Question

The change in the value of g at a height h above the surface of earth is the same as at a depth d below earth. When both d and h are much smaller than the radius of earth, then which one of the following is correct?
A. $d = \dfrac{h}{2}$
B. $d = \dfrac{{3h}}{2}$
C. $d = 2h$
D. $d = h$

Answer 1

Hint: g represents the acceleration due to gravity. It is directly proportional to the force F exerted by the earth on an object and inversely proportionally to the mass of the object m. Change in the physical quantities representing this value, will result in a change in the value of this quantity itself.

Complete step by step answer:
The acceleration due to gravity is given mathematically as:
$g = \dfrac{F}{m}$
where F is the force exerted by the earth on an object of mass m.
If the object is placed at a distance h above the surface of the earth, the force of gravitation acting on it due to earth will be:
$F = \dfrac{{GMm}}{{{{\left( {R + h} \right)}^2}}}$
where M is the mass of earth and R is the radius of earth.
Thus,
$g = \dfrac{F}{m} = \dfrac{{GM}}{{{{\left( {R + h} \right)}^2}}}$
We can clearly see that the value of g decreases as we move upwards, away from the surface of earth.
We can write the above equation as:
$g = \dfrac{{GM}}{{{R^2}{{\left( {1 + \dfrac{h}{R}} \right)}^2}}} = \dfrac{{{g_ \circ }}}{{{{\left( {1 + \dfrac{h}{R}} \right)}^2}}}$
where ${g_ \circ } = \dfrac{{GM}}{{{R^2}}}$is the value of g at the surface of earth.
If h<$g = {g_ \circ }{\left( {1 + \dfrac{h}{R}} \right)^{ - 2}} \approx {g_ \circ }\left( {1 - \dfrac{{2h}}{R}} \right) \cdots \cdots \cdots \cdots \left( 1 \right)$
Similarly, if one goes a distance d inside the earth such as in mines, the value of g again decreases. The force by the earth inside its surface is given by:
$F = \dfrac{{GMm}}{{{R^3}}}\left( {R - d} \right)$
Or
$\eqalign{
  & g = \dfrac{F}{m} = \dfrac{{GM}}{{{R^3}}}\left( {R - d} \right) \cr
  & \Rightarrow g = {g_ \circ }\left( {1 - \dfrac{d}{R}} \right) \cdots \cdots \cdots \cdots \left( 2 \right) \cr} $
Now, according to the question, equation 1 and 2 are equal,
So equating equation 1 and 2, we get:
$\eqalign{
  & \Rightarrow {g_ \circ }\left( {1 - \dfrac{{2h}}{R}} \right) = {g_ \circ }\left( {1 - \dfrac{d}{R}} \right) \cr
  & \Rightarrow 2h = d \cr} $
Therefore, the correct option is C. i.e. , $d = 2h$

Note: The value of g is maximum at earth’s surface and decreases with increase in height h as well as decrease in depth d. The value of g is affected by other factors also like rotation of earth, non-uniformity of earth and nonsphericity of earth.