Rankers Physics

Gravitation: Practice Problem & Solution

The acceleration due to gravity at a height $1 \text{ km}$ above the earth is the same as at a depth d below the surface of earth. Then: (2017-Delhi)
$d = 1 \text{ km}$
$d = \frac{3}{2} \text{ km}$
$d = 2 \text{ km}$
$d = \frac{1}{2} \text{ km}$

Solution Explained:

To solve this problem, we apply the core principles of Gravitation. Understanding the underlying formula is key to arriving at the correct answer below:

For heights much smaller than the radius ($h \ll R$), $g_h \approx g(1 - \frac{2h}{R})$.nFor depth $d$, $g_d = g(1 - \frac{d}{R})$. Equating the two gives $1 - \frac{2h}{R} = 1 - \frac{d}{R}$.nThus $d = 2h$. Since $h = 1 \text{ km}$, $d = 2 \times 1 = 2 \text{ km}$.

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