Answer
$v=4300\frac{m}{s}$
Work Step by Step
According to law of conservation of energy
$\frac{1}{2}mv^2=Gm_Em(\frac{1}{r_1}-\frac{1}{r_2})$
This simplifies to:
$v=\sqrt{2Gm_E(\frac{1}{r_1}-\frac{1}{r_2})}$
We plug in the known values to obtain:
$v=\sqrt{(2)(6.67\times 10^{-11})(5.97\times 10^{24})(\frac{1}{6.37\times 10^6}-\frac{1}{6.37\times 10^6+1100\times 10^3})}$
$v=4300\frac{m}{s}$
