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)/(sqrt2)`
C. `(R)/(2)`
D. `sqrt2 R`

Answer

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

### Step 2: Set up the equation for gravity at height \( h \)
At a height \( h \) above the Earth's surface, the distance from the center of the Earth becomes \( R + h \). Therefore, the acceleration due to gravity at this height is:
\[ g' = \frac{GM}{(R + h)^2} \]

### Step 3: Set the equation for \( g' \) equal to \( \frac{g}{9} \)
We know that the acceleration due to gravity on the surface of the Earth is:
\[ g = \frac{GM}{R^2} \]
We want to find the height where:
\[ g' = \frac{g}{9} \]
Substituting the expressions for \( g' \) and \( g \):
\[ \frac{GM}{(R + h)^2} = \frac{1}{9} \cdot \frac{GM}{R^2} \]

### Step 4: Cancel \( GM \) from both sides
Since \( GM \) is present in both sides of the equation, we can cancel it out:
\[ \frac{1}{(R + h)^2} = \frac{1}{9R^2} \]

### Step 5: Cross-multiply to solve for \( R + h \)
Cross-multiplying gives us:
\[ 9R^2 = (R + h)^2 \]

### Step 6: Expand the right side of the equation
Expanding the right side:
\[ 9R^2 = R^2 + 2Rh + h^2 \]

### Step 7: Rearrange the equation
Rearranging gives:
\[ 9R^2 - R^2 = 2Rh + h^2 \]
\[ 8R^2 = 2Rh + h^2 \]

### Step 8: Rearrange into standard quadratic form
Rearranging this into a standard quadratic equation:
\[ h^2 + 2Rh - 8R^2 = 0 \]

### Step 9: Solve the quadratic equation
Using the quadratic formula \( h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = 2R, c = -8R^2 \):
\[ h = \frac{-2R \pm \sqrt{(2R)^2 - 4 \cdot 1 \cdot (-8R^2)}}{2 \cdot 1} \]
\[ h = \frac{-2R \pm \sqrt{4R^2 + 32R^2}}{2} \]
\[ h = \frac{-2R \pm \sqrt{36R^2}}{2} \]
\[ h = \frac{-2R \pm 6R}{2} \]

### Step 10: Calculate the possible values for \( h \)
Calculating the two possible solutions:
1. \( h = \frac{4R}{2} = 2R \)
2. \( h = \frac{-8R}{2} = -4R \) (not physically meaningful)

Thus, the height \( h \) at which the acceleration due to gravity becomes \( \frac{g}{9} \) is:
\[ h = 2R \]

### Final Answer
The height at which the acceleration due to gravity becomes \( \frac{g}{9} \) is \( 2R \).

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