## Essential University Physics: Volume 1 (3rd Edition)

a) $9\times10^{10} \ m$ b) $.53\times10^{12}J$ c) $38, 369 \ m/s$
a) We use the equation for orbital radius: $-.53\times10^{12}J = \frac{GMm}{2r}$ $-.53\times10^{12}J = \frac{(6.67\times10^{-11})(1.99\times10^{30})(720)}{2r}=9\times10^{10} \ m$ b) The kinetic energy is equal to the total energy times -1, which is: $.53\times10^{12}J$. c) We find: $.53\times10^{12}J =\frac{1}{2}(720)v^2 \\ =38, 369 \ m/s$