What is the value of g at a height 8848 m above sea level. Given g on the surface of the earth is `9.8 ms^(-2)`. Mean radius of the earth `= 6.37 xx 10^(6) m`.

Explanation

To find the value of \( g \) at a height of 8848 m above sea level, we can use the formula for the variation of gravitational acceleration with height:

\[ g' = g \left(1 - \frac{2h}{R}\right) \]

Where:
- \( g' \) is the acceleration due to gravity at height \( h \)
- \( g \) is the acceleration due to gravity at the surface of the Earth (given as \( 9.8 \, \text{m/s}^2 \))
- \( h \) is the height above sea level (given as \( 8848 \, \text{m} \))
- \( R \) is the mean radius of the Earth (given as \( 6.37 \times 10^6 \, \text{m} \))

### Step 1: Identify the known values
- \( g = 9.8 \, \text{m/s}^2 \)
- \( h = 8848 \, \text{m} \)
- \( R = 6.37 \times 10^6 \, \text{m} \)

### Step 2: Substitute the values into the formula
Substituting the known values into the formula:

\[ g' = 9.8 \left(1 - \frac{2 \times 8848}{6.37 \times 10^6}\right) \]

### Step 3: Calculate \( \frac{2h}{R} \)
First, calculate \( \frac{2h}{R} \):

\[ \frac{2 \times 8848}{6.37 \times 10^6} = \frac{17696}{6.37 \times 10^6} \approx 0.00277 \]

### Step 4: Substitute back into the equation
Now substitute this value back into the equation for \( g' \):

\[ g' = 9.8 \left(1 - 0.00277\right) \]

### Step 5: Calculate \( g' \)
Now calculate \( g' \):

\[ g' = 9.8 \times (1 - 0.00277) = 9.8 \times 0.99723 \approx 9.77 \, \text{m/s}^2 \]

### Final Answer
Thus, the value of \( g \) at a height of 8848 m above sea level is approximately:

\[ \boxed{9.77 \, \text{m/s}^2} \]