Problem 11.22

11.22. a)

For the average acceleration, let's look at a point halfway along the space ship.
$a = \frac{Gm}{r^{2}}$
$a = \frac{G(100 \times 1.99 \times 10^{30})}{(10000 + 50)^{2}}$
$a = 1.31 \times 10^{14} m/s^{2}$

b)
For the change in acceleration, we look at the acceleration at the front and the rear of the ship.
$\Delta a = \frac{Gm}{r_{min}^{2}} - \frac{Gm}{r_{max}^{2}}$
if we sub in the fact that
r_min = 10000 m
and r_max = 10000m + 100m
we find that
$\Delta a ~=~ 2.62 \times 10^{12} m/s^{2}$