Question

The value of acceleration due to gravity at a height of $10\,km$ from the surface of earth is $x$. At what depth inside the earth is the value of the acceleration due to gravity has the same value $x$?

• A. 5 km
• B. 20 km
• C. 10 km
• D. 15 km

Answer 1

The Correct Option is B

$g_{h}=g\left(1-\frac{2 h}{R}\right)$
$g_{d}=g\left(1-\frac{d}{R}\right) $
$\Rightarrow g_{h}=g_{d}$
$g\left(1-\frac{2 h}{R}\right)=g\left(1-\frac{d}{R}\right) $
$\Rightarrow d=2 h=2 \times 10=20\, km$

Answer 2

The acceleration due to gravity at a height \(h\) above the Earth's surface is given by the equation: \[ g_h = g_0 \left( \frac{R}{R + h} \right)^2 \] Where: - \(g_h\) is the acceleration due to gravity at height \(h\), - \(g_0\) is the acceleration due to gravity at the Earth's surface, - \(R\) is the radius of the Earth. At a depth \(d\) inside the Earth, the acceleration due to gravity is given by: \[ g_d = g_0 \left( 1 - \frac{d}{R} \right) \] For the gravity at height \(h = 10 \, km\) and at depth \(d\) to be equal, we set the two equations equal to each other: \[ g_0 \left( \frac{R}{R + h} \right)^2 = g_0 \left( 1 - \frac{d}{R} \right) \] Canceling out \(g_0\) and solving for \(d\): \[ \left( \frac{R}{R + h} \right)^2 = 1 - \frac{d}{R} \] Substitute \(h = 10 \, km\), which corresponds to 10,000 m, and solve for \(d\). After simplifying, we find that \(d = 20 \, km\).

Thus, the depth inside the Earth where the acceleration due to gravity is the same as at a height of 10 km is 20 km.