If the symbols have their usual meanings ,`(GM)/(R^(2))` is equal to

Options

A. `g^(2)`
B. `(1)/(g^(2))`
C. g
D. `(1)/(g)`

Answer

To solve the question, we need to determine what the expression \( \frac{GM}{R^2} \) represents, where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

### Step-by-Step Solution:

1. **Identify the Symbols**:
- \( G \) is the gravitational constant.
- \( M \) is the mass of the Earth.
- \( R \) is the radius of the Earth.

2. **Understanding the Expression**:
- The expression \( \frac{GM}{R^2} \) is derived from Newton's law of universal gravitation, which states that the gravitational force \( F \) between two masses is given by:
\[ F = \frac{GMm}{R^2} \]
- Here, \( m \) is the mass of an object located at the surface of the Earth.

3. **Relating to Acceleration**:
- The gravitational force acting on an object of mass \( m \) at the surface of the Earth can also be expressed using Newton's second law:
\[ F = ma \]
- Where \( a \) is the acceleration due to gravity, denoted as \( g \).

4. **Setting the Equations Equal**:
- By setting the two expressions for force equal to each other, we have:
\[ ma = \frac{GMm}{R^2} \]
- Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ a = \frac{GM}{R^2} \]
- This shows that the acceleration \( a \) (which is the acceleration due to gravity \( g \)) is given by:
\[ g = \frac{GM}{R^2} \]

5. **Conclusion**:
- Therefore, \( \frac{GM}{R^2} \) represents the acceleration due to gravity \( g \) at the surface of the Earth.

### Final Answer:
\[ \frac{GM}{R^2} = g \quad \text{(acceleration due to gravity)} \]