Answer
The difference between the gravitational acceleration at the top and at the bottom of the apple is approximately $~~5.0\times 10^6~m/s^2$
Work Step by Step
We can find the acceleration due to gravity at the bottom of the apple which is on the surface:
$g_b = \frac{GM}{R^2}$
$g_b = \frac{(6.67\times 10^{-11}~N~m^2/kg^2)(1.5)(1.99\times 10^{30}~kg)}{(2.0\times 10^4~m)^2}$
$g_b = 4.9774875\times 10^{11}~m/s^2$
We can find the acceleration due to gravity at the top of the apple of height $10~cm$:
$g_t = \frac{GM}{(R+0.10~m)^2}$
$g_t = \frac{(6.67\times 10^{-11}~N~m^2/kg^2)(1.5)(1.99\times 10^{30}~kg)}{(2.0\times 10^4~m+0.10~m)^2}$
$g_t = 4.97743773\times 10^{11}~m/s^2$
We can find the difference:
$g_b-g_t = (4.9774875\times 10^{11}~m/s^2)-(4.97743773\times 10^{11}~m/s^2)$
$g_b-g_t = 5.0\times 10^6~m/s^2$
The difference between the gravitational acceleration at the top and at the bottom of the apple is approximately $~~5.0\times 10^6~m/s^2$
