Question

Acceleration due to gravity as a function of $r$ is given by:
A. $\dfrac{4}{3}\pi Gr\left( {A - Br} \right)$
B. $4\pi Gr\left( {A - Br} \right)$
C. $\dfrac{4}{3}\pi Gr\left( {A - \dfrac{3}{4}Br} \right)$
D. $\dfrac{4}{3}\pi Gr\left( {A - \dfrac{3}{2}Br} \right)$

Answer 1

Assume the sphere of earth is made up of infinite, concentric shells of small thickness. Calculate the mass of each shell and integrate the mass to compute the mass of earth. Use the formula for acceleration due to gravity as function of distance from the centre and substitute the expression for mass of earth to determine the acceleration due to gravity as a function of r.

Formula used:
$\rho = \dfrac{M}{V}$
Here, M is the mass of earth and V is the volume of earth.

Complete step by step answer:
We know that the density of earth is not uniform. It is denser at the centre and least dense as we move towards the surface. The approximation of the density of earth is given as,
$\rho = \left( {A - Br} \right)$, where, $A = 12,700\,km/{m^3}$, $B = 1.5 \times {10^{ - 3}}\,kg/{m^3}$ and r is the distance from the centre of earth.
Now, consider the earth as concentric shells of thickness $dr$ and volume of each shell as $dV$. The total mass of the earth can be found by integrating the mass of each shell from centre to surface of earth that is from 0 to r.
Therefore, the mass of earth is,
$M = \int\limits_0^r {\rho \left( r \right)} \,dV$
Here, $\rho \left( r \right)$ is the density of each shell.
The volume of the concentric shell is $4\pi {r^2}$ and the density of the earth as a function of r is $\rho = \left( {A - Br} \right)$.
Therefore, the above equation becomes,
$M = \int\limits_0^r {\left( {A - Br} \right)} \,\left( {4\pi {r^2}} \right)dr$
$\Rightarrow M = \int\limits_0^r {\left( {4\pi A{r^2} - 4\pi B{r^3}} \right)} dr$
$\Rightarrow M = 4\pi \left( {A\dfrac{{{r^3}}}{3} - B\dfrac{{{r^4}}}{4}} \right)_0^r$
$\Rightarrow M = 4\pi \left[ {\left( {A\dfrac{{{r^3}}}{3} - B\dfrac{{{r^4}}}{4}} \right) - \left( {A\dfrac{{{{\left( 0 \right)}^3}}}{3} - B\dfrac{{{{\left( 0 \right)}^4}}}{4}} \right)} \right]$
$\Rightarrow M = \dfrac{4}{3}\pi {r^3}\left( {A - \dfrac{3}{4}Br} \right)$
The acceleration due to gravity of earth is given as,
$g = \dfrac{{GM\left( r \right)}}{{{r^2}}}$ …… (1)
Here, G is the gravitational constant and $M\left( r \right)$ is the mass of earth as a function of r.
Substitute the expression for mass of earth in the above equation.
$g = \dfrac{{G\left( {\dfrac{4}{3}\pi {r^3}\left( {A - \dfrac{3}{4}Br} \right)} \right)}}{{{r^2}}}$
$\Rightarrow g = \dfrac{4}{3}\pi Gr\left( {A - \dfrac{3}{4}Br} \right)$

So, the correct answer is “Option C”.

Note:
Acceleration due to gravity at the centre of earth becomes zero. The acceleration due to gravity given by equation (1) gives acceleration due to gravity infinite at $r = 0$, that is at the centre of earth. But the mass of earth as a function of distance from the centre becomes zero at the centre, therefore, by substituting 0 for $M\left( r \right)$ in equation (1) gives acceleration due to gravity zero at the centre of earth.