Answer
a) $W$ = $Gm_1m_2(\frac{1}{a}-\frac{1}{b})$
b) $\approx$ $8.50{\times}10^{9}$ $J$
Work Step by Step
a)
$W$ = $\int_a^bF(r)dr$ = $\int_a^bG\frac{m_1m_2}{r^2}dr$ = $Gm_1m_2\left[-\frac{1}{r}\right]_a^b$ = $Gm_1m_2\left(\frac{1}{a}-\frac{1}{b}\right)$
b)
$W$ = $GMm(\frac{1}{R}-\frac{1}{R+1,000,000})$
$M$ = mass of earth in kg
$m$ = mass of satellite in kg
$R$ = radius of the earth in m
so
$W$ =$(6.67{\times}10^{-11})(5.98{\times}10^{24})(1000)\left(\frac{1}{6.73{\times}10^{6}}-\frac{1}{7.37{\times}10^{6}}\right)$
$\approx$ $8.50{\times}10^{9}$ $J$
