Percentage Change in Acceleration Due to Gravity Below and Above Earth Surface

To find the percentage change in the acceleration due to gravity ($g$) when moving from a point $16\text{ km}$ below the Earth's surface to a point $16\text{ km}$ above the Earth's surface, we use the standard formulas for variation of $g$ with depth and height.

1. Acceleration due to gravity at depth $d$:

The acceleration due to gravity at a depth $d = 16\text{ km}$ below the Earth's surface is given by: $g_d = g_0 \left(1 - \frac{d}{R}\right)$ where $g_0$ is the acceleration due to gravity on the Earth's surface and $R = 6400\text{ km}$ is the radius of the Earth.

2. Acceleration due to gravity at height $h$:

Since $h = 16\text{ km} \ll R$, the acceleration due to gravity at a height $h$ above the Earth's surface can be approximated using the binomial expansion: $g_h = g_0 \left(1 - \frac{2h}{R}\right)$

3. Change in $g$:

The magnitude of the change in $g$ when moving from depth $d$ to height $h$ is: $\Delta g = g_d - g_h = g_0 \left(1 - \frac{d}{R}\right) - g_0 \left(1 - \frac{2h}{R}\right)$ $\Delta g = g_0 \left(\frac{2h - d}{R}\right)$

Substitute the given values $d = 16\text{ km}$, $h = 16\text{ km}$, and $R = 6400\text{ km}$: $\Delta g = g_0 \left(\frac{2(16) - 16}{6400}\right) = g_0 \left(\frac{16}{6400}\right) = \frac{g_0}{400}$

4. Percentage change in $g$:

The fractional change in $g$ with respect to the surface value $g_0$ is: $\alpha\% = \frac{\Delta g}{g_0} \times 100\%$ $\alpha\% = \frac{1}{400} \times 100\% = 0.25\%$

Therefore, the value of $\alpha$ is $0.25$.

Correct Answer: B ($0.25$)