Question

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.)

• A. 0.12
• B. 0.25
• C. 0.50
• D. 0.75

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to calculate the value of acceleration due to gravity at two points (depth and height) and find the percentage difference between them relative to the surface gravity.
Step 2: Key Formula or Approach:
At depth $d$: $g_d = g \left( 1 - \frac{d}{R} \right)$.
At height $h$ ($h \ll R$): $g_h = g \left( 1 - \frac{2h}{R} \right)$.
Step 3: Detailed Explanation:
Radius $R = 6400$ km.
At point $d = 16$ km below surface:
$g_{below} = g \left( 1 - \frac{16}{6400} \right) = g \left( 1 - \frac{1}{400} \right)$.
At point $h = 16$ km above surface:
$g_{above} = g \left( 1 + \frac{16}{6400} \right)^{-2} \approx g \left( 1 - \frac{2 \times 16}{6400} \right)$ (since $16 \ll 6400$).
$g_{above} = g \left( 1 - \frac{32}{6400} \right) = g \left( 1 - \frac{1}{200} \right)$.
The "change in $g$" usually refers to the magnitude of the difference relative to the surface $g$.
Change $|\Delta g| = |g_{above} - g_{below}| = |g(1 - \frac{1}{200}) - g(1 - \frac{1}{400})|$.
$|\Delta g| = g \cdot |-\frac{1}{200} + \frac{1}{400}| = g \cdot |-\frac{1}{400}| = \frac{g}{400}$.
Percentage change $\alpha = \frac{|\Delta g|}{g} \times 100 = \frac{1}{400} \times 100 = 0.25 %$.
Thus, $\alpha = 0.25$.
Step 4: Final Answer:
The value of $\alpha$ is 0.25.