The height of any point P above the surface of earth is equal to diameter of earth. The value of acceleration due to gravity at point P will be : (Given g = acceleration due to gravity at the surface of earth).

Concept:

Acceleration due to Gravity: 

• The acceleration by which an object falls on earth under influence of gravity is called acceleration due to Gravity.
• The Acceleration due to Gravity is given as:

\(g = G\frac{M}{R^{2}}\) -- (1)

• Where G is Universal Gravitational Constant, M is the mass of the earth, R is the radius of the earth.
• The acceleration due to the gravity of the earth varies with 

1. Altitude
2. Radius
3. Rotation of Earth
4. Mass of Earth

R here represents the distance from the center of the earth. 

Calculation:

We have to find the value of g at a distance of Diameter i.e. 2R from the surface. 

So the effective R or R' will be 

R' = R + 2R = 3R

The acceleration due to gravity at this point g' will be

\(g' = G\frac{M}{R^{2}}\)

\(\implies g' = G\frac{M}{(R+2R)^{2}}\)

\(\implies g' = G\frac{M}{({3R)}^2}\)

\(\implies g' = G\frac{M}{9R^{2}} \)

\(\implies g' = \frac{g}{9} \) (Using Equation (1))

So, the correct option is \(\frac{g}{9} \)