When one moves from a point \(16\) km below the earth's surface to a point \(16\) km above the earth's surface. The change in \(g\) is approximately \(\alpha\) %. The value of \(\alpha\) is _____.
(Take radius of the earth \(= 6400\) km.)

The acceleration due to gravity at a depth \(d\) below the earth's surface is given by:

\(g_d = g\left(1 - \dfrac{d}{R}\right)\)

The acceleration due to gravity at a height \(h\) above the earth's surface (for \(h \ll R\)) is given by:

\(g_h = g\left(1 - \dfrac{2h}{R}\right)\)

Given \(d = 16\) km and \(h = 16\) km, we can find the values of \(g_d\) and \(g_h\):

\(g_d = g\left(1 - \dfrac{16}{6400}\right) = g\left(1 - \dfrac{1}{400}\right)\)

\(g_h = g\left(1 - \dfrac{2 \times 16}{6400}\right) = g\left(1 - \dfrac{2}{400}\right)\)

The magnitude of the change in acceleration due to gravity when moving from the depth \(d\) to the height \(h\) is:

\(\Delta g = g_d - g_h = g\left(1 - \dfrac{1}{400}\right) - g\left(1 - \dfrac{2}{400}\right)\)

\(\Delta g = g\left(\dfrac{2}{400} - \dfrac{1}{400}\right) = \dfrac{g}{400}\)

The percentage change in \(g\) is approximately:

\(\dfrac{\Delta g}{g} \times 100 = \dfrac{\dfrac{g}{400}}{g} \times 100 = \dfrac{100}{400} = 0.25\%\)

Thus, the value of \(\alpha\) is \(0.25\).

Answer: \(0.25\)