Calculate the surface potentail of the earth. Gravitational constant G, mass of the earth and the radius of the earth are given ?

Explanation

To calculate the surface potential of the Earth, we can use the formula for gravitational potential \( V \) at a distance \( r \) from a mass \( M \):

\[ V = -\frac{G \cdot M}{r} \]

Where:
- \( V \) is the gravitational potential,
- \( G \) is the universal gravitational constant (\( 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)),
- \( M \) is the mass of the Earth (\( 6 \times 10^{24} \, \text{kg} \)),
- \( r \) is the radius of the Earth (\( 6400 \, \text{km} \) or \( 6400 \times 10^3 \, \text{m} \)).

### Step-by-step Solution:

1. **Identify the values:**
- Gravitational constant \( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)
- Mass of the Earth \( M = 6 \times 10^{24} \, \text{kg} \)
- Radius of the Earth \( r = 6400 \, \text{km} = 6400 \times 10^3 \, \text{m} \)

2. **Substitute the values into the formula:**
\[ V = -\frac{G \cdot M}{r} = -\frac{(6.67 \times 10^{-11}) \cdot (6 \times 10^{24})}{6400 \times 10^3} \]

3. **Calculate the numerator:**
\[ G \cdot M = 6.67 \times 10^{-11} \cdot 6 \times 10^{24} = 4.002 \times 10^{14} \, \text{N m}^2/\text{kg} \]

4. **Calculate the denominator:**
\[ r = 6400 \times 10^3 = 6.4 \times 10^6 \, \text{m} \]

5. **Now, divide the numerator by the denominator:**
\[ \frac{4.002 \times 10^{14}}{6.4 \times 10^6} \approx 6.25 \times 10^{7} \, \text{J/kg} \]

6. **Apply the negative sign:**
\[ V = -6.25 \times 10^{7} \, \text{J/kg} \]

### Final Answer:
The surface potential of the Earth is approximately:
\[ V \approx -6.25 \times 10^{7} \, \text{J/kg} \]