GravitationMedium
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 $= 6400\text{ }km$.)
Options
A.0.12
B.0.25
C.0.50
D.0.75
Solution
Sol. [Key Concept: $g_{above} = g(1 - \frac{2h}{R})$ and $g_{below} = g(1 - \frac{h}{R})$, so $\frac{\Delta\ g}{g} = \frac{h}{R}$.]
$${\ g_{above} = g - \frac{2hg}{R} = 10 - \frac{2\text{ }h}{R} }{g_{\text{below}\text{~}} = g - \frac{hg}{R} }{\Delta g = g_{\text{above}\text{~}} - g_{\text{below}\text{~}} = g - \frac{2hg}{R} - g + \frac{hg}{R} }{\Delta g = \frac{hg}{R} }{\frac{\Delta g}{g} = \frac{h}{R} = \frac{16 \times 100}{6400} = 0.25\% }$$
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