Question

A tunnel is dug along a diameter of the earth. If ${M_e}$ and ${R_e}$ are the mass and radius, respectively, of the earth, then the force on a particle of mass $m$ placed in the tunnel at a distance $r$ from the centre is :
A. $\dfrac{{G{M_e}m}}{{{R_e}^3}}r$
B. $\dfrac{{G{M_e}m}}{{{R_e}^3r}}$
C. $\dfrac{{G{M_e}mR_e^3}}{r}$
D. $\dfrac{{G{M_e}m}}{{{R_e}^2}}r$

Answer 1

When a tunnel is dug, the value of acceleration due to gravity will vary at different points from the centre of the earth. Calculate mass of the inner sphere of radius $r$ and then calculate acceleration due to gravity, $g$ at a distance $r$ from centre of the earth.
Force on a particle of mass $m$ where acceleration due to gravity is $g$ is given by $F = mg$.

Complete solution:
As given in the question, a tunnel is dug along the diameter of the earth and we are asked to find the force on a particle of mass $m$ placed in the tunnel at a distance $r$ from the centre.
So, let us first calculate the mass of the inner solid sphere of radius $r$.
Total volume of the earth is $\dfrac{4}{3}\pi R_e^3$ where ${R_e}$ is the radius of the earth.
Volume of the inner sphere of radius $r$ is $\dfrac{4}{3}\pi {r^3}$
So, mass of the inner solid sphere of radius $r$ is given by
$M = \dfrac{{{M_e}}}{{\dfrac{4}{3}\pi R_e^3}} \times \dfrac{4}{3}\pi {r^3} = \dfrac{{{M_e}{r^3}}}{{R_e^3}}$
Now, we calculate acceleration due to gravity, $g$ at a distance $r$ from centre of the earth by using the formula $\dfrac{{GM}}{{{R^2}}}$ where $G$ is Gravitational constant whose value is $6.67 \times {10^{ - 11}}{m^3}k{g^{ - 1}}{s^{ - 2}}$ and $R = r$ .
So,$g=\dfrac{{G{M_e}{r^3}}}{{R_e^3}} \times \dfrac{1}{{{r^2}}} = \dfrac{{G{M_e}r}}{{R_e^3}}$
Now we know that the force on a particle of mass $m$ where acceleration due to gravity is $g$ is given by $F = mg$ .
Then, $F = m \times \dfrac{{G{M_e}r}}{{R_e^3}} = \dfrac{{G{M_e}mr}}{{R_e^3}}$
Hence, the force on a particle of mass $m$ placed in the tunnel at a distance $r$ from the centre is $\dfrac{{G{M_e}m}}{{R_e^3}}r$ .

So, option A is correct.

Note: When a body is falling freely, it acquires an acceleration equal to the acceleration due to gravity. Remember that at a different location from the earth the value of g is different and can be calculated by the formula $\dfrac{{GM}}{{{R^2}}}$ by carefully calculating the mass. This formula also clarifies that the acceleration due to gravity doesn’t depend upon the mass of the body. It means when a big rock and a small ball will drop from the same height, both will reach the ground at the same time.