The ratio of acceleration due to gravity at a height 3R above earth 's surface to the acceleration due to gravity on the surface of the earth is (where R=radius of earth)

Options

A. `1/9`
B. `1/4`
C. `1/16`
D. `1/3`

Explanation

To find the ratio of acceleration due to gravity at a height \(3R\) above the Earth's surface to the acceleration due to gravity on the surface 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 on the surface 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 radius of the Earth.

2. **Acceleration due to Gravity at Height \(h\)**:
When we move to a height \(h\) above the Earth's surface, the acceleration due to gravity \(g'\) at that height is given by:
\[ g' = \frac{GM}{(R + h)^2} \]

3. **Substituting the Height**:
In this case, we are interested in the height \(h = 3R\). Therefore, we substitute \(h\) into the equation:
\[ g' = \frac{GM}{(R + 3R)^2} = \frac{GM}{(4R)^2} = \frac{GM}{16R^2} \]

4. **Finding the Ratio**:
Now, we need to find the ratio of \(g'\) (acceleration due to gravity at height \(3R\)) to \(g\) (acceleration due to gravity at the surface):
\[ \text{Ratio} = \frac{g'}{g} = \frac{\frac{GM}{16R^2}}{\frac{GM}{R^2}} \]

5. **Simplifying the Ratio**:
The \(GM\) terms cancel out:
\[ \text{Ratio} = \frac{1}{16} \]

6. **Conclusion**:
Therefore, the ratio of acceleration due to gravity at a height \(3R\) above the Earth’s surface to the acceleration due to gravity on the surface of the Earth is:
\[ \frac{g'}{g} = \frac{1}{16} \]

### Final Answer:
The ratio is \(\frac{1}{16}\).

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