The mass of the Earth is `6 xx 10^(24)` kg and its radius is 6400 km. Find the acceleration due to gravity on the surface of the Earth.

Options

A. `9.77 m s^(-2)`
B. `8.77 m s^(-2)`
C. `10.77 m s^(-2)`
D. `7.77 m s^(-2)`

Explanation

To find the acceleration due to gravity on the surface of the Earth, we can use the formula:

\[ g = \frac{GM}{R^2} \]

where:
- \( g \) is the acceleration due to gravity,
- \( G \) is the gravitational constant, approximately \( 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \),
- \( M \) is the mass of the Earth, given as \( 6 \times 10^{24} \, \text{kg} \),
- \( R \) is the radius of the Earth, which we need to convert from kilometers to meters.

### Step-by-step Solution:

1. **Convert the radius of the Earth from kilometers to meters**:
\[ R = 6400 \, \text{km} = 6400 \times 1000 \, \text{m} = 6.4 \times 10^6 \, \text{m} \]

2. **Substitute the values into the formula**:
\[ g = \frac{GM}{R^2} \]
Here, \( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \) and \( M = 6 \times 10^{24} \, \text{kg} \).

3. **Calculate \( R^2 \)**:
\[ R^2 = (6.4 \times 10^6 \, \text{m})^2 = 40.96 \times 10^{12} \, \text{m}^2 = 4.096 \times 10^{13} \, \text{m}^2 \]

4. **Now plug in the values into the equation**:
\[ g = \frac{(6.67 \times 10^{-11}) \times (6 \times 10^{24})}{4.096 \times 10^{13}} \]

5. **Calculate the numerator**:
\[ 6.67 \times 10^{-11} \times 6 \times 10^{24} = 40.02 \times 10^{13} = 4.002 \times 10^{14} \]

6. **Now divide the numerator by the denominator**:
\[ g = \frac{4.002 \times 10^{14}}{4.096 \times 10^{13}} \approx 9.77 \, \text{m/s}^2 \]

7. **Round off the final answer**:
The standard value for the acceleration due to gravity on the surface of the Earth is approximately:
\[ g \approx 9.8 \, \text{m/s}^2 \]

### Final Answer:
The acceleration due to gravity on the surface of the Earth is approximately \( 9.8 \, \text{m/s}^2 \).