Question

The value of acceleration due to gravity (g) at earth’s surface is $10m{s^{ - 2}}$. If the earth is assumed to be a sphere of radius R meters and uniform mass density then Its value in $m{s^{ - 2}}$ at the centre of the earth is?
A. $5$
B. $\dfrac{{10}}{R}$
C. $\dfrac{{10}}{{2R}}$
D. Zero

Answer 1

Hint:-Here we are going to use the various formulas related to acceleration due to gravity mainly in the depth of the earth, there conceptual and mathematical approach to solve the questions
$
  g = \dfrac{{G{M^2}}}{R} \\
  {g_d} = g\left( {1 - \dfrac{d}{R}} \right) \\
 $

Complete step-by-step solution:Let is first discuss the formula of acceleration due to gravity at the surface of earth
$g = \dfrac{{G{M^2}}}{R}$
Where G is the universal constant, M is the mass of the object and R is the radius of Earth.
Now coming to the solution,
There is no mass exactly at centre of the Earth to provide a gravitational pull and thus a gravitational acceleration that is acceleration due to gravity $\dfrac{{G{M^2}}}{R}$​ is 0 at centre.
According to shell theorem-no outside mass provides gravitational pull at the centre of Earth.
Also, at centre all the forces act on the object spherical symmetrically and hence cancel out
Hence, net force acting on an object placed at the centre of the earth is zero
Since there is no force at the centre of earth, there would be no acceleration
Hence the correct option is (D).

Note:- Let us do this with a proper mathematical formula
${g_d} = g\left( {1 - \dfrac{d}{R}} \right)$
Where ${g_d}$ is the acceleration due to gravity at a depth d from the earth’s surface, g is the acceleration due to gravity on the earth’s surface, d is the distance from the earth’s surface and R is the radius of Earth which is fixed.
For calculating acceleration due to gravity at the centre put d=R,
Then we get ${g_d} = 0$
Hence correct option is (D)