The height of the point vertically above the earth's surface, at which acceleration due to gravity becomes 1% of its value at the surface is (Radius of the earth =R)

Options

A. `8R`
B. `9R`
C. `10R`
D. `20R`

Explanation

To solve the problem of finding the height at which the acceleration due to gravity becomes 1% of its value at the Earth's surface, we can follow these steps:

### Step 1: Understand the relationship between gravity at height and at the surface
The acceleration due to gravity at a height \( h \) above the Earth's surface can be expressed as:
\[ g' = \frac{g R^2}{(R + h)^2} \]
where:
- \( g' \) is the acceleration due to gravity at height \( h \),
- \( g \) is the acceleration due to gravity at the Earth's surface,
- \( R \) is the radius of the Earth.

### Step 2: Set up the equation for 1% of gravity
We need to find the height \( h \) where \( g' \) is 1% of \( g \):
\[ g' = \frac{g}{100} \]
Substituting this into the equation gives:
\[ \frac{g}{100} = \frac{g R^2}{(R + h)^2} \]

### Step 3: Cancel \( g \) from both sides
Since \( g \) is present on both sides of the equation, we can cancel it out (assuming \( g \neq 0 \)):
\[ \frac{1}{100} = \frac{R^2}{(R + h)^2} \]

### Step 4: Cross-multiply to eliminate the fraction
Cross-multiplying gives:
\[ (R + h)^2 = 100 R^2 \]

### Step 5: Take the square root of both sides
Taking the square root of both sides results in:
\[ R + h = 10R \]

### Step 6: Solve for \( h \)
Rearranging the equation to solve for \( h \):
\[ h = 10R - R = 9R \]

### Conclusion
Thus, the height \( h \) at which the acceleration due to gravity becomes 1% of its value at the surface is:
\[ h = 9R \]