If g is the acceleration due to gravity on the surface of the earth , its value at a height equal to double the radius of the earth is

Options

A. g
B. `g/2`
C. `g/3`
D. `g/9`

Answer

To find the value of the acceleration due to gravity \( g' \) at a height equal to double the radius of the Earth, we can follow these steps:

### Step 1: Understand the Problem
We know that the acceleration due to gravity at the surface of the Earth is given by \( g = \frac{GM}{R^2} \), where:
- \( G \) is the gravitational constant,
- \( M \) is the mass of the Earth,
- \( R \) is the radius of the Earth.

### Step 2: Identify the Height
The question states that the height \( h \) is equal to double the radius of the Earth. Therefore:
\[ h = 2R \]

### Step 3: Calculate the Distance from the Center of the Earth
The distance \( d \) from the center of the Earth to the point where we want to find \( g' \) is the sum of the radius of the Earth and the height:
\[ d = R + h = R + 2R = 3R \]

### Step 4: Use the Formula for Gravitational Acceleration
The formula for the acceleration due to gravity at a distance \( d \) from the center of the Earth is:
\[ g' = \frac{GM}{d^2} \]
Substituting \( d = 3R \):
\[ g' = \frac{GM}{(3R)^2} = \frac{GM}{9R^2} \]

### Step 5: Relate \( g' \) to \( g \)
We know that \( g = \frac{GM}{R^2} \). Thus, we can express \( g' \) in terms of \( g \):
\[ g' = \frac{GM}{9R^2} = \frac{1}{9} \cdot \frac{GM}{R^2} = \frac{g}{9} \]

### Conclusion
The acceleration due to gravity at a height equal to double the radius of the Earth is:
\[ g' = \frac{g}{9} \]

### Final Answer
Thus, the value of \( g' \) at a height equal to double the radius of the Earth is \( \frac{g}{9} \).

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