The height at which the acceleration due to gravity becomes `(g)/(9)` (where g =the acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth, is :

Options

A. `2R`
B. `R/sqrt3`
C. `R/2`
D. `sqrt(2R)`

Explanation

To find the height at which the acceleration due to gravity becomes \( \frac{g}{9} \) in terms of \( R \) (the radius of the Earth), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the formula for acceleration due to gravity**:
The acceleration due to gravity \( g \) at a distance \( r \) from the center of the Earth is given by the formula:
\[ g' = \frac{GM}{r^2} \]
where \( G \) is the universal gravitational constant, \( M \) is the mass of the Earth, and \( r \) is the distance from the center of the Earth.

2. **Set up the equation for height \( h \)**:
We need to find the height \( h \) above the Earth's surface where the acceleration due to gravity \( g' \) becomes \( \frac{g}{9} \). At height \( h \), the distance from the center of the Earth is:
\[ r + h \]
Therefore, we can write:
\[ g' = \frac{GM}{(R + h)^2} \]
where \( R \) is the radius of the Earth.

3. **Set the equation equal to \( \frac{g}{9} \)**:
We set the expression for \( g' \) equal to \( \frac{g}{9} \):
\[ \frac{GM}{(R + h)^2} = \frac{g}{9} \]

4. **Substitute \( g \) in terms of \( GM \)**:
From the surface of the Earth, we know:
\[ g = \frac{GM}{R^2} \]
Substituting this into our equation gives:
\[ \frac{GM}{(R + h)^2} = \frac{GM}{9R^2} \]

5. **Cancel \( GM \) from both sides**:
Since \( GM \) is common on both sides, we can cancel it:
\[ \frac{1}{(R + h)^2} = \frac{1}{9R^2} \]

6. **Cross-multiply to solve for \( R + h \)**:
Cross-multiplying gives:
\[ 9R^2 = (R + h)^2 \]

7. **Expand the right side**:
Expanding the right side:
\[ 9R^2 = R^2 + 2Rh + h^2 \]

8. **Rearranging the equation**:
Rearranging gives:
\[ 0 = h^2 + 2Rh + R^2 - 9R^2 \]
Simplifying this results in:
\[ 0 = h^2 + 2Rh - 8R^2 \]

9. **Using the quadratic formula**:
This is a quadratic equation in \( h \). We can use the quadratic formula \( h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 1 \), \( b = 2R \), and \( c = -8R^2 \).
\[ h = \frac{-2R \pm \sqrt{(2R)^2 - 4 \cdot 1 \cdot (-8R^2)}}{2 \cdot 1} \]
\[ h = \frac{-2R \pm \sqrt{4R^2 + 32R^2}}{2} \]
\[ h = \frac{-2R \pm \sqrt{36R^2}}{2} \]
\[ h = \frac{-2R \pm 6R}{2} \]

10. **Calculate the possible values for \( h \)**:
This gives us two possible solutions:
\[ h = \frac{4R}{2} = 2R \quad \text{(taking the positive root)} \]
or
\[ h = \frac{-8R}{2} = -4R \quad \text{(not physically meaningful)} \]

11. **Final answer**:
Therefore, the height at which the acceleration due to gravity becomes \( \frac{g}{9} \) is:
\[ h = 2R \]