Question

The density of mass inside a solid sphere of radius $a$ is given by $\rho =\frac{{\rho }_{0}a}{r}$, where ${\rho }_0$ is the density at the surface and $r$ denotes the distance from the centre. Find the gravitational field due to this sphere at a distance $2a$ from its centre.

• A. $\frac{1}{2}\pi G{\rho }_{0}a$
  
• B. $\pi G{\rho }_{0}a$
  
• C. $2\pi G{\rho }_{0}a$
  
• D. $\frac{1}{4}\pi G{\rho }_{0}a$

Answer 1

The correct option is A $\frac{1}{2}\pi G{\rho }_{0}a$


The field is required at a point outside the sphere. Dividing the sphere into concentric shells. Each shell can be replaced by a point particle at its centre having mass equal to the mass of the shell.

Thus, the whole sphere can be replaced by a point particle at its centre having mass equal to the mass of the given sphere. If the mass of the sphere is $M$.

The gravitational field at given point is

$E=\frac{GM}{{\left(2a\right)}^2}=\frac{GM}{4a^2}....\left(1\right)$


The mass $M$ may be calculated as follows. Consider concentric shell of radius $r$ and thickness $dr$.

Its volume is, $dV=\left(4\pi r^2\right)dr$ and its mass is

$dM=\rho dV$

$\Rightarrow dM=\left({\rho }_0\frac{a}{r}\right)\left(4\pi r^{2}dr\right)=4\pi {\rho }_{0}ardr$

The mass of the whole sphere is

$M={\int }_0^{a}4\pi {\rho }_{0}ardr=2\pi {\rho }_0{a}^3$

Thus by using equation $\left(1\right)$, the gravitational field is

$E=\frac{2\pi G{\rho }_0{a}^3}{4a^2}$

$\therefore E=\frac{1}{2}\pi G{\rho }_{0}a$

Hence, option (a) is the correct answer.