Question

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

Answer 1

Hint:To solve this question one must have an insight of chapter gravitation and its various formulas. By considering sphere as an Earth and putting the formula of Gravitational field strength, g(r) =$\dfrac{{\left( {G{m_E}} \right)}}{{{r^2}}}$having ${m_E}$=$\rho \left( {4\pi {r^2}dr} \right)$ will make easy to solve. The Detailed solution is below with names and their meaning.

Step by step solution:
Step 1:
If we consider the given sphere is Earth then the acceleration due to gravity near the Earth depends on the distance of an object from Earth's center. The gravitational field formula can be used to find the field strength, meaning the acceleration due to gravity at any position around the Earth. The radius of the Earth is 
The gravitational field strength is measured in Newton per kilogram or$\dfrac{N}{{kg}}$ in the same units as acceleration, $\dfrac{m}{{{{\sec }^2}}}$
Gravitational field strength, g(r) =$\dfrac{{\left( {G{m_E}} \right)}}{{{r^2}}}$….(1)
Here, g(r) = Earth's gravitational field strength
G = gravitational constant = $\left( {6.67 \times {{10}^{ - 11}}\dfrac{{N{m^2}}}{{k{g^2}}}} \right)$
${m_E}$ = mass of the Earth ($5.98 \times {10^{24}}kg$)
r = distance from the center of the Earth (m)
Step 2:
Now let us calculate the mass of the sphere is $\rho \left( {4\pi {r^2}dr} \right)$ and we have given $\rho $ =${\rho _o}$ $\dfrac{a}{r}$, where ${\rho _o}$the density at the surface and r is denotes the distance from centre, which is equal to $4\pi {\rho _o}ardr$…. (2)
To know the mass of the sphere we have to integrate the value from 0 to a thus,
m=$\int\limits_0^a {dm} $ , putting the values of dm from (1)$\int\limits_0^a {4\pi {\rho _o}ardr} $
After integration and putting the values from 0 to a then we get $2\pi {\rho _o}{a^3}$…… (3)
 Here we have got the mass. Now putting in eqn(1)
Which implies g(r)=$\dfrac{{Gm}}{{2{a^2}}}$ , the radius is 2a given in question

Through further solving we will get a gravitational field due to this sphere at a distance 2a from its centre is $\dfrac{{\pi G{\rho _o}a}}{2}$ .

Note:
Just because of the gravitational force of the earth, the atmosphere is present around its surface, which is crucial for sustainability of life on earth. We are able to perform motion due to the force of gravity. If an object falls from one point to another point inside a gravitational field, the force of gravity will do positive work on the object, and the gravitational potential energy will decrease by the same amount.