Answer
$v = 1.4\times 10^6~m/s$
Work Step by Step
We can use conservation of energy to find the apple's speed at the surface:
$K_f+U_f = K_i+U_i$
$K_f+U_f = 0+U_i$
$K_f = U_i-U_f$
$\frac{1}{2}mv^2 = (-\frac{GMm}{R+2.0~m})-(-\frac{GMm}{R})$
$\frac{1}{2}mv^2 = \frac{GMm}{R}-\frac{GMm}{R+2~m}$
$v^2 = \frac{2GM}{R}-\frac{2GM}{R+2~m}$
$v = \sqrt{\frac{2GM}{R}-\frac{2GM}{R+2~m}}$
$v = \sqrt{\frac{(2)(6.67\times 10^{-11}~N~m^2/kg^2)(1.5)(1.99\times 10^{30}~kg)}{2.0\times 10^4~m} - \frac{(2)(6.67\times 10^{-11}~N~m^2/kg^2)(1.5)(1.99\times 10^{30}~kg)}{2.0002\times 10^4~m}}$
$v = 1.4\times 10^6~m/s$
