{7}m$.

(c) The speed of the satellite depends on the distance. From the Newton's law

${Gm_{E} \over r}=v^{2}$

we expect that as r decreases the speed should increase. We can use this relationship to find the speed in the new orbit $v_{new}$ or we can use $K=-E$. Remember that for a circular orbit

$K={\frac {1}{2}}{Gm_{E}m_{s} \over r}=\left\vert E\right\vert$

The second relationship is simpler to use: $K=9.4\times 10^{8}J$, which gives a speed of ${\frac {1}{2}}m{v_{new}}^{2}=9.4\times 10^{8}={\frac {1}{2}}(50){v_{new}}^{2}$. This equation yields $v_{new}=6.13km/s$.

The velocity in the initial orbit $v_{old}$ can be determined similarly using ${\frac {1}{2}}m{v_{old}}^{2}=7.8\times 10^{8}$. This equation yields $v_{old}=5.56km/s$. The percentage change in speed is approximately 10%. The speed increased by 10%.