Answer
a) $9\times10^{10} \ m$
b) $.53\times10^{12}J $
c) $38, 369 \ m/s $
Work Step by Step
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 $
