## Essential University Physics: Volume 1 (4th Edition) Clone

$1.42 \times10^3 \ km$
The first rocket must go 10.5 orbits in the time that it takes the other rocket to go 10. Since radius is directly related to the period raised to the 2/3 power, we find: $(\frac{10}{10.5})^{2/3}=\frac{r}{45,800}$ $r = 44,334\ km$ Thus, we find the difference is: $h = 45,800\ km - 44,334 \ km =1,465 \ km$