The acceleration due to gravity at the surface of the earth is g . The acceleration due to gravity at a height `(1)/(100)` times the radius of the earth above the surface is close to :

Options

A. `((96)/(100))g`
B. `((98)/(100))g`
C. `((99)/(100))g`
D. `((101)/(100))g`

Explanation

To find the acceleration due to gravity at a height of \( \frac{1}{100} \) times the radius of the Earth above the surface, we can follow these steps:

### Step 1: Understand the formula for acceleration due to gravity
The acceleration due to gravity \( g_h \) at a height \( h \) above the surface of the Earth can be expressed as:
\[ g_h = \frac{GM}{(R + h)^2} \]
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 surface.

### Step 2: Substitute the height
Given that the height \( h \) is \( \frac{1}{100} \) times the radius of the Earth, we can write:
\[ h = \frac{R}{100} \]

### Step 3: Substitute \( h \) into the formula
Now substituting \( h \) into the equation for \( g_h \):
\[ g_h = \frac{GM}{(R + \frac{R}{100})^2} \]
This simplifies to:
\[ g_h = \frac{GM}{(R(1 + \frac{1}{100}))^2} \]
\[ g_h = \frac{GM}{(R \cdot \frac{101}{100})^2} \]
\[ g_h = \frac{GM}{R^2 \cdot \frac{10201}{10000}} \]

### Step 4: Relate to \( g \) at the surface
The acceleration due to gravity at the surface of the Earth is given by:
\[ g = \frac{GM}{R^2} \]
Now we can relate \( g_h \) to \( g \):
\[ g_h = g \cdot \frac{10000}{10201} \]

### Step 5: Calculate the value
Now we can compute \( g_h \):
\[ g_h \approx g \cdot \left(1 - \frac{201}{10201}\right) \quad \text{(using the approximation \( \frac{10000}{10201} \approx 1 - \frac{201}{10201} \))} \]
This gives:
\[ g_h \approx g \cdot \left(1 - 0.0197\right) \approx g \cdot 0.9803 \]

### Step 6: Final approximation
Thus, we can approximate:
\[ g_h \approx \frac{98}{100} g \]

### Conclusion
The acceleration due to gravity at a height of \( \frac{1}{100} \) times the radius of the Earth above the surface is approximately:
\[ g_h \approx \frac{98}{100} g \]