The value of acceleration due to gravity at the surface of earth

Options

A. is maximum at the poles
B. is maximum at the equator
C. remains constant everywhere on the surface of the earth
D. is maximum at the international time line

Explanation

To find the value of acceleration due to gravity at the surface of the Earth, we can follow these steps:

### Step 1: Understand the formula for acceleration due to gravity
The acceleration due to gravity (g) at the surface of a celestial body is given by the formula:
\[ g = \frac{G \cdot M}{R^2} \]
where:
- \( G \) is the universal gravitational constant (\( 6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)),
- \( M \) is the mass of the Earth (\( 5.972 \times 10^{24} \, \text{kg} \)),
- \( R \) is the radius of the Earth (approximately \( 6.371 \times 10^6 \, \text{m} \)).

### Step 2: Substitute the known values into the formula
Substituting the values of \( G \), \( M \), and \( R \) into the formula:
\[ g = \frac{(6.674 \times 10^{-11}) \cdot (5.972 \times 10^{24})}{(6.371 \times 10^6)^2} \]

### Step 3: Calculate the value of g
Now, we can calculate the value of \( g \):
1. Calculate \( R^2 \):
\[ R^2 = (6.371 \times 10^6)^2 \approx 4.058 \times 10^{13} \, \text{m}^2 \]
2. Calculate the numerator:
\[ G \cdot M = (6.674 \times 10^{-11}) \cdot (5.972 \times 10^{24}) \approx 3.986 \times 10^{14} \, \text{N m}^2/\text{kg} \]
3. Now, substitute these values into the equation for \( g \):
\[ g = \frac{3.986 \times 10^{14}}{4.058 \times 10^{13}} \approx 9.81 \, \text{m/s}^2 \]

### Step 4: Conclusion
Thus, the value of acceleration due to gravity at the surface of the Earth is approximately:
\[ g \approx 9.81 \, \text{m/s}^2 \]