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

Options

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

Answer

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

### Step 2: Calculate the acceleration due to gravity at the surface of the Earth
At the surface of the Earth, the radius \(r\) is equal to \(R\) (the radius of the Earth). Therefore, the acceleration due to gravity at the surface \(g_s\) is:
\[ g_s = \frac{GM}{R^2} \]

### Step 3: Calculate the acceleration due to gravity at a height of \(3R\)
When we are at a height of \(3R\) above the Earth's surface, the total distance from the center of the Earth becomes:
\[ r = R + 3R = 4R \]
Now, we can find the acceleration due to gravity at this height \(g_h\):
\[ g_h = \frac{GM}{(4R)^2} = \frac{GM}{16R^2} \]

### Step 4: Find the ratio of \(g_h\) to \(g_s\)
Now, we can find the ratio of the acceleration due to gravity at height \(3R\) to that at the surface:
\[ \frac{g_h}{g_s} = \frac{\frac{GM}{16R^2}}{\frac{GM}{R^2}} = \frac{1}{16} \]

### Step 5: Conclusion
Thus, the ratio of the acceleration due to gravity at a height of \(3R\) above the Earth's surface to the acceleration due to gravity on the surface of the Earth is:
\[ \frac{g_h}{g_s} = \frac{1}{16} \]

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

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