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If the change in the value of g at a height h above the earth’s surface is the same as that at depth x (x or $x_e$), then

A. $x = {h^2}$

B. $x = h$

C. $x = \dfrac{h}{2}$

D. $x = 2h$

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Last updated date: 22nd Jun 2024
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Answer
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Hint: As we know that the force of gravity is responsible for the earth’s atmosphere and its value changes with height. As we go to the top of a building the value of gravity decreases and as we lose some height it increases.


Complete answer:

Since we know that the value of g at a height h is defined by:

${g_h} = g\left( {1 - \dfrac{{2h}}{R}} \right)$ …… (I)

Here ${g_h}$ is the acceleration due to gravity at height h and $g$ is the acceleration due to gravity at the surface of the earth, and R is the radius of the earth.


Since we know that the value of g at a depth x is defined by:

${g_x} = g\left( {1 - \dfrac{x}{R}} \right)$ …… (II)

Here ${g_x}$ is the acceleration due to gravity at depth x.

As we are given in the question that the values of gravity at a depth x and height h are equal. So we can equate equation (I) and equation (II).

$\Rightarrow {g_h} = {g_x}$

$\Rightarrow g\left( {1 - \dfrac{{2h}}{R}} \right) = g\left( {1 - \dfrac{x}{R}} \right)$

$\Rightarrow \dfrac{{2h}}{R} = \dfrac{x}{R}$

$\therefore x = 2h$


Therefore, the correct option is (D).


Note: The value of gravity is maximum at the surface of the earth and it decreases with distance above and below the earth’s surface, gravity is zero at the centre of the earth and increases linearly with radial distance till the surface and then decreases(inversely proportional to the square of distance from the surface) as the height increases.