91Ó°ÊÓ

The gravitational acceleration is $a_g/3$

The mass of the sphere is equal to the density multiplied by volume. The gravitational acceleration is expressed in terms of the gravitational constant, the mass of the earth, and the radius of the earth.

Using the formula for gravitational acceleration in which gravitational acceleration is inversely proportional to the square of the distance between the objects, we can find the distance inside and outside the sphere where the gravitational acceleration is ${\mathit{a}}_{\mathbf{g}}\mathbf{/}\mathbf{3}$

Formula:

The volume of sphere,$v=\frac{4}{3}\mathrm{Ï€}{r}^3$ (i)

The density of sphere, $p=\frac{M}{V}$ (ii)

Gravitational acceleration due to free-fall, $g=\frac{GM}{r^2}$ (iii)

As per the given condition,

$a=\frac{a_g}{3}$

Let’s assume that the gravitational acceleration is 1/3^rd at a radius r inside the earth. To write the equation for the gravitational accelerationa, we have to consider the massm enclosed by the sphere of radius r. Therefore,

$a=\frac{GM}{r^2}$

The acceleration due to gravity on the surface of the earth with mass M and radius Ris,

$a_g=\frac{GM}{R^2}$

Substituting the values in the given condition, we get

$$\begin{gathered} \frac{GM}{r^2}=\frac{GM}{3R^2} \\ {r}^2=\frac{3R^{2}m}{M} \end{gathered}$$

The mass is calculated using volume and density. For the calculation purpose, let’s assume that the density of the earth is constant. Now, write the equation for the mass of the sphere of radius r.

$$\begin{gathered} m=p.v \\ =p.\frac{4}{3}{\mathrm{Ï€°ù}}^3 \end{gathered}$$

p is the density of the earth and is the volume of a sphere of radiusr .

And, the mass of the earth with radius R is,

$$\begin{gathered} M=p.V \\ =p.\frac{4}{3}{\mathrm{Ï€¸é}}^3 \end{gathered}$$

p Is the density of the earth and V is the volume of earth with radius R .

Now, substitute the equations for m and M in the above equation for $r^2$

$$\begin{gathered} r^2=\frac{3R^2\left(p.{\displaystyle \frac{4}{3}}{\mathrm{Ï€°ù}}^3\right)}{p.{\displaystyle \frac{4}{3}}{\mathrm{Ï€¸é}}^3} \\ \mathrm{r}=\frac{\mathrm{R}}{3} \end{gathered}$$

Therefore, at $R/3$ inside the earth the gravitational acceleration will be$a_g/3$.

Outside the Earth’s sphere, the mass of the earth under consideration will not change. So gravitational acceleration will depend only on the distance from the center of the earth.

The gravitational acceleration on the surface of the earth is,

$a_g=\frac{GM}{R^2}$

The gravitational acceleration at a distance r outside the earth is,

$a=\frac{GM}{r^2}$

As per the given condition,

$a=\frac{a_g}{3}$

Therefore,

$$\begin{gathered} \frac{GM}{r^2}=\frac{GM}{3R^2} \\ r=\sqrt{3}.R \end{gathered}$$

Therefore, at $\sqrt{3}$outside the earth the gravitational acceleration will be $a_g/3$.