Question

The value of g
(a). Increases with increase in depth
(b). Decreases with increase in depth
(c). Remains the same regardless of the depth
(d). Depends upon the mass of the object

Answer 1

- Hint: Here g is the acceleration due to gravity. When a body falls freely, then the acceleration acquired by it is known as g i.e. gravitational acceleration. This acceleration is due to the gravitational pull of the earth. According to the Gravitational law, it has an inverse relationship with the distance between the objects that are participating in the attraction.

Complete step-by-step solution -
Acceleration due to gravity is the acceleration attained by a body or an object when subjected to gravitational force. Its SI unit is the same as that of acceleration, i.e., \[m/{{s}^{2}}\].
The standard value of g used on the surface of the earth is \[9.8m/{{s}^{2}}\].
To find out the formula for g, we will go through simple mathematical formulation. We will Start from the formula describing the law of Gravitation.
 $F=G\dfrac{{{M}_{E}}m}{{{R}^{2}}}$, where ${{M}_{E}}$ is the mass of earth, $m$ is the mass of an object, $R$ is the distance between the object and surface of earth and $G$ is the gravitational constant.
Whereas, \[F=ma\]
Here we are considering acceleration due to gravity as the acceleration involved in the force.
So the equation can be written as,
\[mg=G\dfrac{{{M}_{E}}m}{{{R}^{2}}}\]
$g=G\dfrac { { M }_{ E } }{ { R }^{ 2 } }$
Now we can substitute mass in terms of density. So the equation can be written as,
$g=G\dfrac{\dfrac{4}{3}\pi {{R}^{3}}\rho }{{{R}^{2}}}$, where $\rho $ is the density and the earth is considered as a sphere of radius $R$.
$g=G\dfrac{4}{3}\pi R\rho $
If we are considering the situation of depth in the earth, then the radius of the earth has to be taken as the difference between the original radius and the depth $(d)$. So the equation will be,
$g=G\dfrac{4}{3}\pi (R-d)\rho $
So from this formula we can easily conclude that the value of g decreases with increase in depth d.

Note: The value of gravitational acceleration g decreases with an increase in depth, but students should also know that its value decreases with the increase in height also. Moreover, students should keep in mind that you can not apply \[g=9.8m/{{s}^{2}}\]everywhere because this value is reliable on the surface of earth.