The acceleration of a body due to the attraction of the earth (radius R) at a distance 2R from the surface of the earth is (g = acceleration due to gravity at the surface of the earth)

The correct option is g/9.

To find the acceleration due to gravity at a distance from the Earth's surface, we use the universal law of gravitation. Let us break down the solution step-by-step:

Step 1: Acceleration due to gravity at the surface of the Earth
The acceleration due to gravity $g$ at the surface of the Earth (at a distance $R$ from the center) is given by:
$g=\frac{GM}{R^2}$
where:
• $G$ is the universal gravitational constant,
• $M$ is the mass of the Earth, and
• $R$ is the radius of the Earth.

Step 2: Distance of the body from the center of the Earth
The body is at a distance of $2R$ from the surface of the Earth. Therefore, the total distance $r$ of the body from the center of the Earth is:
$r=R+2R=3R$

Step 3: Acceleration due to gravity at distance $r$
Let the acceleration due to gravity at this point be $g^{\prime }$. The formula is:
$g^{\prime }=\frac{GM}{r^2}$

Step 4: Substitute the value of $r$
Substitute $r=3R$ into the formula:
$g^{\prime }=\frac{GM}{{\left(3R\right)}^2}$
$g^{\prime }=\frac{GM}{9R^2}$
$g^{\prime }=\frac{1}{9}(\frac{GM}{R^2})$

Since we know that $g=\frac{GM}{R^2}$, we can write:
$g^{\prime }=\frac{g}{9}$

Thus, the acceleration of the body at a distance of $2R$ from the surface of the Earth is $\frac{g}{9}$.