
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 the earth. When both $d$ and $h$ are much smaller than the radius of earth, then which one of the following is correct?
\[
A.{\text{ }}d = \dfrac{h}{2} \\
B.{\text{ }}d = \dfrac{{3h}}{2} \\
C.{\text{ }}d = 2h \\
D.{\text{ }}d = h \\
\]
Answer
611.7k+ views
- Hint- Here we will proceed further by using the expression for acceleration due to gravity at height h, then we will use a concept that the acceleration due to gravity at height and depth is the same for the same gravitational constant due to acceleration.
Complete step-by-step solution -
Given that
The height above the surface of the earth is $h$
And the depth below the surface of the earth is acceleration due to gravity at height $h$
The expression for acceleration due to gravity at height $h$ is:
${g_h} = g\left( {1 - \dfrac{{2h}}{r}} \right)..........(1)$
Here, r is the radius of the earth
$g$ is the acceleration due to gravity
And ${g_h}$ is the acceleration due to gravity at height $h$
The expression for acceleration due to gravity at depth $d$ is,
${g_d} = g\left( {1 - \dfrac{d}{r}} \right)..........(2)$
The acceleration due to gravity at height and depth is the same for the same gravitational constant due to acceleration.
So, let us equate equation (1) and (2).
$
\because {g_h} = {g_d} \\
\Rightarrow g\left( {1 - \dfrac{{2h}}{r}} \right) = g\left( {1 - \dfrac{d}{r}} \right) \\
$
Now, let us simplify the above equation by cancelling the common terms and taking LCM in order to find the relation between $d$ and $h$
\[
\Rightarrow \left( {1 - \dfrac{{2h}}{r}} \right) = \left( {1 - \dfrac{d}{r}} \right) \\
\Rightarrow \left( {\dfrac{{r - 2h}}{r}} \right) = \left( {\dfrac{{r - d}}{r}} \right) \\
\Rightarrow r - 2h = r - d \\
\Rightarrow r - 2h - r + d = 0 \\
\Rightarrow - 2h + d = 0 \\
\Rightarrow d = 2h \\
\]
Hence, the relation between $d$ and $h$ is \[d = 2h\]
So, the correct option is option C.
Note- The heavy matter object feels more gravity; if there is a mass, it will also have gravity. The value of gravity is more at equators and less at poles. The force that is having a tendency to approach all the objects towards the center of the earth is known as the gravitational force. Students must remember the formula for gravity at height h above the ground and at depth d below the surface to solve such problems.
Complete step-by-step solution -
Given that
The height above the surface of the earth is $h$
And the depth below the surface of the earth is acceleration due to gravity at height $h$
The expression for acceleration due to gravity at height $h$ is:
${g_h} = g\left( {1 - \dfrac{{2h}}{r}} \right)..........(1)$
Here, r is the radius of the earth
$g$ is the acceleration due to gravity
And ${g_h}$ is the acceleration due to gravity at height $h$
The expression for acceleration due to gravity at depth $d$ is,
${g_d} = g\left( {1 - \dfrac{d}{r}} \right)..........(2)$
The acceleration due to gravity at height and depth is the same for the same gravitational constant due to acceleration.
So, let us equate equation (1) and (2).
$
\because {g_h} = {g_d} \\
\Rightarrow g\left( {1 - \dfrac{{2h}}{r}} \right) = g\left( {1 - \dfrac{d}{r}} \right) \\
$
Now, let us simplify the above equation by cancelling the common terms and taking LCM in order to find the relation between $d$ and $h$
\[
\Rightarrow \left( {1 - \dfrac{{2h}}{r}} \right) = \left( {1 - \dfrac{d}{r}} \right) \\
\Rightarrow \left( {\dfrac{{r - 2h}}{r}} \right) = \left( {\dfrac{{r - d}}{r}} \right) \\
\Rightarrow r - 2h = r - d \\
\Rightarrow r - 2h - r + d = 0 \\
\Rightarrow - 2h + d = 0 \\
\Rightarrow d = 2h \\
\]
Hence, the relation between $d$ and $h$ is \[d = 2h\]
So, the correct option is option C.
Note- The heavy matter object feels more gravity; if there is a mass, it will also have gravity. The value of gravity is more at equators and less at poles. The force that is having a tendency to approach all the objects towards the center of the earth is known as the gravitational force. Students must remember the formula for gravity at height h above the ground and at depth d below the surface to solve such problems.
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