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's surface, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the given altitude**:
The altitude is equal to one Earth diameter. The diameter of the Earth is twice the radius (\( R \)), so:
\[ \text{Altitude} = 2R \]

2. **Determine the distance from the center of the Earth**:
The distance from the center of the Earth to the point at this altitude is the radius of the Earth plus the altitude:
\[ \text{Distance from center} = R + 2R = 3R \]

3. **Use the formula for gravitational acceleration**:
The formula for gravitational acceleration at a distance \( r \) from the center of the Earth is given by:
\[ g' = \frac{GM}{r^2} \]
where \( G \) is the gravitational constant and \( M \) is the mass of the Earth. Here, \( r = 3R \).

4. **Substitute the distance into the formula**:
Substitute \( r = 3R \) into the formula:
\[ g' = \frac{GM}{(3R)^2} = \frac{GM}{9R^2} \]

5. **Relate it to the standard value of \( g \)**:
We know that the standard value of \( g \) at the surface of the Earth is:
\[ g = \frac{GM}{R^2} \]
Therefore, we can express \( g' \) in terms of \( g \):
\[ g' = \frac{1}{9} \cdot \frac{GM}{R^2} = \frac{1}{9}g \]

6. **Calculate the approximate value of \( g' \)**:
Using the standard value of \( g \approx 9.8 \, \text{m/s}^2 \):
\[ g' = \frac{1}{9} \cdot 9.8 \approx 1.09 \, \text{m/s}^2 \]

7. **Round off the result**:
Rounding \( 1.09 \) gives us approximately:
\[ g' \approx 1.1 \, \text{m/s}^2 \]

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