The approximate value of g (in `m//s^2`) at an altitude above Earth equal to one Earth diameter is:

Options

A. 9.8
B. 4.9
C. 2.5
D. 1.1

Explanation

To find the approximate value of \( g \) (acceleration due to gravity) at an altitude equal to one Earth diameter above the Earth, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**:
- We need to calculate \( g \) at an altitude equal to one Earth diameter above the Earth's surface. The radius of the Earth is denoted as \( R_E \).

2. **Determine the Altitude**:
- The altitude \( h \) is equal to the diameter of the Earth, which is \( 2R_E \).

3. **Calculate the Distance from the Center of the Earth**:
- The distance from the center of the Earth to the point at altitude \( h \) is:
\[ r = R_E + h = R_E + 2R_E = 3R_E \]

4. **Use the Formula for Gravitational Acceleration**:
- The formula for gravitational acceleration \( g \) at a distance \( r \) from the center of the Earth is given by:
\[ g' = \frac{GM}{r^2} \]
- Here, \( G \) is the gravitational constant, and \( M \) is the mass of the Earth.

5. **Substitute for \( r \)**:
- Substituting \( r = 3R_E \) into the formula gives:
\[ g' = \frac{GM}{(3R_E)^2} = \frac{GM}{9R_E^2} \]

6. **Relate to Earth's Surface Gravity**:
- The gravitational acceleration at the surface of the Earth is:
\[ g = \frac{GM}{R_E^2} \]
- Therefore, we can express \( g' \) in terms of \( g \):
\[ g' = \frac{1}{9} \cdot g \]

7. **Calculate the Value of \( g' \)**:
- Given that \( g \approx 9.8 \, \text{m/s}^2 \):
\[ g' = \frac{1}{9} \cdot 9.8 \approx 1.1 \, \text{m/s}^2 \]

### Conclusion:
The approximate value of \( g \) at an altitude equal to one Earth diameter above the Earth is:
\[ \boxed{1.1 \, \text{m/s}^2} \]