Answer
a) $K.E=1.69\times 10^{10}J$
b) $\Delta E=8.45\times 10^9J$
Work Step by Step
(a) We can find the required kinetic energy as follows:
$K.E=\frac{1}{2}mv^2=G\frac{M_Em}{2r}$
We plug in the known values to obtain:
$K.E=\frac{(6.67\times 10^{-11})(1720)(5.97\times 10^{24})}{2(12,600m\times 1609m/mi)}$
$K.E=1.69\times 10^{10}J$
(b) We know that
$\Delta E=-\frac{1}{2}(\frac{1}{r_f}-\frac{1}{r_i})$
We plug in the known values to obtain:
$\Delta E=-\frac{1}{2}(6.67\times 10^{-11})(5.97\times 10^{24})(1720)\times [\frac{1}{25,200mi}-\frac{1}{12,600mi}](\frac{1}{1609m/mi})$
$\Delta E=8.45\times 10^9J$
