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

Options

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

Explanation

To find the height \( h \) at which the acceleration due to gravity becomes \( \frac{g}{9} \) (where \( g \) is the acceleration due to gravity at the surface of the Earth), we can follow these steps:

### Step 1: Understand the formula for acceleration due to gravity
The acceleration due to gravity \( g' \) at a distance \( r + h \) from the center of the Earth is given by the formula:
\[ g' = \frac{GM}{(r + h)^2} \]
where:
- \( G \) is the universal gravitational constant,
- \( M \) is the mass of the Earth,
- \( r \) is the radius of the Earth,
- \( h \) is the height above the Earth's surface.

### Step 2: Relate \( g' \) to \( g \)
We know that the acceleration due to gravity at the surface of the Earth is:
\[ g = \frac{GM}{r^2} \]
We want to find the height \( h \) where:
\[ g' = \frac{g}{9} \]

### Step 3: Set up the equation
Substituting \( g' \) into the equation:
\[ \frac{GM}{(r + h)^2} = \frac{g}{9} \]
Now, substituting \( g = \frac{GM}{r^2} \) into the equation gives:
\[ \frac{GM}{(r + h)^2} = \frac{1}{9} \cdot \frac{GM}{r^2} \]

### Step 4: Cancel \( GM \) from both sides
Assuming \( GM \neq 0 \), we can cancel \( GM \):
\[ \frac{1}{(r + h)^2} = \frac{1}{9r^2} \]

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

### Step 6: Expand the right side
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 to form a quadratic equation
Rearranging this into standard quadratic form:
\[ h^2 + 2rh - 8r^2 = 0 \]

### 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 since height cannot be negative)

Thus, the valid solution is:
\[ h = 2r \]

### Final Answer:
The height at which the acceleration due to gravity becomes \( \frac{g}{9} \) is:
\[ h = 2R \]
where \( R \) is the radius of the Earth.