What is the acceleration due to gravity on the top of Mount Everest? Mount Everest is the highest mountain peak of the world at the height of 8848 m. The value at see level is `9.80 ms^-2`.

Explanation

To find the acceleration due to gravity at the top of Mount Everest, we can use the formula that relates the acceleration due to gravity at a height \( h \) above the Earth's surface to the acceleration due to gravity at sea level.

### Step-by-Step Solution:

1. **Identify the known values**:
- Acceleration due to gravity at sea level, \( g = 9.8 \, \text{m/s}^2 \)
- Height of Mount Everest, \( h = 8848 \, \text{m} \)
- Radius of the Earth, \( r \approx 6400 \, \text{km} = 6400000 \, \text{m} \)

2. **Use the formula for acceleration due to gravity at height \( h \)**:
The formula for the acceleration due to gravity at a height \( h \) is given by:
\[ g' = \frac{g \cdot r^2}{(r + h)^2} \]
where \( g' \) is the acceleration due to gravity at height \( h \).

3. **Substitute the values into the formula**:
\[ g' = \frac{9.8 \cdot (6400000)^2}{(6400000 + 8848)^2} \]

4. **Calculate \( r + h \)**:
\[ r + h = 6400000 + 8848 = 6408848 \, \text{m} \]

5. **Calculate \( (r + h)^2 \)**:
\[ (6408848)^2 \approx 4.103 \times 10^{13} \, \text{m}^2 \]

6. **Calculate \( r^2 \)**:
\[ (6400000)^2 = 4.096 \times 10^{13} \, \text{m}^2 \]

7. **Substitute these values back into the formula**:
\[ g' = \frac{9.8 \cdot 4.096 \times 10^{13}}{4.103 \times 10^{13}} \]

8. **Calculate \( g' \)**:
\[ g' \approx 9.8 \cdot \frac{4.096}{4.103} \approx 9.8 \cdot 0.998 \approx 9.773 \, \text{m/s}^2 \]

### Final Answer:
The acceleration due to gravity at the top of Mount Everest is approximately \( 9.773 \, \text{m/s}^2 \).

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