Answer
The magnitude of the shuttle's acceleration is
$a = 115668$ $km/{h^2}$
Work Step by Step
The trajectory of the shuttle is a circle of radius $R=6378+400=6778$ km. From Example 6, we obtain the normal component of the acceleration:
${a_{\bf{N}}} = \frac{{{v^2}}}{R} = \frac{{{{\left( {28,000} \right)}^2}}}{{6778}} \simeq 115668$ $km/{h^2}$
Since the speed is constant, the tangential acceleration equals to zero. Thus, the magnitude of the shuttle's acceleration is
$a = ||{\bf{a}}|| = \sqrt {{a_{\bf{T}}}^2 + {a_{\bf{N}}}^2} = {a_{\bf{N}}} = 115668$ $km/{h^2}$
