Answer
a) $v=2590m/s$
b) $K.E_f=4.7\times 10^{19}J$
Work Step by Step
(a) We can find the required speed as
$\frac{1}{2}m_Av^2=-(\frac{-Gm_AM_E}{r})$
This simplifies to:
$v=\sqrt{\frac{2GM_E}{r}}$
We plug in the known values to obtain:
$v=\sqrt{\frac{2(6.67\times 10^{-11})(5.9\times 10^{24})}{(73,600mi\times 1609m/mi)}}$
$v=2590m/s$
(b) We can find the required kinetic energy as follows:
$K.E_f=\frac{1}{2}mv^2$
$K.E_f=\frac{1}{2}\rho Vv^2$
$K.E_f=\frac{1}{2}\rho(\frac{4}{3}\pi R^3)v^2$
We plug in the known values to obtain:
$K.E_f=\frac{1}{2}(3330)(\frac{4}{3}\pi)(\frac{1}{2}\times 2000)^3(2590)^2$
$K.E_f=4.7\times 10^{19}J$
