If G is universal gravitational constant and g is acceleration due to gravity then the unit of the quantity `(G)/(g)` is

Options

A. `kg//m^(2)`
B. `m^(2)//kg`
C. kg/m
D. m/kg

Answer

To find the unit of the quantity \( \frac{G}{g} \), where \( G \) is the universal gravitational constant and \( g \) is the acceleration due to gravity, we can follow these steps:

### Step 1: Understand the Definitions
- **Universal Gravitational Constant (G)**: This constant describes the strength of the gravitational force between two masses. Its value is approximately \( 6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \).
- **Acceleration due to Gravity (g)**: This is the acceleration experienced by an object due to the gravitational force exerted by the Earth. Its approximate value at the surface of the Earth is \( 9.81 \, \text{m/s}^2 \).

### Step 2: Write the Units
- The unit of \( G \) is \( \text{N m}^2/\text{kg}^2 \).
- The unit of \( g \) is \( \text{m/s}^2 \).

### Step 3: Convert Units
- We know that \( 1 \, \text{N} = 1 \, \text{kg m/s}^2 \). Therefore, we can rewrite the unit of \( G \):
\[ G = \frac{\text{N m}^2}{\text{kg}^2} = \frac{\text{kg m/s}^2 \cdot \text{m}^2}{\text{kg}^2} = \frac{\text{kg m}^3}{\text{s}^2 \cdot \text{kg}^2} = \frac{\text{m}^3}{\text{s}^2 \cdot \text{kg}} \]

### Step 4: Calculate the Unit of \( \frac{G}{g} \)
Now, we can find the unit of \( \frac{G}{g} \):
\[ \frac{G}{g} = \frac{\frac{\text{m}^3}{\text{s}^2 \cdot \text{kg}}}{\frac{\text{m}}{\text{s}^2}} = \frac{\text{m}^3}{\text{s}^2 \cdot \text{kg}} \cdot \frac{\text{s}^2}{\text{m}} = \frac{\text{m}^2}{\text{kg}} \]

### Step 5: Conclusion
The unit of the quantity \( \frac{G}{g} \) is \( \text{m}^2/\text{kg} \).

### Final Answer:
The unit of the quantity \( \frac{G}{g} \) is \( \text{m}^2/\text{kg} \).

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