What is the effect on the acceleration due to gravity due to increase in the depth below the Earth?

Question

Accepted Answer

The correct option is (2):

To understand the effect of increasing depth below the Earth's surface on the acceleration due to gravity, let's first recall the formula for acceleration due to gravity at the surface of the Earth. If $R$ is the radius of the Earth and $M$ is its total mass, the acceleration due to gravity on the surface, $g$, is given by:
$g = \frac{GM}{R^2}$

Now, consider a point at a depth $d$ below the Earth's surface. At this depth, the effective radius from the center of the Earth to the point is $(R-d)$. More importantly, the mass contributing to the gravitational acceleration at this point is only the mass of the spherical shell below this point, with radius $(R-d)$. If we assume the Earth has a uniform density $ho$, then the mass of the Earth, $M$, can be expressed as $M = ho 	imes \frac{4}{3}\pi R^3$.

At depth $d$, the effective mass $M\_d$ responsible for gravity at that point is the mass of the sphere of radius $(R-d)$, so $M\_d = ho 	imes \frac{4}{3}\pi (R-d)^3$.
The acceleration due to gravity at depth $d$, denoted as $g\_d$, is given by:
$g\_d = \frac{GM\_d}{(R-d)^2} = \frac{G 	imes ho 	imes \frac{4}{3}\pi (R-d)^3}{(R-d)^2} = G ho \frac{4}{3}\pi (R-d)$

We can also express this in terms of $g$:
Since $g = G ho \frac{4}{3}\pi R$, we can write $\frac{g}{R} = G ho \frac{4}{3}\pi$.
Substituting this into the expression for $g\_d$:
$g\_d = \frac{g}{R}(R-d) = g \left(1 - \frac{d}{R}ight)$

From this formula, it is clear that as the depth $d$ increases, the term $\frac{d}{R}$ increases, and therefore $1 - \frac{d}{R}$ decreases. This implies that $g_d$ decreases as $d$ increases. When $d = R$ (at the center of the Earth), $g_d = 0$. So, as you go deeper into the Earth, the acceleration due to gravity continuously decreases.

The incorrect option is (1) Increases: This is incorrect because as we go below the Earth's surface, the mass of the Earth attracting the object decreases, leading to a reduction in gravitational acceleration. The effective mass pulling the object is only the portion of the Earth's mass that is closer to the center than the object's current position.

The incorrect option is (3) No change: This is incorrect. The acceleration due to gravity is dependent on the mass attracting the object and the distance from the center of that mass. Since both of these effectively change with depth, the acceleration due to gravity cannot remain constant.

The incorrect option is (4) Increases then decreases: This is incorrect. The acceleration due to gravity consistently decreases as you go deeper into the Earth, reaching zero at the Earth's center. It does not increase at any point while moving from the surface to the center.