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 solve the problem of finding 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-by-Step Solution:

1. **Understanding the Formula for Gravity**:
The acceleration due to gravity \( g \) at the surface of the Earth is given by the formula:
\[ g = \frac{GM}{R^2} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

2. **Height Above the Surface**:
We need to find the acceleration due to gravity at a height \( h \) above the surface of the Earth. According to the problem, this height is:
\[ h = \frac{1}{100} R \]

3. **New Radius at Height**:
The distance from the center of the Earth to the point at height \( h \) is:
\[ R' = R + h = R + \frac{1}{100} R = R \left(1 + \frac{1}{100}\right) = R \left(\frac{101}{100}\right) \]

4. **Applying the Gravity Formula at Height**:
The acceleration due to gravity \( g' \) at the height \( h \) can be expressed as:
\[ g' = \frac{GM}{(R')^2} = \frac{GM}{\left(R \left(\frac{101}{100}\right)\right)^2} \]

5. **Substituting for \( g' \)**:
Substituting \( R' \) into the formula gives:
\[ g' = \frac{GM}{R^2 \left(\frac{101}{100}\right)^2} = g \cdot \frac{1}{\left(\frac{101}{100}\right)^2} \]

6. **Calculating \( g' \)**:
Simplifying further:
\[ g' = g \cdot \frac{100^2}{101^2} = g \cdot \frac{10000}{10201} \]

7. **Approximation**:
To find an approximate value, we can calculate:
\[ \frac{10000}{10201} \approx 0.9802 \]
Thus, we can say:
\[ g' \approx 0.98g \]

8. **Final Answer**:
Therefore, the acceleration due to gravity at a height of \( \frac{1}{100} \) times the radius of the Earth above the surface is approximately:
\[ g' \approx \frac{98}{100} g \]

### Conclusion:
The final answer is \( \frac{98}{100} g \), which corresponds to option B.