What is the depth at which the value of acceleration due to gravity becomes 1 n times the value that the surface of earth?

Answer

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Hint: Gravity is the force with which the earth attracts every object towards the centre. We are about to use the gravitational equation and solve this question. It is denoted by ‘g’ and its value at the surface of earth is of about 9.8 $m{s}^{-2}$.Formula used: $g={{g}_{e}}\left( 1-\dfrac{d}{R} \right)$

Complete step by step answer:

Acceleration due to gravity is the acceleration possessed by an object because of the gravity. It has both magnitude and direction, which means it is a vector quantity. The formula is given by the combination of Newton’s second law and gravitational law of forces.Let us assume that acceleration due to gravity at earth’s surface is $g_e$,So, the acceleration due to gravity for a depth ‘d’ from earth’s surface can be written as $g={{g}_{e}}\left( 1-\dfrac{d}{R} \right)$ Where, ‘d’ is the depth given and ‘R’ is the radius of the earth.Now considering the question, the depth at which the value of acceleration due to gravity becomes 1/n times the value of acceleration at the earth’s surface can be written as,$g={{g}_{e}}\dfrac{1}{n}$On equating both the equations, we get${{g}_{_{e}}}\left( 1-\dfrac{d}{R} \right)={{g}_{_{e}}}\dfrac{1}{n}$After cancelling the like terms, $\left( 1-\dfrac { d }{ R } \right) =\dfrac { 1 }{ n }$Which is further solved as$\dfrac { d }{ R } =1-\dfrac { 1 }{ n }$ The required value of depth is derived,$d=R\left( 1-\dfrac { 1 }{ n } \right)$After taking LCM in the right-hand side, we get$d=R\left( \dfrac { n-1 }{ n } \right)$Therefore, the correct answer for the given question is option (B).Note: The SI unit of acceleration due to gravity is measured in metres per second square ($m{s}^{-2}$). There is a chance of mistake in choosing the options (A) and (B). The calculations and substitution of values must be done correctly to avoid errors in the derivation.